RBSE Solutions Class 11 Physics Chapter 7 Rigid Body Dynamics

RBSE Class 11 Physics Chapter 7 Very Short Answer Type Questions

Question 1. What is a rigid body?
Answer: A rigid body is one that does not change its shape even when external forces are applied to it. This means the distances between the tiny particles inside the body stay exactly the same. Imagine a very stiff block of wood that won't bend or break. This is a key concept in understanding how objects move and interact without deforming.
In simple words: A rigid body keeps its shape and size, no matter how much force you push or pull it with.

🎯 Exam Tip: When defining a rigid body, always mention both "no change in shape" and "molecular distance remains constant" to score full marks.

Question 2. Define center of mass.
Answer: The center of mass is a special point in a body where we can imagine all its mass is concentrated. When external forces act on the body, it behaves as if these forces are acting directly at this single point. This simplifies how we describe the motion of complex objects. It's like finding the balance point for an object.
In simple words: The center of mass is like the average position of all the mass in an object, where external forces seem to act.

🎯 Exam Tip: Remember to include both aspects in your definition: where the mass is "supposed to be contained" and where "all external forces act" for a complete answer.

Question 4. Write the relation between linear velocity and angular velocity.
Answer: The relationship between linear velocity \(v\) and angular velocity \(\omega\) for a particle moving in a circular path is given by the formula:
\(v = r\omega\)
Here, \(r\) represents the radius of the circular path. This formula shows how quickly a point on a rotating object moves in a straight line compared to how fast the object spins.
In simple words: Linear velocity (how fast something moves in a line) equals the radius times angular velocity (how fast it spins).

🎯 Exam Tip: Clearly define each variable (v, r, ω) in your answer to ensure clarity and accuracy in physical context.

Question 5. What is the angular speed of wrist watch's minute needle in rad/s?
Answer: The angular speed \(\omega\) of a wrist watch's minute hand is calculated using the formula \( \omega = \frac{2\pi}{T} \), where \(T\) is the time period.
A minute hand completes one full rotation (360 degrees or \(2\pi\) radians) in 1 hour.
First, convert 1 hour to seconds:
1 hour \( = 3600 \text{ s} \).
Now, substitute the values into the formula:
\( \omega = \frac{2\pi}{3600} \text{ rad/s} \)
\( \implies \omega = \frac{\pi}{1800} \text{ rad/s} \)
This means the minute hand slowly sweeps out an angle of \(\pi\) radians every 1800 seconds.
In simple words: The minute hand of a watch moves at an angular speed of pi divided by 1800 radians per second.

🎯 Exam Tip: Always remember to convert the time period to seconds when calculating angular speed in rad/s, as SI units are critical.

Question 6. What do you mean by the moment of inertia?
Answer: The moment of inertia describes how resistant a body is to changes in its rotational motion. It's like the rotational version of mass in linear motion. It is measured by multiplying the mass of the body by the square of its distance from the axis of rotation. A heavier object or one with mass further from its center is harder to spin.
In simple words: Moment of inertia tells you how hard it is to make something spin or stop spinning.

🎯 Exam Tip: To define moment of inertia clearly, mention both its role as "rotational inertness" and its calculation involving "mass" and "square of distance from axis of rotation."

Question 7. On what factors does the moment of inertia of an object depend?
Answer: The moment of inertia of an object depends on the following factors:
- On the position of the axis of rotation.
- On the mass of the object.
- On the distance of the mass from the axis of rotation.
These factors determine how easily an object can be rotated or how much force is needed to change its rotational speed. For example, a figure skater spins faster by pulling their arms in, changing the mass distribution.
In simple words: How hard it is to spin an object depends on where you spin it from, how heavy it is, and how far its mass is from the spin point.

🎯 Exam Tip: List these three factors clearly using bullet points for a concise and complete answer.

Question 9. The moment of inertia of a disc is minimum around which axis?
Answer: The moment of inertia of a disc is at its minimum when it rotates about its diameter. Its value is given by:
\( I_d = \frac{1}{4} MR^2 \)
Here, \(M\) is the mass of the disc and \(R\) is its radius. This means a disc is easiest to spin when it is rotated around a line that passes straight through its center and lies flat across its surface.
In simple words: A disc is easiest to spin around an axis that goes straight through its middle, like a line across its flat surface.

🎯 Exam Tip: State both the axis ("its diameter") and the formula for minimum moment of inertia to provide a comprehensive answer.

Question 10. Write the value of radius of gyration of a solid sphere relative to its tangent.
Answer: For a solid sphere, if \(I\) is the moment of inertia about its tangent and \(K\) is the radius of gyration, then:
\( I = MK^2 \)
We also know that the moment of inertia of a solid sphere about a tangent is \( \frac{7}{5} MR^2 \).
So,
\( MK^2 = \frac{7}{5} MR^2 \)
\( \implies K^2 = \frac{7}{5} R^2 \)
\( \implies K = R \sqrt{\frac{7}{5}} \)
The radius of gyration tells us how the mass is distributed around the axis of rotation, effectively giving us an equivalent distance where all the mass could be concentrated.
In simple words: For a solid sphere, the radius of gyration relative to a tangent line is the sphere's radius multiplied by the square root of 7/5.

🎯 Exam Tip: Remember to relate the moment of inertia about the tangent to \(MK^2\) and then solve for K. Always include the square root in the final answer.

Question 11. Write the unit of angular momentum.
Answer: Angular momentum, often denoted by \(J\) or \(L\), is calculated as \(J = I\omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
The units for moment of inertia are \( \text{kg} \cdot \text{m}^2 \) and for angular velocity are \( \text{rad/s} \).
Therefore, the unit of angular momentum \(J\) is \( \text{kg} \cdot \text{m}^2/\text{s} \).
Another equivalent unit for angular momentum is Joule-second (\(J \cdot s\)). This unit connects angular momentum to energy and time.
In simple words: Angular momentum is measured in kilogram-meter-squared per second, or also as Joule-seconds.

🎯 Exam Tip: It is good practice to provide both common units (\( \text{kg} \cdot \text{m}^2/\text{s} \) and \( J \cdot s \)) for angular momentum, as both are correct and widely used.

Question 12. A man is sitting on a moving table stretching his arms, if he brings his arms closer, what would be its effect on the moment of inertia?
Answer: If a man sitting on a moving table pulls his arms closer to his body, his moment of inertia will decrease. This happens because the mass of his arms, which was farther from the axis of rotation (his body), is now brought closer to it. The moment of inertia depends on how mass is distributed around the axis. This is why ice skaters spin faster when they pull their arms in.
In simple words: When the man brings his arms in, his body's moment of inertia gets smaller because the mass is closer to the center.

🎯 Exam Tip: Explain that moment of inertia decreases because the mass distribution changes, bringing mass closer to the axis of rotation.

Question 13. If the rate of change of angular momentum is zero then what would be the value of torque acting on the body?
Answer: According to the principle of rotational dynamics, torque (\(\tau\)) is defined as the rate of change of angular momentum (\(J\)) with respect to time (\(t\)). The formula is:
\( \tau = \frac{\Delta J}{\Delta t} \)
If the rate of change of angular momentum is zero, this means \(\frac{\Delta J}{\Delta t} = 0\).
Therefore, the value of torque acting on the body would be zero. This is the rotational equivalent of Newton's first law, where no change in motion means no net force.
In simple words: If angular momentum isn't changing, then the twisting force (torque) on the body is zero.

🎯 Exam Tip: Clearly state the relationship \(\tau = \frac{\Delta J}{\Delta t}\) and explain that if \(\frac{\Delta J}{\Delta t} = 0\), then \(\tau\) must also be zero.

Question 15. Write the relation between torque, moment of inertia and angular acceleration.
Answer: The relationship between torque (\(\tau\)), moment of inertia (\(I\)), and angular acceleration (\(\alpha\)) is given by:
\( \tau = I\alpha \)
This equation is the rotational equivalent of Newton's second law (\(F = ma\)). It shows that a larger torque is needed to produce the same angular acceleration if the object has a greater moment of inertia. This explains why it's harder to get a massive flywheel spinning than a light one.
In simple words: Torque (twisting force) is equal to moment of inertia (how hard it is to spin something) multiplied by angular acceleration (how quickly it speeds up its spin).

🎯 Exam Tip: Remember to correctly identify each variable and present the formula clearly. You can also mention it's the rotational analogue of Newton's second law.

Question 16. Why is the handle of screw driver large in width?
Answer: The handle of a screwdriver is made large in width to increase the effective distance (\(r\)) from the axis of rotation where the force (\(F\)) is applied. This larger distance helps to generate a greater torque (\(\tau\)) for a given force, because torque is calculated as \( \tau = F \times r \). By increasing \(r\), we can apply a small force \(F\) to produce a much larger torque, making it easier to turn screws. This leverage makes twisting tasks much simpler.
In simple words: A wide screwdriver handle helps you create more twisting power (torque) with less effort, because your hand pushes farther from the center.

🎯 Exam Tip: Focus your explanation on the direct relationship between force, distance (width of the handle), and the resulting torque, using the formula \(\tau = F \times r\).

Question 17. Write the value of total kinetic energy if an object is rotating along its axis and also moving in a straight line.
Answer: If an object is both rotating along its axis and moving in a straight line (translating), its total kinetic energy (\(E_t\)) is the sum of its translational kinetic energy and its rotational kinetic energy.
The formula for total kinetic energy in this case is:
\( E_t = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2 \)
Here, \(M\) is the mass, \(v\) is the linear velocity, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. This principle applies to things like rolling wheels, where both types of motion are present.
In simple words: If an object is rolling, its total energy is found by adding the energy from its straight movement and the energy from its spinning movement.

🎯 Exam Tip: Remember to include both translational (\( \frac{1}{2} Mv^2 \)) and rotational (\( \frac{1}{2} I\omega^2 \)) kinetic energy components when the body has both types of motion.

Question 18. Write the formula of velocity of a body rolling down on an inclined plane.
Answer: The formula for the velocity (\(v\)) of a body rolling down an inclined plane is:
\[ v = \sqrt{\frac{2gh}{\left(1+\frac{K^2}{R^2}\right)}} \]
In this formula, \(h\) is the height of the inclined plane, \(R\) is the radius of the rolling body, and \(K\) is the radius of gyration. This equation helps us understand how quickly different shapes of objects (like spheres or cylinders) will reach the bottom of a ramp, as \(K^2/R^2\) depends on the object's shape.
In simple words: The speed of a rolling object down a ramp depends on the ramp's height, the object's radius, and its radius of gyration.

🎯 Exam Tip: Precisely write the formula and clearly define each variable (h, R, K) to ensure full accuracy in your answer.

Question 19. Write the conditions for mechanical equilibrium of rigid bodies.
Answer: For a rigid body to be in mechanical equilibrium, it must satisfy two conditions simultaneously: one for linear motion and one for rotational motion.
These conditions are:
1. The net external force acting on the body must be zero (for linear equilibrium):
\( \Sigma F = 0 \)
2. The net external torque acting on the body must be zero (for rotational equilibrium):
\( \Sigma \tau = 0 \)
If both of these conditions are met, the body will either remain at rest or continue moving with constant linear and angular velocities. This ensures that the object is not accelerating in any way.
In simple words: For a rigid body to be still or move steadily, all forces pushing and pulling on it must balance out to zero, and all twisting forces (torques) must also balance out to zero.

🎯 Exam Tip: Clearly state both conditions: zero net force for linear equilibrium and zero net torque for rotational equilibrium.

Question 21. If the diameter of the earth reduces and becomes half then what would be the duration of a day?
Answer: If the diameter of the Earth were to reduce and become half, while its mass remained the same, the duration of a day would become 6 hours. This change is due to the conservation of angular momentum. As the Earth shrinks, its moment of inertia decreases, and to keep angular momentum constant, its angular speed must increase, making the day shorter. Imagine a spinning ice skater pulling in their arms; they spin faster.
In simple words: If Earth's size became half, a day would be only 6 hours long because it would spin much faster to keep its momentum.

🎯 Exam Tip: The core concept here is the conservation of angular momentum. Mentioning this principle strengthens your answer, even if you don't show the full calculation.

RBSE Class 11 Physics Chapter 7 Short Answer Type Questions

Question 1. Establish relation between mass and distance of center of mass for a two particles system.
Answer: Let's consider a system with two particles, \(m_1\) and \(m_2\), with position vectors \( \vec{r}_1 \) and \( \vec{r}_2 \) respectively, from an origin O. The position vector of the center of mass (\( \vec{r}_{\mathrm{cm}} \)) for this system is given by the formula:
\[ \vec{r}_{\mathrm{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \quad \text{...(1)} \]
This equation shows that the center of mass is a weighted average of the positions of the individual particles. If external forces \( \vec{F}_{1 \mathrm{ext}} \) and \( \vec{F}_{2 \mathrm{ext}} \) act on these particles, along with internal forces \( \vec{F}_{12} \) and \( \vec{F}_{21} \), the net force on particle 1 is \( \vec{F}_1 = \vec{F}_{1 \mathrm{ext}} + \vec{F}_{12} = m_1 \vec{a}_1 \), and on particle 2 is \( \vec{F}_2 = \vec{F}_{2 \mathrm{ext}} + \vec{F}_{21} = m_2 \vec{a}_2 \).
Adding these two equations:
\( \vec{F}_1 + \vec{F}_2 = \vec{F}_{1 \mathrm{ext}} + \vec{F}_{12} + \vec{F}_{2 \mathrm{ext}} + \vec{F}_{21} = m_1 \vec{a}_1 + m_2 \vec{a}_2 \)
According to Newton's third law, \( \vec{F}_{12} = -\vec{F}_{21} \), which means \( \vec{F}_{12} + \vec{F}_{21} = 0 \).
So, the total external force \( \vec{F}_{\mathrm{ext}} = \vec{F}_{1 \mathrm{ext}} + \vec{F}_{2 \mathrm{ext}} \) is:
\( \vec{F}_{\mathrm{ext}} = m_1 \vec{a}_1 + m_2 \vec{a}_2 \)
If \( M = m_1 + m_2 \) is the total mass of the system and \( \vec{a}_{\mathrm{cm}} \) is the acceleration of the center of mass, then \( \vec{F}_{\mathrm{ext}} = M \vec{a}_{\mathrm{cm}} \).
So, \( M \vec{a}_{\mathrm{cm}} = m_1 \vec{a}_1 + m_2 \vec{a}_2 \).
Since \( \vec{a}_1 = \frac{d\vec{v}_1}{dt} \) and \( \vec{a}_2 = \frac{d\vec{v}_2}{dt} \), we have:
\( M \vec{a}_{\mathrm{cm}} = m_1 \frac{d\vec{v}_1}{dt} + m_2 \frac{d\vec{v}_2}{dt} = \frac{d}{dt}(m_1 \vec{v}_1 + m_2 \vec{v}_2) \)
And since \( \vec{v}_{\mathrm{cm}} = \frac{d\vec{r}_{\mathrm{cm}}}{dt} \), then \( \vec{a}_{\mathrm{cm}} = \frac{d\vec{v}_{\mathrm{cm}}}{dt} = \frac{d^2\vec{r}_{\mathrm{cm}}}{dt^2} \).
This means \( M \frac{d\vec{v}_{\mathrm{cm}}}{dt} = \frac{d}{dt}(m_1 \vec{v}_1 + m_2 \vec{v}_2) \).
Integrating with respect to time, we get:
\( M \vec{v}_{\mathrm{cm}} = m_1 \vec{v}_1 + m_2 \vec{v}_2 \)
And further integrating, we get:
\( M \vec{r}_{\mathrm{cm}} = m_1 \vec{r}_1 + m_2 \vec{r}_2 \)
Therefore, the position vector of the center of mass for a two-particle system is:
\[ \vec{r}_{\mathrm{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \]
This point, the center of mass, is where all external forces are effectively considered to act.


In simple words: For two particles, the center of mass is like a balance point. It's found by taking the mass of each particle and its position, multiplying them, adding them up, and then dividing by the total mass.

🎯 Exam Tip: Always show the derivation from Newton's laws to fully establish the relationship. Ensure the vector notation is correct for positions and forces.

Question 2. Establish vector relation between linear velocity and angular velocity.
Answer: Let's find the vector relationship between linear velocity and angular velocity. When a rigid body spins around a fixed axis, every particle in it moves in a circular path. Each circular path is in a plane perpendicular to the axis of rotation, with its center on the axis. For a pure rotational motion, all particles share the same angular velocity. The speed of any particle is directly proportional to its angular velocity and its distance from the axis.
Consider a particle P on a rigid body moving in a circular path of radius \(r\). If its angular displacement at any moment is \(\theta\) radians, then the arc length \(s\) covered is \(s = r\theta\).
Differentiating this equation with respect to time (\(t\)), assuming \(r\) is constant for a given particle:
\( \frac{ds}{dt} = r \frac{d\theta}{dt} \)
Here, \( \frac{ds}{dt} \) is the linear speed \(v\), and \( \frac{d\theta}{dt} \) is the angular velocity \(\omega\).
So, \( v = r\omega \).
This is the scalar relationship.
To represent this in vector form, let the particle P have a position vector \( \vec{R} = \overrightarrow{OP} \) from the origin O. The linear velocity vector \( \vec{v} \) is tangential to the circular path, and the angular velocity vector \( \vec{\omega} \) points along the axis of rotation (using the right-hand rule).
The vector relation is given by the cross product:
\( \vec{v} = \vec{\omega} \times \vec{R} \)
This means the linear velocity of any point on a rotating body is the cross product of the angular velocity vector and the position vector of that point from the axis of rotation. This relationship is crucial for analyzing complex rotational motion.


In simple words: The linear speed of a point on a spinning object is found by crossing the angular velocity (how fast it spins) with its position vector (where it is from the center).

🎯 Exam Tip: Ensure you show both the scalar relationship \(v = r\omega\) and the vector relationship \( \vec{v} = \vec{\omega} \times \vec{R} \), explaining the direction of each vector.

Question 3. Write the three equations of rotational motion.
Answer: The three equations of rotational motion are very similar to the equations of linear motion, but they use angular quantities instead. These equations are used when a body rotates with a constant angular acceleration.
1. First equation:
\( \omega = \omega_0 + \alpha t \)
2. Second equation:
\( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
3. Third equation:
\( \omega^2 = \omega_0^2 + 2 \alpha \theta \)
Here, \(\omega\) is the final angular velocity, \(\omega_0\) is the initial angular velocity, \(\alpha\) is the constant angular acceleration, \(t\) is the time, and \(\theta\) is the angular displacement. These formulas help predict how a spinning object will move over time. For instance, they can tell you how many rotations a flywheel makes when accelerating.
In simple words: There are three main formulas to describe how things spin, just like how things move in a straight line. They tell you about final spin speed, total spin angle, and how spin speed changes over distance.

🎯 Exam Tip: Clearly write out all three equations and define each variable used for a complete and accurate answer.

Question 4. Give the statement for theorem of perpendicular axis of moment of inertia.
Answer: The Theorem of Perpendicular Axes states that for a plane lamina (a flat, thin object), the moment of inertia about an axis perpendicular to its plane is equal to the sum of the moments of inertia about two mutually perpendicular axes lying in the plane of the lamina and intersecting at the point where the perpendicular axis passes.
Mathematically, this is expressed as:
\( I_z = I_x + I_y \)
Here, \(I_x\) and \(I_y\) are the moments of inertia about the x and y axes in the plane, and \(I_z\) is the moment of inertia about the z-axis, which is perpendicular to the plane. This theorem simplifies finding the moment of inertia for flat objects without having to do complex calculations.
In simple words: For a flat object, the spinning resistance around an axis sticking out of its flat surface is the sum of its spinning resistances around two axes that cross each other on that flat surface.

🎯 Exam Tip: Ensure your statement explicitly mentions a "plane lamina" and "mutually perpendicular axes" that "lie in the plane" and "intersect at the same point" as the perpendicular axis.

Question 5. Write the theorem of parallel axis of moment of inertia.
Answer: The Theorem of Parallel Axes states that the moment of inertia (\(I\)) of a body about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its center of mass (\(I_{\mathrm{CM}}\)) and the product of the mass of the body (\(M\)) and the square of the perpendicular distance (\(h\)) between the two parallel axes.
Mathematically, this is expressed as:
\( I = I_{\mathrm{CM}} + Mh^2 \)
This theorem is very useful because if you know the moment of inertia about the center of mass, you can easily find it for any other parallel axis. For example, it helps calculate the moment of inertia of a spinning door around its hinges, knowing its moment around its center.
In simple words: The spinning resistance around any axis is equal to the spinning resistance around a parallel axis through the center, plus the object's mass times the distance between the two axes squared.

🎯 Exam Tip: Clearly state the formula and define all its components (\(I\), \(I_{\mathrm{CM}}\), \(M\), \(h\)) to ensure a comprehensive answer.

Question 6. Calculate the moment of inertia of a ring along an axis passing through the center and perpendicular to the plane.
Answer: To calculate the moment of inertia of a ring about an axis passing through its center and perpendicular to its plane:
Imagine a uniform thin ring of mass \(M\) and radius \(R\). Let's place this ring in the X-Y plane with its center at the origin. The Z-axis will be the required axis of rotation, which passes through the center and is perpendicular to the plane of the ring.
The entire mass of the ring is uniformly spread out along its circumference. We can think of the ring as being made up of many small mass elements \(m_1, m_2, m_3, \ldots, m_n\). Each of these tiny mass elements is at the same distance \(R\) from the Z-axis.
The moment of inertia of each small element \(m_i\) about the Z-axis is \(m_i R^2\).
To find the total moment of inertia (\(I\)) of the ring, we sum up the moments of inertia of all these small elements:
\( I = I_1 + I_2 + I_3 + \ldots + I_n \)
\( I = m_1 R^2 + m_2 R^2 + m_3 R^2 + \ldots + m_n R^2 \)
\( I = (m_1 + m_2 + m_3 + \ldots + m_n) R^2 \)
Since the sum of all the small masses is the total mass of the ring (\(M = m_1 + m_2 + m_3 + \ldots + m_n\)), we get:
\( I = MR^2 \)
This formula shows that for a thin ring, all its mass is effectively at the radius \(R\), making its moment of inertia relatively large for its mass compared to a solid disk.


In simple words: The moment of inertia for a ring spinning through its center and perpendicular to its flat surface is found by simply multiplying its total mass by its radius squared.

🎯 Exam Tip: Clearly state the assumption of uniform mass distribution and that all mass elements are at the same distance R from the axis.

Question 7. Calculate the moment of inertia of a solid cylinder about its axis.
Answer: To calculate the moment of inertia of a solid cylinder about its own geometrical axis (the axis passing through its center and along its length):
Imagine a solid cylinder of mass \(M\), radius \(R\), and length \(L\). Let's say the axis of rotation is the ZZ' axis, which runs along the center of the cylinder. We can think of this solid cylinder as being made up of many thin, coaxial discs stacked one on top of another. Each of these discs has a very small thickness and the same radius \(R\).
The moment of inertia of a single thin disc of mass \(dm\) and radius \(R\) about an axis passing through its center and perpendicular to its plane is given by \( dI = \frac{1}{2} dm R^2 \).
For the solid cylinder, all the mass \(dm\) for each disc is at varying distances from the central axis. However, since each disc is centered on the axis, its moment of inertia is simply:
\( dI = \frac{1}{2} dm R^2 \)
To find the total moment of inertia (\(I\)) of the cylinder, we integrate this expression over the entire mass of the cylinder.
\( I = \int dI = \int \frac{1}{2} dm R^2 \)
Since \(R\) is constant for the entire cylinder, we can take it out of the integral:
\( I = \frac{1}{2} R^2 \int dm \)
The integral of \(dm\) over the entire cylinder is simply the total mass \(M\) of the cylinder.
So,
\( I = \frac{1}{2} MR^2 \)
This formula is important for understanding how cylindrical objects rotate, from gears to rolling barrels.


In simple words: The moment of inertia for a solid cylinder spinning around its long central axis is half its total mass multiplied by its radius squared.

🎯 Exam Tip: Explain that a solid cylinder can be thought of as many thin discs, and then sum their moments of inertia. The key is \(\frac{1}{2}MR^2\).

Question 8. Establish a relation between torque and moment of inertia for a body.
Answer: The relationship between torque, moment of inertia, and angular acceleration is fundamental to rotational dynamics.
Consider a rigid body rotating about an axis (say, through a constant point O) with a constant angular acceleration \(\alpha\). All particles in the body will have this same angular acceleration \(\alpha\), but their linear accelerations will differ depending on their distance from the axis.
Let's focus on a particle with mass \(m_1\) at a distance \(r_1\) from the rotational axis. Its linear acceleration is \(a_1 = r_1\alpha\).
The force acting on this particle is \(F_1 = m_1 a_1 = m_1 r_1 \alpha\).
The torque \(\tau_1\) produced by this force about the rotational axis is \( \tau_1 = F_1 \times r_1 = (m_1 r_1 \alpha) \times r_1 = m_1 r_1^2 \alpha \).
If the body consists of many particles with masses \(m_2, m_3, \ldots\) at distances \(r_2, r_3, \ldots\), the total torque (\(\tau\)) acting on the body is the sum of the torques on all individual particles, assuming all torques are along the same axis:
\( \tau = \tau_1 + \tau_2 + \tau_3 + \ldots \)
\( \tau = m_1 r_1^2 \alpha + m_2 r_2^2 \alpha + m_3 r_3^2 \alpha + \ldots \)
\( \tau = (m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + \ldots) \alpha \)
The term \( (m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + \ldots) \) is the sum of \(mr^2\) for all particles, which is the definition of the moment of inertia \(I\) about the rotational axis (\(I = \Sigma mr^2\)).
Thus, the relationship is:
\( \tau = I\alpha \)
This equation is the rotational equivalent of Newton's second law for linear motion, \(F=ma\). It tells us that the torque required to produce a certain angular acceleration depends directly on the object's moment of inertia. This is why a larger engine is needed to accelerate a heavier car (more moment of inertia in its wheels).
If we set \( \alpha = 1 \text{ rad/s}^2 \), then \( \tau = I \). This means the moment of inertia of a body about a rotational axis is equal to the torque required to generate unit angular acceleration in the body.
In simple words: The twisting force (torque) that makes an object spin faster is equal to its spinning resistance (moment of inertia) multiplied by how quickly its spin speed changes (angular acceleration).

🎯 Exam Tip: Derive the relation step-by-step from individual particle torques to the total torque, clearly defining moment of inertia as \( \Sigma mr^2 \).

Question 10. Calculate the kinetic energy of a body, rotating about an axis with uniform angular velocity.
Answer: To calculate the rotational kinetic energy of a body spinning with uniform angular velocity:
Consider a body rotating with a uniform angular velocity \(\omega\) about an axis (say, YY'). Every small particle within this body will have the same angular velocity \(\omega\), but their linear velocities will differ based on their distance from the axis.
Let's consider an individual particle of mass \(m_1\) located at a distance \(r_1\) from the axis of rotation. Its linear velocity is \(v_1 = r_1\omega\).
The kinetic energy of this single particle is \( KE_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 (r_1\omega)^2 = \frac{1}{2} m_1 r_1^2 \omega^2 \).
If the body is made up of many particles with masses \(m_2, m_3, \ldots\) at distances \(r_2, r_3, \ldots\), their individual kinetic energies would be \( \frac{1}{2} m_2 r_2^2 \omega^2 \), \( \frac{1}{2} m_3 r_3^2 \omega^2 \), and so on.
Since kinetic energy is a scalar quantity, the total rotational kinetic energy (\(E_R\)) of the body is the sum of the kinetic energies of all its particles:
\( E_R = \frac{1}{2} m_1 r_1^2 \omega^2 + \frac{1}{2} m_2 r_2^2 \omega^2 + \frac{1}{2} m_3 r_3^2 \omega^2 + \ldots \)
We can factor out \( \frac{1}{2} \omega^2 \):
\( E_R = \frac{1}{2} (m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + \ldots) \omega^2 \)
The term \( (m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + \ldots) \) is the sum of \(mr^2\) for all particles, which is the definition of the moment of inertia \(I\) about the rotational axis (\(I = \Sigma mr^2\)).
Therefore, the total rotational kinetic energy is:
\( E_R = \frac{1}{2} I\omega^2 \)
This formula is analogous to the linear kinetic energy formula \( \frac{1}{2} mv^2 \), with \(I\) replacing \(m\) and \(\omega\) replacing \(v\). This energy is stored in the spinning motion of the object, like in a flywheel.
If \(\omega = 1 \text{ rad/s}\), then \(I = 2E_R\). This means if a body rotates with unit angular velocity, its moment of inertia is twice its rotational kinetic energy.


In simple words: The energy of a spinning body is found by taking half of its moment of inertia (how hard it is to spin) multiplied by its angular velocity (how fast it spins) squared.

🎯 Exam Tip: Derive the formula by summing the kinetic energies of individual particles, and clearly define moment of inertia in the process.

Question 11. A 1 kg mass ball is moving on a horizontal surface with a velocity 20 m/s and reaches a plane which makes an angle of 30° with the horizontal. If friction is negligible then how much distance vertically will the ball reach?
Answer:
Given:
Mass of the ball \( M = 1 \, \text{kg} \)
Initial velocity \( u = 20 \, \text{m/s} \)
Angle of inclination \( \theta = 30^\circ \)

Since the ball is rolling, its total kinetic energy is the sum of its translational and rotational kinetic energy.
The formula for the velocity of a rolling body at the base of an inclined plane is:
\[ v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}} \]
Where \( h \) is the vertical height reached, \( g \) is acceleration due to gravity (\( 10 \, \text{m/s}^2 \)), \( K \) is the radius of gyration, and \( R \) is the radius of the ball.
For a solid sphere (ball), \( \frac{K^2}{R^2} = \frac{2}{5} \).
So, the velocity becomes:
\[ v = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{2gh}{\frac{7}{5}}} = \sqrt{\frac{10gh}{7}} \]
From this, we can find the height \( h \):
\( v^2 = \frac{10gh}{7} \)
\( h = \frac{7v^2}{10g} \)

Substituting the given values:
\( h = \frac{7 \times (20)^2}{10 \times 10} = \frac{7 \times 400}{100} = 7 \times 4 = 28 \, \text{m} \)

The ball travels a vertical distance of 28 meters. In this scenario, all its kinetic energy at the start is converted into potential energy at its highest point.
In simple words: The ball uses its initial speed to roll up the slope. Since there's no friction, all its moving energy changes into height energy. This means it can go up to a height of 28 meters before stopping.

🎯 Exam Tip: Always specify the type of rolling body (e.g., solid sphere, cylinder, ring) to correctly use the \( \frac{K^2}{R^2} \) value, as this ratio changes for different shapes.

Question 12. What is the physical importance of moment of inertia?
Answer: The moment of inertia is very important in our everyday lives. It helps us understand how things resist changes in their spinning motion. For example, to make the wheels of scooters, motorcycles, bicycles, or carts have a high moment of inertia, most of their material is placed around the outer edge, connected to the center by spokes. This design allows wheels to continue rolling for some distance even after the driving force is removed, because they have a greater resistance to stopping.

Moment of inertia is also crucial in automotive engines, where it is used in the flywheel. A flywheel stores rotational energy. If the engine's torque changes suddenly, the high moment of inertia of the flywheel helps to keep the shaft rotating smoothly, protecting the engine from sudden shocks. Even toy cars use flywheels; when you rub them on the ground, the flywheel spins rapidly, storing energy that keeps the toy moving for a while.
In simple words: Moment of inertia tells us how hard it is to start or stop something that is spinning. Things like bicycle wheels or car flywheels are designed with high moment of inertia so they can keep spinning easily or smooth out uneven forces.

🎯 Exam Tip: When asked about physical importance, always provide real-world examples to illustrate the concept clearly, showing how it affects motion and design.

Question 13. Calculate the moment of inertia of a thin rod about an axis perpendicular to the length and passing through one edge.
Answer:
Let's consider a uniform thin rod AB with mass \( M \) and length \( L \). We want to find its moment of inertia about an axis perpendicular to its length and passing through one of its ends, say point A.

First, we know the moment of inertia of a thin rod about an axis passing through its center of mass (midpoint, YY') and perpendicular to its length is \( I_{CM} = \frac{ML^2}{12} \).

Now, we can use the theorem of parallel axes to find the moment of inertia about an axis passing through one end (point A). The distance \( d \) between the center of mass (midpoint) and one end (A) is \( d = \frac{L}{2} \).

According to the parallel axis theorem: \( I_A = I_{CM} + Md^2 \)
Substitute the values:
\( I_A = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 \)
\( I_A = \frac{ML^2}{12} + \frac{ML^2}{4} \)
To add these, find a common denominator (12):
\( I_A = \frac{ML^2}{12} + \frac{3ML^2}{12} \)
\( I_A = \frac{4ML^2}{12} \)
\( I_A = \frac{ML^2}{3} \)

Therefore, the moment of inertia of a thin rod about an axis perpendicular to its length and passing through one of its ends is \( \frac{ML^2}{3} \). This calculation is fundamental in understanding how mass distribution affects rotational motion. The moment of inertia is greater when the axis is at the end, making it harder to rotate than around the center.
In simple words: Imagine a thin stick. If you want to spin it around an axis right in the middle, it's easier. But if you try to spin it around an axis at one of its ends, it's harder. The formula \( \frac{ML^2}{3} \) tells us exactly how much harder it is when spinning it from the end, compared to \( \frac{ML^2}{12} \) from the middle.

🎯 Exam Tip: Remember to clearly state the parallel axis theorem and define all variables used. Ensure the distance \( d \) is correctly identified as the distance between the center of mass and the new axis of rotation.

Question 14. Derive the formula of total kinetic energy of a body at the bottom rolling down on an inclined plane.
Answer: When a body rolls down an inclined plane, its total kinetic energy \( E \) is the sum of two types of kinetic energy: translational kinetic energy and rotational kinetic energy. This happens because the body is both moving forward (translating) and spinning (rotating).

Translational kinetic energy is the energy due to the body's linear motion. It is given by:
\[ KE_{translational} = \frac{1}{2} Mv^2 \]
where \( M \) is the mass of the body and \( v \) is its linear velocity.

Rotational kinetic energy is the energy due to the body's spinning motion. It is given by:
\[ KE_{rotational} = \frac{1}{2} I\omega^2 \]
where \( I \) is the moment of inertia of the body and \( \omega \) is its angular velocity.

For a rolling body, there is a relationship between its linear velocity and angular velocity: \( v = R\omega \), or \( \omega = \frac{v}{R} \), where \( R \) is the radius of the body. Also, the moment of inertia \( I \) can be expressed in terms of the radius of gyration \( K \) as \( I = MK^2 \).

Substitute these into the rotational kinetic energy formula:
\[ KE_{rotational} = \frac{1}{2} (MK^2) \left(\frac{v}{R}\right)^2 = \frac{1}{2} MK^2 \frac{v^2}{R^2} = \frac{1}{2} Mv^2 \frac{K^2}{R^2} \]

Now, the total kinetic energy \( E \) is:
\[ E = KE_{translational} + KE_{rotational} \]
\[ E = \frac{1}{2} Mv^2 + \frac{1}{2} Mv^2 \frac{K^2}{R^2} \]
Factor out \( \frac{1}{2} Mv^2 \):
\[ E = \frac{1}{2} Mv^2 \left(1 + \frac{K^2}{R^2}\right) \]
This formula shows that the total kinetic energy of a rolling body depends on its mass, linear velocity, and how its mass is distributed around its axis of rotation (represented by \( \frac{K^2}{R^2} \)). Different shapes (like spheres, cylinders, or rings) will have different \( \frac{K^2}{R^2} \) values, affecting their total kinetic energy and how they roll.
In simple words: When a ball rolls down a slope, it does two things at once: it moves forward and it spins. Its total energy from moving is a mix of the energy from moving forward (translational) and the energy from spinning (rotational). The final formula \( E = \frac{1}{2} Mv^2 \left(1 + \frac{K^2}{R^2}\right) \) combines both these energies into one, showing how the total movement energy depends on its speed and shape.

🎯 Exam Tip: Clearly define all terms like translational kinetic energy, rotational kinetic energy, moment of inertia, and radius of gyration. Remember to show the relationship between linear and angular velocity for rolling without slipping.

RBSE Class 11 Physics Chapter 7 Long Answer Type Questions

Question 1. What do you mean by center of mass ? Calculate the position coordinates of the center of mass of two particles in a system.
Answer: The center of mass is a specific point in a body or system of particles where all its mass is considered to be concentrated. This point behaves as if all the external forces acting on the system are applied directly to it, making it easier to analyze the overall motion of complex objects. Even if the body is rotating or deforming, the center of mass will move according to Newton's laws as if it were a single particle.

To calculate the position coordinates of the center of mass for a two-particle system:
Suppose we have two particles with masses \( m_1 \) and \( m_2 \). Let their position vectors be \( \vec{r}_1 \) and \( \vec{r}_2 \) respectively. The position vector of the center of mass, denoted as \( \vec{r}_{\text{cm}} \), for this two-particle system is given by:
\[ \vec{r}_{\text{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \]
This formula calculates a weighted average of the positions of the two particles, where the weights are their respective masses. If the masses are equal, the center of mass will be exactly in the middle of the line connecting the two particles.

If we consider the components of these vectors in a Cartesian coordinate system, the coordinates of the center of mass \( (X_{\text{cm}}, Y_{\text{cm}}, Z_{\text{cm}}) \) are:
\[ X_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
\[ Y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]
\[ Z_{\text{cm}} = \frac{m_1 z_1 + m_2 z_2}{m_1 + m_2} \]
This can be generalized for a system of \( n \) particles.

This concept is crucial because it simplifies the study of the motion of complex systems. Instead of tracking every single particle, we can just track the center of mass to understand the system's overall translational motion.
In simple words: The center of mass is like the balance point of an object. For two objects, you find this point by considering both their weight and where they are located. The formula helps you find this exact spot, which is very useful for figuring out how the whole system will move.

🎯 Exam Tip: For the definition, emphasize that the center of mass is a point where the entire mass is *imagined* to be concentrated and where *all external forces* act. For calculations, remember it's a weighted average of positions.

Question 4. Give the statement for theorem of perpendicular axis of moment of inertia.
Answer: The theorem of perpendicular axes is a fundamental principle used to determine the moment of inertia for a flat, thin body (a plane lamina).

Statement of the Theorem: "The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia of two perpendicular axes lying in the plane of the lamina and meeting at the scheduled axis."
Mathematically, if \( I_x \) and \( I_y \) are the moments of inertia about two mutually perpendicular axes (x-axis and y-axis) lying in the plane of the lamina, and \( I_z \) is the moment of inertia about an axis (z-axis) perpendicular to the plane and passing through the point of intersection of the x and y axes, then:
\[ I_z = I_x + I_y \]

Proof:
Consider a plane lamina lying in the XY-plane. Let the origin O be the point where the three axes (X, Y, and Z) intersect. Consider a tiny particle of mass \( m \) located at a point P(x, y) in the lamina.

The perpendicular distance of this particle from the x-axis is \( y \). So, its moment of inertia about the x-axis is \( I_x = \sum m y^2 \).
The perpendicular distance of this particle from the y-axis is \( x \). So, its moment of inertia about the y-axis is \( I_y = \sum m x^2 \).

The perpendicular distance of this particle from the z-axis (which is perpendicular to the XY-plane) is \( r \), where \( r \) is the distance of P from the origin O. From Pythagoras theorem, \( r^2 = x^2 + y^2 \).
So, its moment of inertia about the z-axis is \( I_z = \sum m r^2 \).

Now, substitute \( r^2 = x^2 + y^2 \) into the expression for \( I_z \):
\[ I_z = \sum m (x^2 + y^2) \]
\[ I_z = \sum m x^2 + \sum m y^2 \]
By comparing this with the expressions for \( I_x \) and \( I_y \), we get:
\[ I_z = I_y + I_x \]
This proves the theorem of perpendicular axes. This theorem is extremely useful for calculating the moment of inertia of flat objects when the moments of inertia about two perpendicular in-plane axes are known. For example, if you know how easy it is to spin a flat sheet around its length and its width, you can easily find how easy it is to spin it like a pizza base.
In simple words: For a flat object, if you know how hard it is to spin it around two straight lines that cross each other on its surface, then it's easy to find how hard it is to spin it around a line that sticks straight out from where those two lines cross. You just add the "hardnesses" of the first two spins.

🎯 Exam Tip: When proving this theorem, remember to use a small mass element at a point (x,y) and relate its distances from the X, Y, and Z axes using the Pythagorean theorem.

Question 5. Write the theorem of parallel axis of moment of inertia.
Answer: The theorem of parallel axes is a powerful tool used to calculate the moment of inertia of a rigid body about any axis, provided its moment of inertia about a parallel axis passing through its center of mass is known.

Statement of the Theorem: "The moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis passing through its center of mass plus the product of the total mass of the body and the square of the perpendicular distance between the two axes."
Mathematically, if \( I_{\text{CM}} \) is the moment of inertia about an axis passing through the center of mass (CM), \( M \) is the total mass of the body, and \( d \) is the perpendicular distance between the center of mass axis and the new parallel axis, then the moment of inertia \( I \) about the new axis is:
\[ I = I_{\text{CM}} + Md^2 \]

Proof:
Consider a rigid body with its center of mass at C. Let \( AB \) be an axis passing through the center of mass C, and let \( AB' \) be another axis parallel to \( AB \) at a perpendicular distance \( d \).

Let the origin be at C. Consider a small particle of mass \( m_i \) at position \( \vec{r}_i \) relative to the center of mass C. The moment of inertia of the body about the axis through C is \( I_{\text{CM}} = \sum m_i |\vec{r}_i|^2 \).

The position vector of the particle \( m_i \) relative to the axis \( AB' \) (which is at distance \( d \) from C) can be written as \( \vec{R}_i = \vec{r}_i + \vec{d} \), where \( \vec{d} \) is the vector from C to the axis \( AB' \).

The moment of inertia about the axis \( AB' \) is:
\[ I = \sum m_i |\vec{R}_i|^2 = \sum m_i (\vec{r}_i + \vec{d}) \cdot (\vec{r}_i + \vec{d}) \]
\[ I = \sum m_i (|\vec{r}_i|^2 + |\vec{d}|^2 + 2\vec{r}_i \cdot \vec{d}) \]
\[ I = \sum m_i |\vec{r}_i|^2 + \sum m_i |\vec{d}|^2 + \sum m_i (2\vec{r}_i \cdot \vec{d}) \]
\[ I = I_{\text{CM}} + d^2 \sum m_i + 2\vec{d} \cdot \left(\sum m_i \vec{r}_i\right) \]

Since \( \sum m_i \vec{r}_i \) represents the first moment of mass about the center of mass, and the center of mass is the reference point, this term is zero. Therefore, \( \sum m_i \vec{r}_i = 0 \).
Also, \( \sum m_i = M \), the total mass of the body.

So, the equation simplifies to:
\[ I = I_{\text{CM}} + Md^2 \]
This completes the proof. This theorem is very useful because it allows us to calculate the moment of inertia about any axis once we know it for the center of mass, which is often easier to find. For example, knowing how a wheel spins around its central axle lets you figure out how it would spin if it was off-center.
In simple words: If you know how hard it is to spin an object around its balance point (center of mass), this theorem helps you find how hard it is to spin it around any other line parallel to the first one. You just need to add the object's total mass multiplied by the square of the distance between the two parallel lines.

🎯 Exam Tip: When using the parallel axis theorem, always ensure the two axes are parallel, and \( d \) is the *perpendicular* distance between them. The term \( \sum m_i \vec{r}_i \) being zero for the center of mass is a crucial step in the proof.

Question 3. Calculate the moment of inertia of a disc along the axis and perpendicular to the axis passing through the center.
Answer: Let's calculate the moment of inertia of a circular disc about an axis that passes through its center and is perpendicular to its plane. Imagine a disc with mass \( M \) and radius \( R \). The mass per unit area (surface density) of the disc, \( \sigma \), can be expressed as:
\[ \sigma = \frac{M}{\text{Area}} = \frac{M}{\pi R^2} \]

We can imagine this disc as being made up of many thin, concentric circular rings. Let's consider one such ring with radius \( x \) and a very small width \( dx \).

The area of this small ring is its circumference multiplied by its width:
Area of ring \( = (2\pi x) dx \)

The mass of this small ring \( dm \) can be found by multiplying its area by the surface density \( \sigma \):
\( dm = \sigma (2\pi x) dx = \frac{M}{\pi R^2} (2\pi x) dx = \frac{2M x}{R^2} dx \)

The moment of inertia of this thin ring about an axis passing through its center and perpendicular to its plane is \( dm \times (\text{radius})^2 \). Since all parts of this ring are at the same distance \( x \) from the axis:
\( dI = dm \cdot x^2 = \left(\frac{2M x}{R^2} dx\right) x^2 = \frac{2M}{R^2} x^3 dx \)

To find the total moment of inertia \( I \) of the disc, we sum up the moments of inertia of all such concentric rings from radius \( x = 0 \) to \( x = R \). This is done by integrating \( dI \):
\[ I = \int_{0}^{R} dI = \int_{0}^{R} \frac{2M}{R^2} x^3 dx \]
\[ I = \frac{2M}{R^2} \int_{0}^{R} x^3 dx \]
\[ I = \frac{2M}{R^2} \left[\frac{x^4}{4}\right]_{0}^{R} \]
\[ I = \frac{2M}{R^2} \left(\frac{R^4}{4} - \frac{0^4}{4}\right) \]
\[ I = \frac{2M}{R^2} \frac{R^4}{4} = \frac{2MR^2}{4} = \frac{1}{2} MR^2 \]
So, the moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is \( \frac{1}{2} MR^2 \). This value is smaller than that of a ring of the same mass and radius because the disc's mass is distributed closer to the axis, making it easier to rotate. This formula is vital for calculations involving rotating platforms or gears.
In simple words: Imagine a flat circular plate. To find how hard it is to spin it around a central rod standing straight up, we break it into many tiny rings. We add up the "spinning hardness" of all these tiny rings from the center to the edge. The final answer is \( \frac{1}{2} MR^2 \), which means it's easier to spin than a ring because the disc has more mass closer to the middle.

🎯 Exam Tip: When calculating moment of inertia by integration, correctly define the mass element \( dm \) and its distance from the axis of rotation. Ensure the integration limits span the entire object.

Question 4. Calculate the moment of inertia of a solid cylinder along the axis passing through the center of mass and perpendicular to its length.
Answer: To find the moment of inertia of a solid cylinder about an axis passing through its center of mass and perpendicular to its length, we can imagine the cylinder as being made up of many thin circular discs stacked together. Let the cylinder have mass \( M \), length \( L \), and radius \( R \).

First, determine the density of the cylinder. If the volume of the cylinder is \( V = \pi R^2 L \), then its density \( \rho = \frac{M}{V} = \frac{M}{\pi R^2 L} \).

Now, consider a thin disc of thickness \( dx \) located at a distance \( x \) from the center of the cylinder. The radius of this disc is \( R \).
The volume of this elementary disc is \( dV = \pi R^2 dx \).
The mass of this elementary disc \( dm \) is \( dm = \rho \cdot dV = \frac{M}{\pi R^2 L} (\pi R^2 dx) = \frac{M}{L} dx \).

The moment of inertia of a single disc about its own diameter (which is parallel to the axis of the cylinder's rotation in this case) is \( \frac{1}{4} dm R^2 \). This is because for a disc rotating about its diameter, the moment of inertia is half of what it would be if rotating about an axis perpendicular to its plane. However, the axis we are interested in (perpendicular to the cylinder's length) is parallel to the diameter of each disc. So we use the parallel axis theorem.

The moment of inertia of this elementary disc about the central axis of the cylinder (its own diameter) is \( dI_{diam} = \frac{1}{4} dm R^2 \).

Now, apply the parallel axis theorem. The axis of rotation for the cylinder passes through its center of mass and is perpendicular to its length. The axis for the elementary disc (its diameter) is parallel to this axis. The distance between the cylinder's central axis (perpendicular to its length) and the center of the elementary disc is \( x \).

So, the moment of inertia of this elementary disc about the desired axis of the cylinder is:
\[ dI = dI_{diam} + dm \cdot x^2 \]
\[ dI = \frac{1}{4} dm R^2 + dm \cdot x^2 \]
Substitute \( dm = \frac{M}{L} dx \):
\[ dI = \frac{1}{4} \left(\frac{M}{L} dx\right) R^2 + \left(\frac{M}{L} dx\right) x^2 \]
\[ dI = \frac{M}{L} \left(\frac{R^2}{4} + x^2\right) dx \]

To find the total moment of inertia \( I \) of the cylinder, we integrate \( dI \) from \( x = -\frac{L}{2} \) to \( x = \frac{L}{2} \):
\[ I = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{M}{L} \left(\frac{R^2}{4} + x^2\right) dx \]
\[ I = \frac{M}{L} \left[ \frac{R^2}{4} x + \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} \]
\[ I = \frac{M}{L} \left[ \left(\frac{R^2}{4} \frac{L}{2} + \frac{(L/2)^3}{3}\right) - \left(\frac{R^2}{4} \left(-\frac{L}{2}\right) + \frac{(-L/2)^3}{3}\right) \right] \]
\[ I = \frac{M}{L} \left[ \left(\frac{R^2 L}{8} + \frac{L^3}{24}\right) - \left(-\frac{R^2 L}{8} - \frac{L^3}{24}\right) \right] \]
\[ I = \frac{M}{L} \left[ \frac{R^2 L}{8} + \frac{L^3}{24} + \frac{R^2 L}{8} + \frac{L^3}{24} \right] \]
\[ I = \frac{M}{L} \left[ \frac{2R^2 L}{8} + \frac{2L^3}{24} \right] \]
\[ I = \frac{M}{L} \left[ \frac{R^2 L}{4} + \frac{L^3}{12} \right] \]
\[ I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \]
Therefore, the moment of inertia of a solid cylinder about an axis passing through its center of mass and perpendicular to its length is \( M \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \). This formula helps explain why a long, thin rod or a wide, short disc (which are special cases of a cylinder) behave differently when rotated about a central perpendicular axis. For instance, a long pencil will be harder to rotate about its center when held horizontally than a short, thick hockey puck.
In simple words: To find how hard it is to spin a solid cylinder sideways through its middle, we think of it as many thin disks stacked up. We find how hard it is to spin each disk and then add all these "hardnesses" together. The final formula \( M \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \) shows that the spinning difficulty depends on both how wide the cylinder is and how long it is.

🎯 Exam Tip: Remember to use the parallel axis theorem correctly for each elementary disc. The moment of inertia of a disc about its diameter is \( \frac{1}{4} dm R^2 \), and the distance \( x \) from the cylinder's center of mass to the disc's center must be included in the parallel axis calculation.

Question 5. Calculate the moment of inertia of a solid sphere about its diameter.
Answer: To calculate the moment of inertia of a solid sphere about its diameter, we can imagine the sphere being made up of a series of thin circular discs. These discs would be stacked along the diameter (let's say the x-axis) of the sphere.

Consider a solid sphere with total mass \( M \) and radius \( R \). The density of the sphere \( \rho \) is \( \frac{M}{\frac{4}{3}\pi R^3} \).

Let's take a thin disc of thickness \( dx \) at a distance \( x \) from the center of the sphere. The radius of this disc, \( y \), can be related to \( x \) and \( R \) by the Pythagorean theorem: \( y^2 = R^2 - x^2 \).

The volume of this elementary disc is \( dV = \pi y^2 dx = \pi (R^2 - x^2) dx \).
The mass of this elementary disc, \( dm \), is \( dm = \rho \cdot dV = \frac{M}{\frac{4}{3}\pi R^3} \cdot \pi (R^2 - x^2) dx = \frac{3M}{4R^3} (R^2 - x^2) dx \).

The moment of inertia of a single thin disc about an axis perpendicular to its plane (its own axis, which is the diameter of the sphere in this case) is \( dI = \frac{1}{2} dm y^2 \).

Substitute \( dm \) and \( y^2 \):
\[ dI = \frac{1}{2} \left( \frac{3M}{4R^3} (R^2 - x^2) dx \right) (R^2 - x^2) \]
\[ dI = \frac{3M}{8R^3} (R^2 - x^2)^2 dx \]
\[ dI = \frac{3M}{8R^3} (R^4 - 2R^2 x^2 + x^4) dx \]

To find the total moment of inertia \( I \) of the solid sphere, we integrate \( dI \) from \( x = -R \) to \( x = R \):
\[ I = \int_{-R}^{R} \frac{3M}{8R^3} (R^4 - 2R^2 x^2 + x^4) dx \]
Since the integrand is an even function, we can simplify this to:
\[ I = 2 \int_{0}^{R} \frac{3M}{8R^3} (R^4 - 2R^2 x^2 + x^4) dx \]
\[ I = \frac{3M}{4R^3} \int_{0}^{R} (R^4 - 2R^2 x^2 + x^4) dx \]
\[ I = \frac{3M}{4R^3} \left[ R^4 x - \frac{2R^2 x^3}{3} + \frac{x^5}{5} \right]_{0}^{R} \]
\[ I = \frac{3M}{4R^3} \left[ R^4 (R) - \frac{2R^2 (R)^3}{3} + \frac{(R)^5}{5} \right] \]
\[ I = \frac{3M}{4R^3} \left[ R^5 - \frac{2R^5}{3} + \frac{R^5}{5} \right] \]
Find a common denominator for the terms in the bracket (15):
\[ I = \frac{3M}{4R^3} \left[ \frac{15R^5 - 10R^5 + 3R^5}{15} \right] \]
\[ I = \frac{3M}{4R^3} \left[ \frac{8R^5}{15} \right] \]
\[ I = \frac{24M R^5}{60 R^3} = \frac{2}{5} MR^2 \]
Thus, the moment of inertia of a solid sphere about its diameter is \( \frac{2}{5} MR^2 \). This value is crucial in astrophysics for understanding planetary and stellar rotation, and in mechanics for analyzing spherical objects in motion. It tells us that a solid sphere is harder to rotate than a thin disc of the same mass and radius about an axis perpendicular to its plane because its mass is spread further from the center in three dimensions.
In simple words: To figure out how hard it is to spin a solid ball around its center, like spinning a globe, we imagine it's made of many thin, flat disks stacked up. We add up the "spinning hardness" of all these disks. The final formula \( \frac{2}{5} MR^2 \) shows that a solid ball is harder to spin than a flat disc, but easier than a hollow ball of the same size.

🎯 Exam Tip: When integrating for a solid sphere, remember to express the radius of each elementary disc (\(y\)) in terms of its position (\(x\)) and the sphere's radius (\(R\)) using \(y^2 = R^2 - x^2\). Also, apply the correct moment of inertia formula for a disc about its central axis.

Question 6. Establish the formula of moment of inertia of a rectangular rod along the axis passing through the center of mass and perpendicular to the length of the rod.
Answer: Let's find the moment of inertia for a uniform rectangular rod (or lamina) about an axis passing through its center of mass and perpendicular to its length. Consider a rectangular rod with total mass \( M \), length \( L \), and breadth \( B \). We'll set up a coordinate system where the center of mass is at the origin (0,0). Let the length \( L \) be along the X-axis and the breadth \( B \) along the Y-axis. The axis of rotation passes through the center and is perpendicular to the plane (Z-axis).

First, calculate the surface density (mass per unit area) of the rod: \( \sigma = \frac{M}{\text{Area}} = \frac{M}{LB} \).

Imagine the rod as being made up of many small thin strips, each of length \( L \) and width \( dy \), located at a distance \( y \) from the X-axis.
The mass of such a strip, \( dm \), is \( dm = \sigma \cdot (L \cdot dy) = \frac{M}{LB} (L \cdot dy) = \frac{M}{B} dy \).

The moment of inertia of this strip about the X-axis (which is along its length) would be negligible if we treat it as a thin line. However, we are rotating it about the Z-axis (perpendicular to its plane).

Let's consider the moment of inertia about the Y-axis first (an axis passing through the center of mass and parallel to the length \(L\)).
Consider a thin strip of thickness \( dx \) at a distance \( x \) from the Y-axis, parallel to the Y-axis. The mass of this strip is \( dm = \sigma (B dx) = \frac{M}{L} dx \).
The moment of inertia of this strip about the Y-axis (the axis through its center, parallel to its length) is \( dI_y = dm \cdot x^2 = \frac{M}{L} x^2 dx \).
Integrating from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ I_y = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{M}{L} x^2 dx = \frac{M}{L} \left[\frac{x^3}{3}\right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{M}{L} \left(\frac{(L/2)^3}{3} - \frac{(-L/2)^3}{3}\right) \]
\[ I_y = \frac{M}{L} \left(\frac{L^3}{24} + \frac{L^3}{24}\right) = \frac{M}{L} \frac{2L^3}{24} = \frac{ML^2}{12} \]

Similarly, for the moment of inertia about the X-axis (an axis passing through the center of mass and parallel to the breadth \(B\)), by symmetry:
\[ I_x = \frac{MB^2}{12} \]

Now, we use the theorem of perpendicular axes. The moment of inertia about an axis perpendicular to the plane of the lamina (Z-axis) and passing through the intersection of the X and Y axes (the center of mass) is the sum of \( I_x \) and \( I_y \).
\[ I_z = I_x + I_y \]
\[ I_z = \frac{MB^2}{12} + \frac{ML^2}{12} \]
\[ I_z = \frac{M(L^2 + B^2)}{12} \]
Therefore, the moment of inertia of a rectangular rod about an axis passing through its center of mass and perpendicular to its length is \( \frac{M(L^2 + B^2)}{12} \). This formula demonstrates how a rectangular plate's resistance to rotation depends on both its length and width. For example, a square plate will have a simpler rotational inertia than a long, thin rectangular plate, but both are derived from this general formula.
In simple words: To find how hard it is to spin a flat, rectangular block (like a book) around an axis that goes straight through its center and pops out of its surface, we first figure out how hard it is to spin it around lines that go through its center along its length and width. Then, we simply add those two "hardnesses" together to get the final answer: \( \frac{M(L^2 + B^2)}{12} \).

🎯 Exam Tip: For problems involving rectangular laminas, first calculate the moments of inertia about axes parallel to the sides passing through the center of mass (\( I_x \) and \( I_y \)), then apply the perpendicular axis theorem to find \( I_z \).

Question 6. Establish the formula of moment of inertia of a rectangular rod along the axis passing through its center of mass and perpendicular to the length of the rod.
Answer:
Moment of Inertia of a Solid Rod of Rectangular Cross Section about the Axis Perpendicular to its Length and Passing through its Center
Imagine a uniform thin rod AB with mass \( M \) and length \( L \). Let the YY' axis pass through the center of the rod and be perpendicular to its length. We need to calculate the moment of inertia about this YY' axis.
If we take a small piece \( dx \) at a distance \( x \) from the YY' axis, its mass will be \( \left(\frac{M}{L}\right) dx \).
The moment of inertia of this small piece about the YY' axis is \( dl = \left[\left(\frac{M}{L}\right) dx\right] x^{2} \).
To find the total moment of inertia for the entire rod about the YY' axis, we integrate from \( -L/2 \) to \( L/2 \):
\( I = \int dI = \int_{-L/2}^{L/2} \left(\frac{M}{L}\right) x^{2} dx \)
\( I = \frac{M}{L} \int_{-L/2}^{L/2} x^{2} dx \)
\( I = \frac{M}{L} \left[ \frac{x^{3}}{3} \right]_{-L/2}^{L/2} \)
\( I = \frac{M}{L} \left[ \frac{(L/2)^{3}}{3} - \frac{(-L/2)^{3}}{3} \right] \)
\( I = \frac{M}{L} \left[ \frac{L^{3}}{24} - \frac{-L^{3}}{24} \right] \)
\( I = \frac{M}{L} \left[ \frac{L^{3}}{24} + \frac{L^{3}}{24} \right] \)
\( I = \frac{M}{L} \left[ \frac{2L^{3}}{24} \right] \)
\( I = \frac{M}{L} \left[ \frac{L^{3}}{12} \right] \)
\( I = \frac{ML^{2}}{12} \)
This is the formula for the moment of inertia of a rectangular rod about an axis perpendicular to its length and passing through its center. The distribution of mass affects how much force is needed to make an object spin.
In simple words: For a straight rod spinning around its middle, the moment of inertia is found by a formula that uses its total mass and how long it is. This formula helps us understand how hard it is to make the rod start or stop spinning.

🎯 Exam Tip: Remember to use the correct limits of integration (from \( -L/2 \) to \( L/2 \)) for an axis passing through the center of mass to ensure accuracy.

Question 8. Suppose the mass of a hollow sphere is M, \( \rho \) is the density, inner radius \( R_2 \) and outer radius \( R_1 \). Calculate the moment of inertia of the hollow sphere about its diameter.
Answer:
To find the moment of inertia of a hollow sphere, we can think of it as a large solid sphere with radius \( R_1 \) from which a smaller solid sphere with radius \( R_2 \) has been removed. The moment of inertia of a solid sphere about its diameter is \( \frac{2}{5} MR^2 \).
The density \( \rho \) of the sphere is given by \( \rho = \frac{\text{Mass}}{\text{Volume}} \). For a sphere, the volume is \( \frac{4}{3} \pi R^3 \).
So, \( M = \rho \times \text{Volume} \).
Mass of the large solid sphere (if it were completely solid) with radius \( R_1 \) would be \( M_1 = \rho \frac{4}{3} \pi R_1^3 \).
Mass of the smaller solid sphere (that is removed) with radius \( R_2 \) would be \( M_2 = \rho \frac{4}{3} \pi R_2^3 \).
The moment of inertia of the hollow sphere \( I \) is the difference between the moment of inertia of the large sphere \( I_1 \) and the small sphere \( I_2 \):
\( I = I_1 - I_2 \)
\( I = \frac{2}{5} M_1 R_1^2 - \frac{2}{5} M_2 R_2^2 \)
Substitute the expressions for \( M_1 \) and \( M_2 \):
\( I = \frac{2}{5} \left( \rho \frac{4}{3} \pi R_1^3 \right) R_1^2 - \frac{2}{5} \left( \rho \frac{4}{3} \pi R_2^3 \right) R_2^2 \)
\( I = \frac{2}{5} \rho \frac{4}{3} \pi (R_1^5 - R_2^5) \)
We know the total mass of the hollow sphere is \( M = \rho \frac{4}{3} \pi (R_1^3 - R_2^3) \).
From this, we can express density \( \rho = \frac{M}{\frac{4}{3} \pi (R_1^3 - R_2^3)} \).
Substitute this \( \rho \) back into the equation for \( I \):
\( I = \frac{2}{5} \left( \frac{M}{\frac{4}{3} \pi (R_1^3 - R_2^3)} \right) \frac{4}{3} \pi (R_1^5 - R_2^5) \)
\( I = \frac{2}{5} M \frac{(R_1^5 - R_2^5)}{(R_1^3 - R_2^3)} \)
This formula helps calculate how a hollow sphere resists changes in its rotation. The amount of material further from the center makes it harder to spin.
In simple words: To find how hard it is to spin a hollow ball, you subtract the "spin resistance" of the empty space from the "spin resistance" of a solid ball of the same outer size. The final formula uses the total mass and both inner and outer radii.

🎯 Exam Tip: When calculating moments of inertia for hollow objects, remember to use the principle of superposition: subtract the moment of inertia of the removed part from that of the original solid body.

Question 14. Derive the formula of total kinetic energy of a body at the bottom rolling down on an inclined plane.
Answer:
When a body rolls down an inclined plane, its total kinetic energy is a combination of two types of energy: translational kinetic energy (due to its overall movement) and rotational kinetic energy (due to its spinning motion).
The total kinetic energy \( E \) of the body is given by:
\( E = \text{Translational Kinetic Energy} + \text{Rotational Kinetic Energy} \)
Translational kinetic energy is \( KE_t = \frac{1}{2} Mv^2 \), where \( M \) is the mass of the body and \( v \) is its linear velocity.
Rotational kinetic energy is \( KE_r = \frac{1}{2} I\omega^2 \), where \( I \) is the moment of inertia of the body and \( \omega \) is its angular velocity.
So, the total kinetic energy is \( E = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2 \).
For a rolling body, the linear velocity \( v \) and angular velocity \( \omega \) are related by \( v = R\omega \), or \( \omega = \frac{v}{R} \), where \( R \) is the radius of the body.
Also, the moment of inertia \( I \) can be written as \( I = MK^2 \), where \( K \) is the radius of gyration.
Substitute \( I = MK^2 \) and \( \omega = \frac{v}{R} \) into the rotational kinetic energy term:
\( KE_r = \frac{1}{2} (MK^2) \left(\frac{v}{R}\right)^2 \)
\( KE_r = \frac{1}{2} MK^2 \frac{v^2}{R^2} \)
\( KE_r = \frac{1}{2} Mv^2 \frac{K^2}{R^2} \)
Now, substitute this back into the total kinetic energy equation:
\( E = \frac{1}{2} Mv^2 + \frac{1}{2} Mv^2 \frac{K^2}{R^2} \)
\( E = \frac{1}{2} Mv^2 \left(1 + \frac{K^2}{R^2}\right) \)
This is the formula for the total kinetic energy of a body rolling down an inclined plane. This formula helps us to understand how objects like wheels and balls move and spin at the same time.
In simple words: When an object rolls down a slope, its energy comes from both moving forward and spinning. The total energy formula combines these two parts, showing how its speed, mass, and how its mass is spread out (radius of gyration) all play a role.

🎯 Exam Tip: Remember that for a purely rolling motion, the linear velocity \( v \) and angular velocity \( \omega \) are related by \( v=R\omega \), which is crucial for combining the translational and rotational kinetic energies.

Question 15. Establish the relation between angular momentum and angular velocity.
Answer:
Relation between Angular Momentum, Moment of Inertia, and Angular Velocity
Angular momentum (\( J \)) is a measure of an object's tendency to continue rotating. For a single particle with mass \( m \) moving in a circle of radius \( r \) with linear velocity \( v \), the angular momentum is given by \( J = rmv \).
We know that linear velocity \( v \) is related to angular velocity \( \omega \) by \( v = r\omega \).
Substituting \( v = r\omega \) into the angular momentum equation:
\( J = r m (r\omega) \)
\( J = mr^2 \omega \)
For a rigid body, which is made up of many particles, the total angular momentum is the sum of the angular momenta of all individual particles. If the body rotates about an axis, each particle \( i \) with mass \( m_i \) at a distance \( r_i \) from the axis will have angular momentum \( m_i r_i^2 \omega \).
Since all particles in a rigid body rotating about a fixed axis have the same angular velocity \( \omega \):
\( J_{\text{total}} = \sum (m_i r_i^2 \omega) \)
\( J_{\text{total}} = \left(\sum m_i r_i^2\right) \omega \)
The term \( \sum m_i r_i^2 \) is defined as the moment of inertia (\( I \)) of the body about the axis of rotation.
So, \( I = \sum m_i r_i^2 \).
Substituting \( I \) into the equation for total angular momentum:
\( J = I\omega \)
This is the fundamental relationship between angular momentum, moment of inertia, and angular velocity. This relationship is very much like how linear momentum (\( p = mv \)) works, but for spinning objects. If the body spins faster, its angular momentum increases.
In simple words: Angular momentum is how much an object wants to keep spinning. It's found by multiplying the object's "spin inertia" (moment of inertia) by how fast it's spinning (angular velocity). It's the rotational version of how mass and speed give an object linear momentum.

🎯 Exam Tip: Draw an analogy with linear motion to remember the relationship: force is to torque as mass is to moment of inertia as linear acceleration is to angular acceleration, and linear momentum is to angular momentum.

RBSE Class 11 Physics Chapter 7 Numerical Questions

Question 1. Two particles of masses 5 g and 2 g respectively; have velocity \( (\hat{i}+\hat{j}-\hat{k}) \) and \( (3 \hat{i}-2 \hat{j}+\hat{k}) \) cm/s; and form a two particles system. What will be the velocity of the center of mass of the system?
Answer:
Given:
Mass of first particle \( m_1 = 5 \text{ g} \)
Mass of second particle \( m_2 = 2 \text{ g} \)
Velocity of first particle \( \vec{v_1} = (\hat{i}+\hat{j}-\hat{k}) \text{ cm/s} \)
Velocity of second particle \( \vec{v_2} = (3\hat{i}-2\hat{j}+\hat{k}) \text{ cm/s} \)
We need to find the velocity of the center of mass \( \vec{v}_{\text{cm}} \).
The formula for the velocity of the center of mass for a two-particle system is:
\( \vec{v}_{\text{cm}} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2} \)
Substitute the given values:
\( \vec{v}_{\text{cm}} = \frac{5(\hat{i}+\hat{j}-\hat{k}) + 2(3\hat{i}-2\hat{j}+\hat{k})}{5+2} \)
\( \vec{v}_{\text{cm}} = \frac{(5\hat{i}+5\hat{j}-5\hat{k}) + (6\hat{i}-4\hat{j}+2\hat{k})}{7} \)
Combine the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) components:
\( \vec{v}_{\text{cm}} = \frac{(5+6)\hat{i} + (5-4)\hat{j} + (-5+2)\hat{k}}{7} \)
\( \vec{v}_{\text{cm}} = \frac{11\hat{i} + 1\hat{j} - 3\hat{k}}{7} \)
\( \vec{v}_{\text{cm}} = \frac{11}{7}\hat{i} + \frac{1}{7}\hat{j} - \frac{3}{7}\hat{k} \text{ cm/s} \)
This calculation shows where the combined system would move if all its mass were concentrated at a single point. It's like finding the average velocity, but weighted by mass.
In simple words: To find the speed and direction of the center of mass, you add up each particle's mass multiplied by its velocity, then divide by the total mass. This gives you the overall motion of the whole group of particles.

🎯 Exam Tip: Remember to treat vector components (\( \hat{i}, \hat{j}, \hat{k} \)) separately and sum them correctly. Double-check your arithmetic, especially signs.

Question 2. A wheel which is at rest, rotates with 3.0 rad/s² angular acceleration for 2.0 s. What will be the angular velocity gained and how much would be the displacement in this time interval?
Answer:
Given:
Initial angular velocity \( \omega_0 = 0 \) (since the wheel is at rest)
Angular acceleration \( \alpha = 3.0 \text{ rad/s}^2 \)
Time \( t = 2.0 \text{ s} \)
We need to find the final angular velocity \( \omega \) and the angular displacement \( \theta \).
Using the first equation of rotational motion:
\( \omega = \omega_0 + \alpha t \)
\( \omega = 0 + (3.0 \text{ rad/s}^2)(2.0 \text{ s}) \)
\( \omega = 6.0 \text{ rad/s} \)
Now, using the second equation of rotational motion for angular displacement:
\( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
\( \theta = (0)(2.0 \text{ s}) + \frac{1}{2} (3.0 \text{ rad/s}^2)(2.0 \text{ s})^2 \)
\( \theta = 0 + \frac{1}{2} (3.0)(4.0) \)
\( \theta = \frac{1}{2} (12.0) \)
\( \theta = 6.0 \text{ radians} \)
So, the wheel gains an angular velocity of 6.0 rad/s and rotates through an angle of 6.0 radians in 2.0 seconds. This shows how quickly the wheel speeds up and how far it turns when it starts from rest.
In simple words: The wheel starts still, then speeds up. After two seconds, it spins at 6 radians per second and has turned a total of 6 radians. These calculations use basic spinning motion formulas.

🎯 Exam Tip: Remember to use the correct units (radians for angular displacement and angular velocity, rad/s² for angular acceleration) and ensure consistency in your calculations.

Question 3. A car which is at rest is accelerated with an acceleration of 40 rad/s². In how much time will it get the angular velocity of 800 cycles/minute?
Answer:
Given:
Initial angular velocity \( \omega_0 = 0 \) (since the car starts from rest)
Angular acceleration \( \alpha = 40 \text{ rad/s}^2 \)
Final angular velocity \( \omega = 800 \text{ cycles/minute} \)
First, convert the final angular velocity from cycles/minute to rad/s:
\( 1 \text{ cycle} = 2\pi \text{ radians} \)
\( 1 \text{ minute} = 60 \text{ seconds} \)
\( \omega = 800 \frac{\text{cycles}}{\text{minute}} = 800 \times \frac{2\pi \text{ rad}}{60 \text{ s}} \)
\( \omega = \frac{1600\pi}{60} \text{ rad/s} \)
\( \omega = \frac{160\pi}{6} \text{ rad/s} \)
\( \omega = \frac{80\pi}{3} \text{ rad/s} \)
Using \( \pi \approx 3.14 \):
\( \omega \approx \frac{80 \times 3.14}{3} = \frac{251.2}{3} \approx 83.73 \text{ rad/s} \)
Now, use the first equation of rotational motion:
\( \omega = \omega_0 + \alpha t \)
\( \frac{80\pi}{3} = 0 + (40) t \)
\( t = \frac{80\pi}{3 \times 40} \)
\( t = \frac{2\pi}{3} \text{ s} \)
Using \( \pi \approx 3.14 \):
\( t = \frac{2 \times 3.14}{3} = \frac{6.28}{3} \approx 2.093 \text{ s} \)
It will take approximately 2.093 seconds for the car to reach the desired angular velocity. This calculation helps determine the time needed for an object to spin up to a certain speed under constant acceleration.
In simple words: We need to find how long it takes for a car's wheel to spin at a certain speed, starting from still. First, convert the speed to a standard unit (radians per second). Then, use the spin acceleration to calculate the time needed.

🎯 Exam Tip: Always convert angular velocity units (like cycles/minute or rpm) to standard SI units (rad/s) before applying kinematic equations to avoid errors.

Question 4. Calculate the center of mass of three particles kept on the vertices of an equilateral triangle. The masses of the particles are 100 g, 150 g and 200 g. The length of each side of the triangle is 0.5 m.
Answer:
Given:
Masses: \( m_1 = 100 \text{ g} \), \( m_2 = 150 \text{ g} \), \( m_3 = 200 \text{ g} \)
Side length of equilateral triangle \( L = 0.5 \text{ m} \)
Let's place the triangle in a coordinate system.
Place \( m_1 \) at the origin: \( (x_1, y_1) = (0, 0) \).
Place \( m_3 \) on the x-axis: \( (x_3, y_3) = (L, 0) = (0.5, 0) \).
For an equilateral triangle, the coordinates of \( m_2 \) will be \( (L/2, L\frac{\sqrt{3}}{2}) \).
So, \( x_2 = \frac{0.5}{2} = 0.25 \text{ m} \).
\( y_2 = 0.5 \times \frac{\sqrt{3}}{2} = 0.5 \times \frac{1.732}{2} = 0.5 \times 0.866 = 0.433 \text{ m} \).
So, the coordinates are:
\( m_1 = 100 \text{ g} \) at \( (0, 0) \)
\( m_2 = 150 \text{ g} \) at \( (0.25, 0.433) \)
\( m_3 = 200 \text{ g} \) at \( (0.5, 0) \)
Now, calculate the x-coordinate of the center of mass \( X_{\text{cm}} \):
\( X_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \)
\( X_{\text{cm}} = \frac{(100)(0) + (150)(0.25) + (200)(0.5)}{100 + 150 + 200} \)
\( X_{\text{cm}} = \frac{0 + 37.5 + 100}{450} \)
\( X_{\text{cm}} = \frac{137.5}{450} \approx 0.3056 \text{ m} \)
Now, calculate the y-coordinate of the center of mass \( Y_{\text{cm}} \):
\( Y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \)
\( Y_{\text{cm}} = \frac{(100)(0) + (150)(0.433) + (200)(0)}{100 + 150 + 200} \)
\( Y_{\text{cm}} = \frac{0 + 64.95 + 0}{450} \)
\( Y_{\text{cm}} = \frac{64.95}{450} \approx 0.1443 \text{ m} \)
The center of mass for these three particles is approximately at \( (0.3056 \text{ m}, 0.1443 \text{ m}) \). This point represents the average position of the total mass of the system.


In simple words: We place the triangle on a graph and find the coordinates of each corner where the masses are. Then, we use a special formula that weighs each coordinate by its mass to find the exact point where the overall "balance" of the system would be.

🎯 Exam Tip: When setting up coordinates for geometric shapes, choose an origin and axes that simplify calculations (e.g., placing one particle at the origin and one side along an axis).

Question 5. To generate angular acceleration of 4 rad/s² in a rotating body a torque of \( 2.0 \times 10^{-4} \) Nm is applied. What is the moment of inertia of the body?
Answer:
Given:
Angular acceleration \( \alpha = 4 \text{ rad/s}^2 \)
Torque \( \tau = 2.0 \times 10^{-4} \text{ Nm} \)
We need to find the moment of inertia \( I \).
The relationship between torque, moment of inertia, and angular acceleration is given by:
\( \tau = I \alpha \)
To find \( I \), we can rearrange the formula:
\( I = \frac{\tau}{\alpha} \)
Substitute the given values:
\( I = \frac{2.0 \times 10^{-4} \text{ Nm}}{4 \text{ rad/s}^2} \)
\( I = 0.5 \times 10^{-4} \text{ kg m}^2 \)
\( I = 5.0 \times 10^{-5} \text{ kg m}^2 \)
The moment of inertia of the body is \( 5.0 \times 10^{-5} \text{ kg m}^2 \). This value tells us how much rotational "stubbornness" the body has, meaning how hard it is to change its spinning motion.
In simple words: We know how much twisting force (torque) is applied and how fast it makes an object spin faster (angular acceleration). By dividing the torque by the acceleration, we find the object's "spin inertia" (moment of inertia).

🎯 Exam Tip: This question is a direct application of Newton's second law for rotation (\( \tau = I\alpha \)). Ensure units are consistent: Newton-meters for torque, rad/s² for angular acceleration, and kg m² for moment of inertia.

Question 6. The two atomic molecules have masses \( m_1 \) and \( m_2 \) in a diatomic molecule. The distance between them is \( a \) m. Calculate the moment of inertia of the system along the axis passing through the center of gravity and perpendicular to the line joining the molecules.
Answer:
Given:
Masses of two atoms: \( m_1 \) and \( m_2 \)
Distance between them: \( a \)
We need to calculate the moment of inertia about an axis passing through the center of mass and perpendicular to the line joining the molecules.
Let the center of mass (C.G.) be at a distance \( r_1 \) from \( m_1 \) and \( r_2 \) from \( m_2 \).
For the center of mass, \( m_1 r_1 = m_2 r_2 \).
Also, the total distance between the masses is \( a = r_1 + r_2 \).
From \( r_1 + r_2 = a \implies r_2 = a - r_1 \).
Substitute this into the center of mass equation:
\( m_1 r_1 = m_2 (a - r_1) \)
\( m_1 r_1 = m_2 a - m_2 r_1 \)
\( m_1 r_1 + m_2 r_1 = m_2 a \)
\( r_1 (m_1 + m_2) = m_2 a \)
\( \implies r_1 = \frac{m_2 a}{m_1 + m_2} \)
Now, find \( r_2 \):
\( r_2 = a - r_1 = a - \frac{m_2 a}{m_1 + m_2} = \frac{a(m_1 + m_2) - m_2 a}{m_1 + m_2} = \frac{m_1 a}{m_1 + m_2} \)
The moment of inertia of the system about an axis passing through the center of mass and perpendicular to the line joining the molecules is the sum of the moments of inertia of each particle about that axis:
\( I = m_1 r_1^2 + m_2 r_2^2 \)
Substitute the expressions for \( r_1 \) and \( r_2 \):
\( I = m_1 \left( \frac{m_2 a}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 a}{m_1 + m_2} \right)^2 \)
\( I = m_1 \frac{m_2^2 a^2}{(m_1 + m_2)^2} + m_2 \frac{m_1^2 a^2}{(m_1 + m_2)^2} \)
\( I = \frac{m_1 m_2^2 a^2 + m_2 m_1^2 a^2}{(m_1 + m_2)^2} \)
Factor out \( m_1 m_2 a^2 \) from the numerator:
\( I = \frac{m_1 m_2 a^2 (m_2 + m_1)}{(m_1 + m_2)^2} \)
\( I = \frac{m_1 m_2 a^2}{m_1 + m_2} \)
This formula gives the moment of inertia for a diatomic molecule spinning around its central balance point. It shows how the masses and distance between them affect the molecule's resistance to rotational changes.
In simple words: For two atoms connected together, spinning around their shared balance point, the "spin inertia" (moment of inertia) depends on both of their masses and the distance between them. The formula helps us calculate this spinning property.

🎯 Exam Tip: For systems of discrete particles, always first locate the center of mass, as the moment of inertia calculation will typically be simplest when referred to an axis through the center of mass.

Question 7. A circular disc whose radius is 0.5 m and mass 25 kg is rotating on its axis with 120 revolutions/minute. Calculate the moment of inertia of the disc and rotational kinetic energy.
Answer:
Given:
Radius of the disc \( R = 0.5 \text{ m} \)
Mass of the disc \( M = 25 \text{ kg} \)
Angular speed \( = 120 \text{ revolutions/minute} \)
First, calculate the angular frequency \( f \) in rev/s:
\( f = \frac{120 \text{ revolutions}}{1 \text{ minute}} = \frac{120 \text{ revolutions}}{60 \text{ seconds}} = 2 \text{ rev/s} \)
Now, calculate the angular velocity \( \omega \) in rad/s:
\( \omega = 2\pi f = 2\pi (2) = 4\pi \text{ rad/s} \)
Moment of inertia of a circular disc about its own axis (passing through its center and perpendicular to its plane) is given by:
\( I = \frac{1}{2} MR^2 \)
Substitute the values of \( M \) and \( R \):
\( I = \frac{1}{2} (25 \text{ kg}) (0.5 \text{ m})^2 \)
\( I = \frac{1}{2} (25)(0.25) \)
\( I = \frac{6.25}{2} = 3.125 \text{ kg m}^2 \)
Now, calculate the rotational kinetic energy \( E_r \):
\( E_r = \frac{1}{2} I\omega^2 \)
Substitute the values of \( I \) and \( \omega \):
\( E_r = \frac{1}{2} (3.125 \text{ kg m}^2) (4\pi \text{ rad/s})^2 \)
\( E_r = \frac{1}{2} (3.125) (16\pi^2) \)
\( E_r = (3.125) (8\pi^2) \)
Using \( \pi^2 \approx (3.14)^2 \approx 9.86 \):
\( E_r = 3.125 \times 8 \times 9.86 \)
\( E_r = 25 \times 9.86 \)
\( E_r = 246.5 \text{ J} \)
The moment of inertia of the disc is 3.125 kg m² and its rotational kinetic energy is 246.5 J. This tells us how much energy is stored in the disc due to its spinning motion.
In simple words: First, convert the disc's spinning speed into radians per second. Then, use formulas to find its "spin inertia" (moment of inertia) and the energy it has because it's spinning (rotational kinetic energy).

🎯 Exam Tip: Remember the formula for the moment of inertia of a disc is \( \frac{1}{2}MR^2 \). Be careful to convert angular speed units to rad/s before using them in calculations for rotational kinetic energy.

Question 8. What will be the moment of inertia of a thin rod of mass M and length I along the axis perpendicular to its length through one point from a distance 1/4 from one end?
Answer:
Given:
Mass of the thin rod \( = M \)
Length of the thin rod \( = L \)
The axis is perpendicular to the length of the rod and passes through a point at a distance \( \frac{L}{4} \) from one end.
First, find the moment of inertia about an axis passing through the center of mass and perpendicular to the rod's length. This is \( I_{\text{cm}} = \frac{1}{12} ML^2 \).
The center of mass of the rod is at its midpoint, i.e., at a distance \( \frac{L}{2} \) from either end.
The axis of rotation is at a distance \( \frac{L}{4} \) from one end.
We need to find the distance \( d \) between the center of mass and the new axis.
Distance of the center of mass from the end is \( \frac{L}{2} \).
Distance of the new axis from the same end is \( \frac{L}{4} \).
So, the distance \( d \) between the center of mass and the new axis is:
\( d = \left| \frac{L}{2} - \frac{L}{4} \right| = \left| \frac{2L - L}{4} \right| = \frac{L}{4} \)
Now, we can use the parallel axis theorem, which states that \( I = I_{\text{cm}} + Md^2 \).
Substitute the values of \( I_{\text{cm}} \) and \( d \):
\( I = \frac{1}{12} ML^2 + M\left(\frac{L}{4}\right)^2 \)
\( I = \frac{1}{12} ML^2 + M\frac{L^2}{16} \)
To add these fractions, find a common denominator, which is 48:
\( I = \frac{4}{48} ML^2 + \frac{3}{48} ML^2 \)
\( I = \frac{4+3}{48} ML^2 \)
\( I = \frac{7}{48} ML^2 \)
The moment of inertia of the thin rod about the given axis is \( \frac{7}{48} ML^2 \). This result demonstrates how moving the axis of rotation away from the center of mass increases an object's resistance to spinning.
In simple words: To find how hard it is to spin a rod around a point not in its middle, we first find the spin inertia around its true middle. Then, we use a special rule (the parallel axis theorem) to add an extra amount of spin inertia because the new spin point is off-center.

🎯 Exam Tip: For problems involving an axis not passing through the center of mass, the parallel axis theorem (\( I = I_{\text{cm}} + Md^2 \)) is essential. Be careful in calculating the distance \( d \) between the center of mass and the new axis.

Question 9. Calculate the angular momentum of the spherical Earth rotating along its axis. (Mass of the Earth = \( 6 \times 10^{24} \) kg and radius = \( 6.4 \times 10^6 \) m)
Answer:
Given:
Mass of the Earth \( M = 6 \times 10^{24} \text{ kg} \)
Radius of the Earth \( R = 6.4 \times 10^6 \text{ m} \)
Time period of rotation of the Earth \( T = 24 \text{ hours} \).
First, convert the time period to seconds:
\( T = 24 \text{ hours} \times 3600 \text{ s/hour} = 86400 \text{ s} \)
The Earth is a solid sphere. The moment of inertia of a solid sphere about its diameter is:
\( I = \frac{2}{5} MR^2 \)
Substitute the values of \( M \) and \( R \):
\( I = \frac{2}{5} (6 \times 10^{24} \text{ kg}) (6.4 \times 10^6 \text{ m})^2 \)
\( I = \frac{2}{5} (6 \times 10^{24}) (40.96 \times 10^{12}) \)
\( I = \frac{2}{5} (245.76 \times 10^{36}) \)
\( I = 98.304 \times 10^{36} = 9.8304 \times 10^{37} \text{ kg m}^2 \)
Next, calculate the angular velocity \( \omega \) of the Earth:
\( \omega = \frac{2\pi}{T} \)
\( \omega = \frac{2\pi}{86400 \text{ s}} \text{ rad/s} \)
Now, calculate the angular momentum \( J \) of the Earth:
\( J = I\omega \)
\( J = (9.8304 \times 10^{37} \text{ kg m}^2) \left( \frac{2\pi}{86400} \text{ rad/s} \right) \)
Using \( \pi \approx 3.14159 \):
\( J = \frac{9.8304 \times 10^{37} \times 2 \times 3.14159}{86400} \)
\( J \approx \frac{61.776 \times 10^{37}}{86400} \)
\( J \approx 0.000715 \times 10^{37} \)
\( J \approx 7.15 \times 10^{33} \text{ kg m}^2\text{/s} \)
The angular momentum of the Earth is approximately \( 7.15 \times 10^{33} \text{ kg m}^2\text{/s} \). This huge value shows how the Earth's massive spinning motion keeps it stable in space.
In simple words: To find Earth's spinning momentum, we first calculate how difficult it is to change its spin (moment of inertia) using its mass and radius. Then, we find its spinning speed from how long a day is. Multiplying these two gives us the Earth's total angular momentum.

🎯 Exam Tip: Ensure all units are in SI (kilograms, meters, seconds) before calculation. For celestial bodies, remember to use the moment of inertia formula for a solid sphere unless specified otherwise.

Question 10. The mass of a hollow sphere is 1 kg and its inner and outer radius are 0.1 m and 0.2 m respectively. Calculate the moment of inertia of the sphere along the diameter and radius of gyration.
Answer:
Given:
Total mass of hollow sphere \( M = 1 \text{ kg} \)
Outer radius \( R_1 = 0.2 \text{ m} \)
Inner radius \( R_2 = 0.1 \text{ m} \)
We need to calculate the moment of inertia \( I \) and the radius of gyration \( K \).
The formula for the moment of inertia of a hollow sphere about its diameter is:
\( I = \frac{2}{5} M \frac{(R_1^5 - R_2^5)}{(R_1^3 - R_2^3)} \)
First, calculate \( R_1^5 \), \( R_2^5 \), \( R_1^3 \), and \( R_2^3 \):
\( R_1^5 = (0.2)^5 = 0.00032 \text{ m}^5 = 32 \times 10^{-5} \text{ m}^5 \)
\( R_2^5 = (0.1)^5 = 0.00001 \text{ m}^5 = 1 \times 10^{-5} \text{ m}^5 \)
\( R_1^3 = (0.2)^3 = 0.008 \text{ m}^3 = 8 \times 10^{-3} \text{ m}^3 \)
\( R_2^3 = (0.1)^3 = 0.001 \text{ m}^3 = 1 \times 10^{-3} \text{ m}^3 \)
Now, substitute these values into the formula for \( I \):
\( I = \frac{2}{5} (1 \text{ kg}) \frac{(32 \times 10^{-5} - 1 \times 10^{-5}) \text{ m}^5}{(8 \times 10^{-3} - 1 \times 10^{-3}) \text{ m}^3} \)
\( I = \frac{2}{5} \frac{(31 \times 10^{-5})}{(7 \times 10^{-3})} \)
\( I = \frac{2}{5} \frac{31}{7} \times \frac{10^{-5}}{10^{-3}} \)
\( I = \frac{62}{35} \times 10^{-2} \)
\( I \approx 1.7714 \times 10^{-2} \text{ kg m}^2 \)
Now, calculate the radius of gyration \( K \). The moment of inertia is also defined as \( I = MK^2 \).
So, \( K^2 = \frac{I}{M} \)
\( K^2 = \frac{1.7714 \times 10^{-2} \text{ kg m}^2}{1 \text{ kg}} \)
\( K^2 = 1.7714 \times 10^{-2} \text{ m}^2 \)
\( K = \sqrt{1.7714 \times 10^{-2}} \text{ m} \)
\( K = \sqrt{0.017714} \text{ m} \)
\( K \approx 0.1331 \text{ m} \)
The moment of inertia of the hollow sphere is approximately \( 1.77 \times 10^{-2} \text{ kg m}^2 \), and its radius of gyration is approximately 0.133 m. This means the mass of the sphere acts as if it's all concentrated at this 'gyration radius' from the center for rotational purposes.
In simple words: We find the "spin inertia" (moment of inertia) of the hollow ball using its total mass and both inner and outer radii. Then, we find its radius of gyration, which is like an imaginary radius where all its mass could be placed to give the same spin inertia.

🎯 Exam Tip: Be careful with powers when calculating \( R^3 \) and \( R^5 \). Also, remember that the radius of gyration is a conceptual distance, not a physical radius, and is calculated directly from the moment of inertia and mass.

Question 11. A 1 kg mass ball is moving on a horizontal surface with a velocity 20 m/s and reaches a plane which makes an angle of 30° with the horizontal. If friction is negligible then how much distance vertically will the ball reach?
Answer:
Given:
Mass of the ball \( M = 1 \text{ kg} \)
Initial velocity of the ball \( u = 20 \text{ m/s} \)
Angle of the inclined plane \( \theta = 30^\circ \)
Friction is negligible.
This problem involves the conversion of kinetic energy into potential energy. Since friction is negligible, the total mechanical energy is conserved.
When the ball moves from the horizontal surface to the inclined plane, its kinetic energy will be converted into potential energy as it moves upwards vertically.
Assuming the ball is rolling without slipping on the horizontal surface, its initial kinetic energy \( KE_{initial} \) has both translational and rotational components.
For a solid ball (sphere), the moment of inertia is \( I = \frac{2}{5} MR^2 \).
The radius of gyration relation is \( \frac{K^2}{R^2} = \frac{2}{5} \).
The total kinetic energy of a rolling body is \( KE = \frac{1}{2} Mv^2 \left(1 + \frac{K^2}{R^2}\right) \).
Substitute \( \frac{K^2}{R^2} = \frac{2}{5} \):
\( KE_{initial} = \frac{1}{2} M u^2 \left(1 + \frac{2}{5}\right) \)
\( KE_{initial} = \frac{1}{2} M u^2 \left(\frac{7}{5}\right) \)
Substitute \( M = 1 \text{ kg} \) and \( u = 20 \text{ m/s} \):
\( KE_{initial} = \frac{1}{2} (1 \text{ kg}) (20 \text{ m/s})^2 \left(\frac{7}{5}\right) \)
\( KE_{initial} = \frac{1}{2} (1) (400) \left(\frac{7}{5}\right) \)
\( KE_{initial} = 200 \times \frac{7}{5} = 40 \times 7 = 280 \text{ J} \)
At the maximum vertical height \( h \), the ball momentarily stops, so all its kinetic energy is converted into potential energy \( PE_{final} \).
\( PE_{final} = Mgh \)
By conservation of energy:
\( KE_{initial} = PE_{final} \)
\( 280 \text{ J} = Mgh \)
\( 280 = (1 \text{ kg}) (10 \text{ m/s}^2) h \) (Assuming \( g = 10 \text{ m/s}^2 \))
\( 280 = 10 h \)
\( h = \frac{280}{10} = 28 \text{ m} \)
The ball will reach a vertical height of 28 meters. This calculation shows how energy changes form from motion to height when friction is not present.
In simple words: The ball's initial movement energy (kinetic energy) changes into height energy (potential energy) as it rolls up the slope. Since there's no friction, the total energy stays the same. We use the ball's mass and speed to find its starting energy, then convert that into the height it can reach.

🎯 Exam Tip: For rolling bodies, total kinetic energy includes both translational and rotational components. Ensure you use the correct moment of inertia or radius of gyration for the specific shape (e.g., solid sphere, hollow sphere, disc) to correctly calculate the total kinetic energy.

Question 13. If keeping the mass constant a rotating disc's radius is suddenly reduced to half. Then how much would be its new angular velocity value?
Answer: Let the initial moment of inertia of the disc be \(I_1\) and its initial angular velocity be \(\omega_1\). The moment of inertia is related to mass (M) and radius (R) by \(I = \frac{1}{2}MR^2\).
When the radius (\(R_1\)) is reduced to half (\(R_2 = R_1/2\)), the new moment of inertia \(I_2\) becomes:
\(I_2 = \frac{1}{2} M R_2^2 = \frac{1}{2} M \left(\frac{R_1}{2}\right)^2 = \frac{1}{2} M \frac{R_1^2}{4}\)
\( \implies I_2 = \frac{1}{4} \left(\frac{1}{2} M R_1^2\right) \)
\( \implies I_2 = \frac{1}{4} I_1 \)
According to the law of conservation of angular momentum, the total angular momentum remains constant if no external torque acts on the system. So, the initial angular momentum equals the final angular momentum:
\(I_1 \omega_1 = I_2 \omega_2\)
Now, substitute the value of \(I_2\):
\(I_1 \omega_1 = \frac{1}{4} I_1 \omega_2\)
To find the new angular velocity \(\omega_2\), we can cancel \(I_1\) from both sides:
\(\omega_1 = \frac{1}{4} \omega_2\)
\( \implies \omega_2 = 4 \omega_1 \)
Thus, when the radius is halved, the angular velocity becomes four times its original value. This principle is often seen when figures skaters pull their arms in during a spin.
In simple words: When a spinning disc's size shrinks by half (its radius), it spins much faster. The speed of its spin goes up by four times because the way its mass is spread out changes.

🎯 Exam Tip: Remember to express the new moment of inertia in terms of the initial moment of inertia before applying the conservation of angular momentum. This simplifies the calculation and reduces errors.

Question 15. M and N two wheels are on the same axis. The moment of inertia of M is 6 kg m² and is rotating with 600 cycles/min and N is at rest. Joining both of them together by a clutch they do motion by 400 cycles/min. Calculate the moment of inertia of N.
Answer: We are given the following values:
Moment of inertia of wheel M (\(I_M\)) = 6 kg m\(^2\)
Initial frequency of wheel M (\(n_M\)) = 600 cycles/min
First, convert the initial frequency to cycles per second:
\(n_M = \frac{600 \text{ cycles}}{1 \text{ min}} = \frac{600 \text{ cycles}}{60 \text{ s}} = 10 \text{ cycles/s}\)
Now, calculate the initial angular velocity of wheel M (\(\omega_M\)):
\(\omega_M = 2 \pi n_M = 2 \pi \times 10 = 20 \pi \text{ rad/s}\)
Wheel N is at rest, so its initial angular velocity is 0. Its initial moment of inertia is \(I_N\).
The wheels are joined and rotate together with a final frequency of 400 cycles/min.
Convert the final frequency to cycles per second:
\(n_{\text{final}} = \frac{400 \text{ cycles}}{1 \text{ min}} = \frac{400 \text{ cycles}}{60 \text{ s}} = \frac{20}{3} \text{ cycles/s}\)
Calculate the final angular velocity of the combined system (\(\omega_{\text{final}}\)):
\(\omega_{\text{final}} = 2 \pi n_{\text{final}} = 2 \pi \times \frac{20}{3} = \frac{40}{3} \pi \text{ rad/s}\)
According to the law of conservation of angular momentum, the total initial angular momentum equals the total final angular momentum (since no external torque acts):
\(I_M \omega_M + I_N \omega_N = (I_M + I_N) \omega_{\text{final}}\)
Since wheel N was initially at rest, \(\omega_N = 0\):
\(I_M \omega_M = (I_M + I_N) \omega_{\text{final}}\)
Substitute the known values:
\(6 \text{ kg m}^2 \times 20 \pi \text{ rad/s} = (6 \text{ kg m}^2 + I_N) \times \frac{40}{3} \pi \text{ rad/s}\)
\(120 \pi = (6 + I_N) \frac{40}{3} \pi\)
Divide both sides by \(\pi\):
\(120 = (6 + I_N) \frac{40}{3}\)
Multiply both sides by 3/40:
\(120 \times \frac{3}{40} = 6 + I_N\)
\(3 \times 3 = 6 + I_N\)
\(9 = 6 + I_N\)
\( \implies I_N = 9 - 6 \)
\( \implies I_N = 3 \text{ kg m}^2 \)
The moment of inertia of wheel N is 3 kg m\(^2\). This problem illustrates how angular momentum is conserved even when parts of a system interact and change their motion.
In simple words: Wheel M spins with a certain energy, and wheel N is still. When they join, they spin slower together. By using a rule that says total spinning energy stays the same, we can figure out how "heavy" wheel N is in terms of spinning (its moment of inertia).

🎯 Exam Tip: Always convert frequencies to standard units (rad/s or cycles/s) before using them in angular momentum calculations. Remember that angular momentum is conserved for the entire system, not individual components, when they interact internally.