GSEB Class 11 Physics Solutions Chapter 7 System of Particles and Rotational Motion

Question 1. Give the location of the centre of mass of a (i) sphere (ii) cylinder (iii) ring and (iv) cube each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Answer: The Center of Mass (C.M.) is located at the following points:
1. For a sphere, the C.M. is at its center.
2. For a cylinder, the C.M. is at the midpoint of its axis of symmetry, which is its geometrical center.
3. For a ring, the C.M. is at its center.
4. For a cube, the C.M. is at the point where its diagonals intersect, which is its geometrical center. No, the center of mass of a body does not always lie inside the body. For example, for objects like a ring, a hollow cylinder, a hollow sphere, or a hollow cube, the C.M. may be found outside the physical body.
In simple words: The center of mass is usually at the exact middle of regular shapes. But for hollow shapes, it can be outside the object itself.

Exam Tip: Remember to distinguish between solid and hollow objects when determining if the center of mass lies inside or outside the body. For solid objects, it's always inside; for hollow ones, it can be outside.

Question 2. In the HCl molecule, the separation between the nuclei of the two atoms is about \( 1.27 \text{ Å} \) (\( 1\text{Å} = 10^{-10} \text{ m} \)). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus?
Answer: Let the center of mass (CM) be at a distance \( x \text{ Å} \) from the hydrogen (H) atom. Let \( m \) be the unit mass, which equals the mass of the H atom, so \( m_1 = m \). The mass of the chlorine (Cl) atom is \( m_2 = 35.5m \) units. If \( \vec{r_1} \) and \( \vec{r_2} \) are the position vectors of the H and Cl atoms with respect to the CM as the origin, the position vector of the CM is zero.
So, \( \vec{R_{CM}} = 0 = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} \)
\( \implies m_1 \vec{r_1} + m_2 \vec{r_2} = 0 \) (i)
Here, \( \vec{r_1} = -x\hat{i} \) and \( \vec{r_2} = (1.27 - x)\hat{i} \).
From (i),
\( m(-x\hat{i}) + 35.5m(1.27 - x)\hat{i} = 0 \)
\( \implies -mx + 35.5m(1.27 - x) = 0 \)
\( \implies -x + 35.5(1.27 - x) = 0 \)
\( \implies -x + 35.5 \times 1.27 - 35.5x = 0 \)
\( \implies 35.5x + x = 35.5 \times 1.27 \)
\( \implies 36.5x = 35.5 \times 1.27 \)
\( \implies x = \frac{35.5 \times 1.27}{36.5} \)
\( \implies x = 1.235 \text{ Å} \)
Thus, \( x \approx 1.24 \text{ Å} \) from the H-atom and on the line connecting the H and Cl atoms.
In simple words: The center of mass in an HCl molecule is very close to the heavier chlorine atom, specifically about 1.24 Angstroms away from the hydrogen atom.

Exam Tip: Remember that the center of mass is always closer to the more massive object. In calculations, ensure you assign appropriate signs to position vectors relative to your chosen origin.

Question 3. A child sits stationary at one end of a long trolley moving uniformly with a speed \( V \) on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the C.M. of the (trolley + child) system?
Answer: The speed of the center of mass for the entire system (which includes the trolley and the child) stays unchanged at \( V \), even if the child stands up and moves around on the trolley. This occurs because the state of a system can only alter due to the influence of an external force. In this specific case, no external force is acting on the system. The forces involved when the child runs on the trolley are internal forces, meaning they originate from within the system itself.
In simple words: The center of mass speed stays the same because no outside force is pushing or pulling the trolley-child system. Only internal forces are at play.

Exam Tip: This question highlights the principle of conservation of momentum. If no external forces act on a system, its total momentum (and thus the velocity of its center of mass) remains constant.

Question 4. Show that the area of the triangle contained between the vectors \( \vec{a} \) and \( \vec{b} \) is one half of the magnitude of \( \vec{a} \times \vec{b} \).
Answer: Let the vectors \( \vec{a} \) and \( \vec{b} \) represent two adjacent sides of a triangle \( \Delta AOB \).
Let \( OA = b \) and \( OB = a \).
Let the angle between \( \vec{a} \) and \( \vec{b} \) be \( \theta \), so \( \angle AOB = \theta \).
Also, let \( h \) be the height of the triangle, such that \( h = AC \).

In the right-angled triangle \( \triangle OCA \),
\( \sin \theta = \frac{AC}{OA} \)
\( \implies AC = OA \sin \theta \)
\( \implies h = b \sin \theta \) (1)
The area of \( \triangle AOB \) is calculated as:
Area \( = \frac{1}{2} \times \text{base} \times \text{height} \)
Area \( = \frac{1}{2} \times OB \times AC \)
Area \( = \frac{1}{2} \times a \times h \)
Area \( = \frac{1}{2} \times a \times b \sin \theta \)
Area \( = \frac{1}{2} ab \sin \theta \) (2)
By the definition of the cross-product of two vectors:
\( \vec{a} \times \vec{b} = ab \sin \theta \hat{n} \)
The magnitude is \( |\vec{a} \times \vec{b}| = |ab \sin \theta \hat{n}| \)
\( \implies |\vec{a} \times \vec{b}| = ab \sin \theta |\hat{n}| \)
Since \( |\hat{n}| = 1 \),
\( |\vec{a} \times \vec{b}| = ab \sin \theta \) (3)
From (2) and (3), we can see that:
Area of \( \triangle AOB = \frac{1}{2} |\vec{a} \times \vec{b}| \)
This proves that the area of a triangle formed by two vectors is half the magnitude of their cross product.
In simple words: If you have two vectors forming a triangle, the space inside that triangle is half the size of the cross-product of those two vectors.

Exam Tip: Remember that the cross product of two vectors gives a new vector perpendicular to both, and its magnitude is related to the area of the parallelogram formed by the vectors. A triangle is half of such a parallelogram.

Question 5. Show that \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is equal in magnitude to the volume of the parallelopiped formed on the three vectors \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \).
Answer: Let us consider a parallelopiped \( OABCDEFG \) with its sides \( OA, OC \), and \( OE \) such that \( \overrightarrow{OA} = \vec{b} \), \( \overrightarrow{OC} = \vec{c} \), and \( \overrightarrow{OE} = \vec{a} \).
The vectors \( \vec{b} \) and \( \vec{c} \) represent the two adjacent sides of the parallelogram \( OABC \).
If \( S \) represents the area of the face \( OABC \), which is a parallelogram, then
\( \vec{S} = \vec{b} \times \vec{c} = S\hat{n} \)
where \( \hat{n} \) is a unit vector acting along \( S \) or perpendicular to the plane containing \( \vec{b} \) and \( \vec{c} \). \( S \) is the magnitude of the area of face \( OABC \).

Let \( \theta \) be the angle made by vector \( \vec{a} \) with the vector \( \vec{S} \).
Then \( \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot \vec{S} = a S \cos \theta \).
Since \( |\hat{n}| = 1 \),
\( \vec{a} \cdot (\vec{b} \times \vec{c}) = aS \cos \theta \) (i)
Here, \( h = a \cos \theta \) represents the length of the perpendicular \( EE' \) from the terminus of \( \vec{a} \) onto the surface of the parallelogram \( OABC \). This is essentially the height of the vector \( \vec{a} \) relative to the base.
Also, let \( V \) be the volume of the parallelopiped \( OABCDEFG \).
Volume \( V = \text{area of the face OABC} \times \text{normal distance from E to the face OABC} \)
\( V = S \times h \) (ii)
From equations (i) and (ii), we find that:
\( V = \vec{a} \cdot (\vec{b} \times \vec{c}) \)
Thus, it is proved that the scalar triple product \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is equal in magnitude to the volume of the parallelopiped formed by the three vectors \( \vec{a}, \vec{b} \), and \( \vec{c} \).
In simple words: The volume of a 3D box (a parallelopiped) made by three vectors is found by taking the dot product of one vector with the cross product of the other two.

Exam Tip: The scalar triple product \( (\vec{a} \cdot (\vec{b} \times \vec{c})) \) gives the volume of a parallelopiped, and its sign indicates the handedness of the vectors. Always calculate the cross product first, then the dot product.

Question 6. (a) Find the components along X, Y and Z axis of the angular momentum \( L \) of a particle, whose position vector is \( \vec{r} \) with components \( x, y \) and \( z \) and momentum is \( \vec{P} \) with
(b) Show that if the particle moves only in X - Y plane, the angular momentum has only a z - component?
Answer:
(a) Let \( OX, OY \), and \( OZ \) be the three mutually perpendicular axes. Consider a particle at point P with position vector \( \vec{r} \).
\( \vec{OP} = \vec{r} \) in the X - Y plane.
Let \( \theta \) be the angle made by its linear momentum \( \vec{P} \) with \( \vec{r} \). If \( \vec{L} \) is its angular momentum, then
\( \vec{L} = \vec{r} \times \vec{P} \) (i)
Angular momentum is a vector quantity, and its direction is determined by the right-hand rule for the vector product. As \( \vec{r} \) and \( \vec{P} \) are in the XOY plane, \( \vec{L} \) acts along the Z-axis.
In Cartesian co-ordinates,
\( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \)
and \( \vec{P} = P_x\hat{i} + P_y\hat{j} + P_z\hat{k} \) (ii)
From (i) and (ii), we get:
\( \vec{L} = (x\hat{i} + y\hat{j} + z\hat{k}) \times (P_x\hat{i} + P_y\hat{j} + P_z\hat{k}) \)
Using the determinant form:
\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ P_x & P_y & P_z \end{vmatrix} \]
\( \implies \vec{L} = \hat{i}(yP_z - zP_y) + \hat{j}(zP_x - xP_z) + \hat{k}(xP_y - yP_x) \)
Comparing this with \( \vec{L} = L_x\hat{i} + L_y\hat{j} + L_z\hat{k} \), we get:
\( L_x = yP_z - zP_y \)
\( L_y = zP_x - xP_z \)
\( L_z = xP_y - yP_x \) (iii)
Equation (iii) provides the required components of \( \vec{L} \) along the X, Y, and Z axes.
(b) We know that the torque experienced by a particle moving in the XY plane is given by
\( \tau_z = xF_y - yF_x \) (1)
where \( \tau_z \) is the component of torque acting along the Z-axis on the particle moving in the XY plane.
Let \( m \) be the mass of the particle. Let \( v_x \) and \( v_y \) be its velocity components along the X and Y axes, respectively.
According to Newton's Second Law of Motion:
\( F_x = \frac{d}{dt}(P_x) = \frac{d}{dt}(mv_x) = m\frac{dv_x}{dt} \)
and \( F_y = \frac{d}{dt}(P_y) = \frac{d}{dt}(mv_y) = m\frac{dv_y}{dt} \) (2)
From (1) and (2) we get,
\( \tau_z = x\left(m\frac{dv_y}{dt}\right) - y\left(m\frac{dv_x}{dt}\right) \)
\( \tau_z = m\left(x\frac{dv_y}{dt} - y\frac{dv_x}{dt}\right) \) (3)
Now, let's consider \( \frac{d}{dt}(xv_y - yv_x) \).
\( \frac{d}{dt}(xv_y - yv_x) = x\frac{dv_y}{dt} + v_y\frac{dx}{dt} - \left(y\frac{dv_x}{dt} + v_x\frac{dy}{dt}\right) \)
\( = x\frac{dv_y}{dt} + v_y v_x - y\frac{dv_x}{dt} - v_x v_y \)
\( = x\frac{dv_y}{dt} - y\frac{dv_x}{dt} \) (4)
From (3) and (4), we get
\( \tau_z = m\frac{d}{dt}(xv_y - yv_x) \)
We also know that \( P_x = mv_x \) and \( P_y = mv_y \).
So, \( \tau_z = \frac{d}{dt}(xP_y - yP_x) \) (5)
By using equation (iii) from part (a), we know that \( L_z = xP_y - yP_x \).
Substituting this into (5):
\( \tau_z = \frac{dL_z}{dt} \) (6)
Hence, from equation (6), we conclude that a particle moving in the XY plane will only have one component of angular momentum \( \vec{L} \), which is \( L_z \) acting along the Z-axis.
In simple words: (a) The angular momentum of a particle has three parts (components) along the X, Y, and Z directions, calculated using its position and momentum. (b) If a particle moves only on a flat surface (XY plane), its angular momentum will only have a part that points up or down (along the Z-axis), meaning it's perpendicular to that surface.

Exam Tip: Remember that angular momentum is a vector product, \( \vec{L} = \vec{r} \times \vec{P} \). If the motion is confined to a plane (e.g., XY), both \( \vec{r} \) and \( \vec{P} \) are in that plane, so their cross product ( \( \vec{L} \)) must be perpendicular to it (along the Z-axis). This is a fundamental concept in rotational dynamics.

Question 8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2m long. Calculate the distance of the centre of gravity of the bar from its left end?
Answer: Let \( AB \) be the non-uniform bar of weight \( W \) suspended at rest by two strings \( OA \) and \( O'B \), which make angles \( 36.9^\circ \) and \( 53.1^\circ \) with the vertical, respectively.
\( \angle OAA' = 90^\circ - 36.9^\circ = 53.1^\circ \)
Similarly, \( \angle O'BB' = 90^\circ - 53.1^\circ = 36.9^\circ \)
The length of the bar is \( AB = 2\text{m} \). Let \( AC = d\text{m} \).
Let \( T_1 \) and \( T_2 \) be the tensions in the strings \( OA \) and \( O'B \), respectively. Their rectangular components are shown in the figure.

As the rod is at rest, the vector sum of the forces acting along the A'B' axis and perpendicular to it are zero.
\( -T_1 \cos 53.1^\circ + T_2 \cos 36.9^\circ = 0 \) (i)
\( T_1 \sin 53.1^\circ + T_2 \sin 36.9^\circ - W = 0 \) (ii)
Taking the torques about A and equating the sum of torques to zero, we get:
\( (T_2 \sin 36.9^\circ) \times 2 - Wd = 0 \)
\( \implies T_2 = \frac{Wd}{2 \sin 36.9^\circ} \) (iii)
From (ii) and (iii), we get:
\( T_1 \sin 53.1^\circ = W - T_2 \sin 36.9^\circ \)
\( = W - \frac{Wd}{2 \sin 36.9^\circ} \sin 36.9^\circ \)
\( = W - \frac{Wd}{2} \) (iv)
From (i), (iii), and (iv), we get:
\( T_1 \cos 53.1^\circ = T_2 \cos 36.9^\circ \)
Substituting \( T_1 \) from (iv) and \( T_2 \) from (iii):
\( \left(W - \frac{Wd}{2}\right) \frac{\cos 53.1^\circ}{\sin 53.1^\circ} = \frac{Wd}{2 \sin 36.9^\circ} \cos 36.9^\circ \)
\( W\left(1 - \frac{d}{2}\right) \cot 53.1^\circ = \frac{Wd}{2} \cot 36.9^\circ \)
Dividing by \( W \) (since \( W \neq 0 \)):
\( \left(1 - \frac{d}{2}\right) \cot 53.1^\circ = \frac{d}{2} \cot 36.9^\circ \)
We know that \( \cot 53.1^\circ \approx 0.7508 \) and \( \cot 36.9^\circ \approx 1.3319 \).
\( \left(1 - \frac{d}{2}\right) (0.7508) = \frac{d}{2} (1.3319) \)
\( 0.7508 - 0.7508 \frac{d}{2} = 1.3319 \frac{d}{2} \)
Multiplying by 2:
\( 1.5016 - 0.7508d = 1.3319d \)
\( 1.5016 = 1.3319d + 0.7508d \)
\( 1.5016 = 2.0827d \)
\( d = \frac{1.5016}{2.0827} \approx 0.721 \text{ m} \)
The distance of the center of gravity of the bar from its left end is approximately \( 0.721 \text{ m} \), or \( 72.1 \text{ cm} \).
In simple words: By using the angles of the supporting strings and the length of the bar, we found that the center of gravity is about 72.1 cm from the left end.

Exam Tip: When dealing with equilibrium problems involving torques, choose a convenient pivot point (like one of the ends of the bar) to simplify the torque equation, as the force at that point will produce zero torque. Remember to resolve forces into components for force equilibrium equations.

Question 9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel?
Answer: Given:
Total mass of the car \( M = 1800 \text{ kg} \)
Distance between front and back axles \( L = 1.8 \text{ m} \)
Distance of C.G. from the front axle \( d_F = 1.05 \text{ m} \)
Distance of C.G. from the back axle \( d_B = L - d_F = 1.8 - 1.05 = 0.75 \text{ m} \)
Let \( R_F \) be the total normal force on the two front wheels and \( R_B \) be the total normal force on the two back wheels.
For vertical equilibrium:
\( R_F + R_B = Mg = 1800 \times 9.8 \text{ N} = 17640 \text{ N} \) (1)
Taking moments about the front axle:
\( R_B \times L = Mg \times d_F \)
\( R_B \times 1.8 = 17640 \times 1.05 \)
\( R_B = \frac{17640 \times 1.05}{1.8} = \frac{18522}{1.8} = 10290 \text{ N} \)
Now, substitute \( R_B \) into (1):
\( R_F + 10290 = 17640 \)
\( R_F = 17640 - 10290 = 7350 \text{ N} \)
The force on each back wheel is \( \frac{R_B}{2} = \frac{10290}{2} = 5145 \text{ N} \).
The force on each front wheel is \( \frac{R_F}{2} = \frac{7350}{2} = 3675 \text{ N} \).
In simple words: We calculated how much upward force the ground applies to each wheel of the car. The front wheels each get 3675 N of force, and the back wheels each get 5145 N of force.

Exam Tip: When solving equilibrium problems, always draw a clear diagram and specify your pivot point for torque calculations. Remember that the sum of upward forces must equal the sum of downward forces (weight), and the sum of torques about any point must be zero.

Question 10. (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the M.I. of
(b) Given the M.I. of disc of mass M and radius R about any of its diameters to the \( MR^2/4 \), find its M.I. about an axis normal to the disc and passing through a point on its edge?
Answer:
(a) Let \( I_{AB} \) be the moment of inertia (M.I.) of the given sphere of radius \( R \) and mass \( m \) about its diameter \( AB \).
\( I_{AB} = \frac{2}{5} MR^2 \) (i)
This implies the sphere is solid. Let \( CD \) be a tangent to the sphere, parallel to its diameter \( AB \).
The distance between the two parallel axes (diameter and tangent) is \( R \). If \( I_{CD} \) is its M.I. about the \( CD \) axis, then according to the theorem of parallel axes:
\( I_{CD} = I_{AB} + MR^2 \)
\( I_{CD} = \frac{2}{5} MR^2 + MR^2 \)
\( I_{CD} = \left(\frac{2}{5} + 1\right) MR^2 \)
\( I_{CD} = \frac{7}{5} MR^2 \)
(b) Here, \( AB \) and \( CD \) are diameters of the disc of radius \( R \) and mass \( M \). Let \( EF \) be an axis perpendicular to the plane of the disc and passing through its C.M., i.e., \( O \). Clearly, \( DG \) is the axis normal to the disc and passing through a point \( D \) on its edge.
The moment of inertia of a disc about its diameter is given as \( I_{diameter} = \frac{MR^2}{4} \).
According to the theorem of perpendicular axes, the moment of inertia about an axis perpendicular to the plane of the disc and passing through its center (\( I_{EF} \)) is the sum of the moments of inertia about two perpendicular diameters in its plane.
So, \( I_{EF} = I_{AB} + I_{CD} \)
\( I_{EF} = \frac{MR^2}{4} + \frac{MR^2}{4} = \frac{1}{2} MR^2 \)
Now, the axis \( DG \) is parallel to the axis \( EF \). The perpendicular distance between the \( EF \) axis (passing through the center \( O \)) and the \( DG \) axis (passing through a point \( D \) on the edge) is \( R \).
If \( I_{DG} \) is the M.I. of the disc about the required axis, then according to the theorem of parallel axes:
\( I_{DG} = I_{EF} + MR^2 \)
\( I_{DG} = \frac{1}{2} MR^2 + MR^2 \)
\( I_{DG} = \left(\frac{1}{2} + 1\right) MR^2 \)
\( I_{DG} = \frac{3}{2} MR^2 \)
In simple words: (a) For a solid sphere, if you want to know how hard it is to spin it around a line touching its edge, you add its usual spinning resistance (around its middle) to a value based on its mass and size. (b) For a flat disc, to find how hard it is to spin it around a line sticking out from its edge, you add its spinning resistance around its center to another value based on its mass and size.

Exam Tip: Remember to use the parallel axis theorem when the axis of rotation is parallel to an axis passing through the center of mass. For planar bodies, the perpendicular axis theorem helps find the moment of inertia about an axis normal to the plane if moments about two perpendicular axes in the plane are known.

Question 11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Answer: Let \( M \) and \( R \) be the mass and radius of both the hollow cylinder and the solid sphere.
Let \( I_1 \) be the moment of inertia (M.I.) of the hollow cylinder about its axis of symmetry.
For a hollow cylinder, \( I_1 = MR^2 \) (i)
(Note: If it's a thin hollow cylinder with inner radius \( R_1 \) and outer radius \( R_2 \), its M.I. is \( \frac{1}{2}M(R_1^2 + R_2^2) \). If thin, \( R_1 \approx R_2 \approx R \), so \( I_1 = \frac{1}{2}M(2R^2) = MR^2 \)).
Let \( I_2 \) be the M.I. of the solid sphere about its axis through its center.
For a solid sphere, \( I_2 = \frac{2}{5} MR^2 \) (ii)
Let \( \tau \) be the magnitude of the torque applied to each of them.
If \( \alpha_1 \) and \( \alpha_2 \) are the angular accelerations produced in the cylinder and sphere, respectively, then:
\( \tau = I_1 \alpha_1 \)
\( \tau = I_2 \alpha_2 \)
Therefore, \( I_1 \alpha_1 = I_2 \alpha_2 \)
\( \implies \alpha_1 = \frac{I_2}{I_1} \alpha_2 \)
Substitute the values of \( I_1 \) and \( I_2 \):
\( \alpha_1 = \frac{\frac{2}{5} MR^2}{MR^2} \alpha_2 \)
\( \alpha_1 = \frac{2}{5} \alpha_2 \)
\( \implies \alpha_2 = \frac{5}{2} \alpha_1 = 2.5 \alpha_1 \) (iii)
If \( \omega_1 \) and \( \omega_2 \) are the angular speeds of the cylinder and sphere after a time \( t \), and starting from an initial angular speed \( \omega_0 \), then:
\( \omega_1 = \omega_0 + \alpha_1 t \) (iv)
\( \omega_2 = \omega_0 + \alpha_2 t \)
Substitute \( \alpha_2 = 2.5 \alpha_1 \) from (iii):
\( \omega_2 = \omega_0 + 2.5 \alpha_1 t \) (v)
From (iv) and (v), it is clear that \( \omega_2 > \omega_1 \).
Thus, the sphere will acquire a greater angular speed compared to the cylinder after a given time.
In simple words: When the same push (torque) is given to a hollow cylinder and a solid sphere of the same size and weight, the solid sphere will spin faster. This is because the sphere has less 'resistance to turning' (moment of inertia) than the hollow cylinder.

Exam Tip: This problem emphasizes that objects with smaller moments of inertia (like a solid sphere where mass is closer to the center) will experience greater angular acceleration and thus achieve higher angular speeds for the same applied torque, compared to objects with larger moments of inertia (like a hollow cylinder where mass is farther from the center).

Question 12. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s\( ^{-1} \). The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Answer: Given:
Mass of the solid cylinder \( m = 20 \text{ kg} \)
Angular speed \( \omega = 100 \text{ rad s}^{-1} \)
Radius of the cylinder \( R = 0.25 \text{ m} \)
First, calculate the moment of inertia (M.I.) of the solid cylinder about its axis:
\( I = \frac{1}{2} MR^2 \)
\( I = \frac{1}{2} \times 20 \text{ kg} \times (0.25 \text{ m})^2 \)
\( I = 10 \text{ kg} \times 0.0625 \text{ m}^2 \)
\( I = 0.625 \text{ kg m}^2 \)
Now, calculate the kinetic energy (K.E.) associated with the rotating cylinder:
\( K.E. = \frac{1}{2} I \omega^2 \)
\( K.E. = \frac{1}{2} \times 0.625 \text{ kg m}^2 \times (100 \text{ rad s}^{-1})^2 \)
\( K.E. = \frac{1}{2} \times 0.625 \times 10000 \)
\( K.E. = 0.625 \times 5000 \)
\( K.E. = 3125 \text{ J} \)
Next, calculate the magnitude of the angular momentum \( L \) of the cylinder:
\( L = I \omega \)
\( L = 0.625 \text{ kg m}^2 \times 100 \text{ rad s}^{-1} \)
\( L = 62.5 \text{ kg m}^2 \text{ s}^{-1} \) or \( 62.5 \text{ Js} \)
Alternatively, using the relation \( K.E. = \frac{L^2}{2I} \):
\( L^2 = 2 \times K.E. \times I \)
\( L = \sqrt{2 \times K.E. \times I} \)
\( L = \sqrt{2 \times 3125 \text{ J} \times 0.625 \text{ kg m}^2} \)
\( L = \sqrt{6250 \times 0.625} \)
\( L = \sqrt{3906.25} \)
\( L = 62.5 \text{ Js} \)
In simple words: The spinning cylinder has a rotational energy of 3125 Joules. The amount of "spin" it has, called angular momentum, is 62.5 Js.

Exam Tip: Remember the formulas for rotational kinetic energy \( (K.E. = \frac{1}{2}I\omega^2) \) and angular momentum \( (L = I\omega) \). For a solid cylinder rotating about its axis, the moment of inertia is \( I = \frac{1}{2}MR^2 \). Ensure units are consistent throughout your calculations.

Question 13. (a) A child stands at the centre of a turn table with his two arms outstretched. The turn table is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moments of inertia to 2/5 times the initial value? Assume that the turn table rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Answer:
(a) Let \( I_1 \) and \( I_2 \) be the initial and final moments of inertia of the child, with outstretched and folded arms respectively. We are given that \( I_2 = \frac{2}{5} I_1 \).
The initial angular speed is \( \nu_1 = 40 \text{ rev/min} \). To convert this to radians per second, we use the formula \( \omega = 2\pi\nu \).
\( \omega_1 = 2\pi \left(\frac{40}{60}\right) \text{ rad s}^{-1} = \frac{4\pi}{3} \text{ rad s}^{-1} \)
According to the principle of conservation of angular momentum, the total angular momentum stays constant because there is no external torque. So, the initial angular momentum equals the final angular momentum:
\( I_1\omega_1 = I_2\omega_2 \)
We can determine the final angular speed \( \omega_2 \):
\( \omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{I_1}{\frac{2}{5}I_1}\omega_1 = \frac{5}{2}\omega_1 \)
Substitute the value of \( \omega_1 \):
\( \omega_2 = \frac{5}{2} \times \frac{4\pi}{3} = \frac{10\pi}{3} \text{ rad s}^{-1} \)
To find the frequency of revolution \( \nu_2 \), we use \( \nu_2 = \frac{\omega_2}{2\pi} \):
\( \nu_2 = \frac{10\pi/3}{2\pi} = \frac{5}{3} \text{ rps} \)
Converting revolutions per second to revolutions per minute:
\( \nu_2 = \frac{5}{3} \times 60 = 100 \text{ rev/min} \)
So, when the child folds their arms, the angular speed becomes \( 100 \text{ rev/min} \).

(b) Now let's examine the kinetic energy. The initial kinetic energy of rotation is:
\( K.E._{initial} = \frac{1}{2}I_1\omega_1^2 \)
The new kinetic energy of rotation when the arms are folded is:
\( K.E._{new} = \frac{1}{2}I_2\omega_2^2 \)
Substitute the expressions for \( I_2 \) and \( \omega_2 \) in terms of \( I_1 \) and \( \omega_1 \):
\( K.E._{new} = \frac{1}{2} \left(\frac{2}{5}I_1\right) \left(\frac{5}{2}\omega_1\right)^2 \)
\( K.E._{new} = \frac{1}{2} \left(\frac{2}{5}I_1\right) \left(\frac{25}{4}\omega_1^2\right) \)
\( K.E._{new} = \frac{5}{2} \left(\frac{1}{2}I_1\omega_1^2\right) \)
This shows that \( K.E._{new} = \frac{5}{2} K.E._{initial} \).
The new kinetic energy of rotation is \( \frac{5}{2} \) times larger than the original kinetic energy. This rise in kinetic energy happens because the child uses their own internal muscular energy to fold their arms, which changes the distribution of mass and causes faster rotation.
In simple words: When the child pulls their arms in, they spin faster because their mass is closer to the center. This makes their new spinning energy \( \frac{5}{2} \) times bigger. The extra energy comes from the child's muscles as they pull their arms inwards.

Exam Tip: Remember to apply the conservation of angular momentum when there are no external torques, and understand that internal forces can change kinetic energy while conserving angular momentum.

Question 14. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assuming that there is no slipping?
Answer:
Given information:
Mass of cylinder, \( M = 3 \text{ kg} \)
Radius of cylinder, \( R = 40 \text{ cm} = 0.4 \text{ m} \)
Tangential force, \( F = 30 \text{ N} \)

First, we calculate the moment of inertia (M.I.) for a hollow cylinder about its axis:
\( I = MR^2 = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48 \text{ kg m}^2 \)

Next, we determine the torque \( \tau \) acting on the cylinder:
\( \tau = FR = 30 \times 0.4 = 12 \text{ Nm} \)

Now, we can find the angular acceleration \( \alpha \) using the relation \( \tau = I\alpha \):
\( \alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25 \text{ rad s}^{-2} \)

Finally, we find the linear acceleration \( a \) of the rope using the relation \( a = R\alpha \):
\( a = R\alpha = 0.4 \times 25 = 10 \text{ m s}^{-2} \)
In simple words: First, we find how hard it is to make the cylinder spin (its moment of inertia). Then, we calculate the twist (torque) from the rope's pull. Using these, we figure out how quickly it starts spinning (angular acceleration). Finally, from that, we can tell how fast the rope itself moves (linear acceleration).

Exam Tip: Ensure units are consistent (convert cm to m) and correctly apply formulas for moment of inertia, torque, and the relationship between linear and angular acceleration.

Question 15. To maintain a rotor at a uniform angular speed of 206 rad s¯¹, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note uniform angular velocity in the absence of friction implied zero torque. In practice, applied torque is needed to counter frictional torque.) Assume that the engine is 100% efficient?
Answer:
Given information:
Angular speed, \( \omega = 200 \text{ rad s}^{-1} \)
Torque transmitted, \( \tau = 180 \text{ Nm} \)

The power \( P \) required by the engine can be calculated using the relation \( P = \tau\omega \).
\( P = 180 \text{ Nm} \times 200 \text{ rad s}^{-1} \)
\( P = 36000 \text{ W} \)
Converting watts to kilowatts:
\( P = 36 \text{ kW} \)
In simple words: To find the power needed, you simply multiply the twisting force (torque) by how fast it's spinning (angular speed). This gives the total power output in watts, which can then be converted to kilowatts.

Exam Tip: Remember the formula \( P = \tau\omega \) for rotational power. Pay attention to units and convert to kilowatts if required by the question.

Question 16. From a uniform disk of radius R, a circular section of radius \( \frac{R}{2} \) is cut out. Locate the centre of gravity of the resulting flat body?
Answer:
Given information:
Radius of the original disk \( = R \)
Radius of the removed circular section \( = \frac{R}{2} \)

Let \( \rho \) be the mass per unit area of the disk.
Area of the original disk \( A = \pi R^2 \)
Area of the removed circular section \( a = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \)

Mass of the original disk \( m_1 = \rho A = \rho \pi R^2 \)
Mass of the removed section \( m = \rho a = \frac{\rho \pi R^2}{4} \)
Mass of the remaining portion \( m_2 = m_1 - m = \rho \pi R^2 - \frac{\rho \pi R^2}{4} = \frac{3}{4} \rho \pi R^2 \)

Let's set the origin at the center of the original disk, O.
The center of the removed circular section, \( O_1 \), is located at \( x_1 = \frac{R}{2} \) from O.
Let \( x_2 \) be the location of the center of mass of the remaining portion.

The center of mass of the entire original disk (before the cut) is at O, so its coordinate is 0. We can consider the original disk as a system composed of the removed part and the remaining part.
Therefore, the center of mass of this combined system is:
\( X_{CM, total} = \frac{m x_1 + m_2 x_2}{m + m_2} \)
Since \( X_{CM, total} = 0 \) (center of original disk):
\( m x_1 + m_2 x_2 = 0 \)
Substitute the values for masses and \( x_1 \):
\( \left(\frac{\rho \pi R^2}{4}\right) \left(\frac{R}{2}\right) + \left(\frac{3}{4} \rho \pi R^2\right) x_2 = 0 \)
\( \frac{\rho \pi R^3}{8} + \frac{3}{4} \rho \pi R^2 x_2 = 0 \)
Divide both sides by \( \rho \pi R^2 \):
\( \frac{R}{8} + \frac{3}{4} x_2 = 0 \)
\( \frac{3}{4} x_2 = -\frac{R}{8} \)
\( x_2 = -\frac{R}{8} \times \frac{4}{3} = -\frac{R}{6} \)

The center of mass of the remaining portion is located at a distance \( \frac{R}{6} \) to the left of the original disk's center O.
In simple words: Imagine a whole disk, then a smaller circle is cut out from its side. We want to find the new balance point (center of gravity) for the leftover shape. We treat the original disk as two parts: the cut-out piece and the remaining piece. Since the original disk balanced at its center, the combined balance point of these two parts must also be there. Using this idea, we can calculate where the remaining part must balance to keep the overall balance. It turns out to be a little bit away from the original center, on the opposite side from where the piece was cut.

Exam Tip: When dealing with composite bodies or bodies with parts removed, use the concept of the center of mass of the combined system. Assume the original center of mass for the entire body (if intact) as the reference point to simplify calculations. Remember that the center of mass can lie outside the physical body.

Question 17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Answer:
Given information:
Mass of each coin \( = 5 \text{g} \)
Number of coins \( = 2 \)
Total mass of coins \( m_{coins} = 2 \times 5 \text{g} = 10 \text{g} \)
Position of coins \( = 12 \text{ cm mark} \)
Original balance point of metre stick (its C.G.) \( = 50 \text{ cm mark} \)
New balance point with coins \( = 45 \text{ cm mark} \)

Let \( M_{stick} \) be the mass of the metre stick. The center of gravity of the stick is at the 50 cm mark.
When the stick is balanced at 45 cm, we can apply the principle of moments about this new fulcrum.

The coins are at 12 cm, which is to the left of the fulcrum (45 cm).
Distance of coins from fulcrum \( = 45 \text{ cm} - 12 \text{ cm} = 33 \text{ cm} \).
Moment due to coins \( = m_{coins} \times 33 \text{ cm} = 10 \text{g} \times 33 \text{ cm} = 330 \text{ g cm} \).

The center of gravity of the metre stick is at 50 cm, which is to the right of the fulcrum (45 cm).
Distance of stick's C.G. from fulcrum \( = 50 \text{ cm} - 45 \text{ cm} = 5 \text{ cm} \).
Moment due to metre stick \( = M_{stick} \times 5 \text{ cm} \).

For the stick to be in equilibrium, the clockwise moment must equal the counter-clockwise moment:
\( M_{stick} \times 5 = 10 \times 33 \)
\( M_{stick} \times 5 = 330 \)
\( M_{stick} = \frac{330}{5} = 66 \text{ g} \)

Therefore, the mass of the metre stick is \( 66 \text{ g} \).
In simple words: We are looking for the weight of the metre stick. We know where the stick normally balances (50 cm mark) and where it balances when two small coins are added (45 cm mark). By using the rule of moments (where force times distance from the pivot must balance), we can set up an equation. The coins create a turning force on one side, and the metre stick's own weight creates an opposite turning force. When these balance, we can calculate the stick's mass.

Exam Tip: For equilibrium problems involving moments, choose a convenient pivot point (often the fulcrum) and ensure all distances are measured correctly from that point. Moments on one side must balance moments on the other.

Question 18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination? (a) Will it reach the bottom with the same speed in each case?
Answer: Yes, the sphere will reach the bottom with the same final speed in both cases.
When the sphere rolls down an inclined plane, its initial potential energy (P.E.) at the top is converted into translational kinetic energy and rotational kinetic energy at the bottom.
Initial P.E. \( = mgh \)
Total K.E. at the bottom \( = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
For a solid sphere, the moment of inertia \( I = \frac{2}{5}mR^2 \), and for rolling without slipping, \( \omega = \frac{v}{R} \).
Substituting these into the K.E. equation:
\( \text{Total K.E.} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 \)
\( = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \)
\( = \left(\frac{1}{2} + \frac{1}{5}\right)mv^2 = \left(\frac{5+2}{10}\right)mv^2 = \frac{7}{10}mv^2 \)
By conservation of energy, P.E. = Total K.E.:
\( mgh = \frac{7}{10}mv^2 \)
\( gh = \frac{7}{10}v^2 \)
\( v^2 = \frac{10}{7}gh \)
\( v = \sqrt{\frac{10}{7}gh} \)
Since the final speed \( v \) depends only on \( g \) and \( h \), and both are constant for the two planes, the sphere will attain the same final speed at the bottom regardless of the inclination angle.
In simple words: Yes, it will reach the same speed. The final speed only depends on how high the ramp is, not how steep it is. All the energy from its height turns into movement energy, and this amount is the same no matter the slope.

Exam Tip: Remember that for rolling objects, the final speed depends only on the vertical height dropped, provided there's no energy loss due to friction. The angle affects the acceleration and time, but not the final speed.

Question 18. (b) Will it take longer to roll down one plane than the other? (c) If so which one and why?
Answer: Yes, it will take longer to roll down the plane with the smaller angle of inclination.

To understand why, we first calculate the acceleration \( a \) of a solid sphere rolling down an inclined plane. The acceleration is given by:

\( a = \frac{g\sin\theta}{1 + \frac{K^2}{R^2}} \)

For a solid sphere, the radius of gyration \( K^2 = \frac{2}{5}R^2 \), so \( \frac{K^2}{R^2} = \frac{2}{5} \).

\( a = \frac{g\sin\theta}{1 + \frac{2}{5}} = \frac{g\sin\theta}{\frac{7}{5}} = \frac{5}{7}g\sin\theta \)

This shows that the acceleration \( a \) is directly proportional to \( \sin\theta \).
Now, let \( t \) be the time taken to roll down the plane. The distance traveled along the plane is \( S = \frac{h}{\sin\theta} \). Using the kinematic equation \( S = ut + \frac{1}{2}at^2 \), with initial velocity \( u=0 \):

\( S = \frac{1}{2}at^2 \)

\( t^2 = \frac{2S}{a} = \frac{2(h/\sin\theta)}{(\frac{5}{7}g\sin\theta)} = \frac{14h}{5g\sin^2\theta} \)

\( t = \sqrt{\frac{14h}{5g\sin^2\theta}} = \frac{1}{\sin\theta} \sqrt{\frac{14h}{5g}} \)

This result implies that the time \( t \) is inversely proportional to \( \sin\theta \).
Given that \( \theta_1 > \theta_2 \), it means \( \sin\theta_1 > \sin\theta_2 \).
Therefore, \( \frac{1}{\sin\theta_1} < \frac{1}{\sin\theta_2} \).
This leads to \( t_1 < t_2 \).
So, the sphere will take a longer time to roll down the plane with the smaller angle of inclination (\( \theta_2 \)), because the acceleration down that plane is less.
In simple words: Yes, it will take more time on the gentler slope. This is because the acceleration, which makes the sphere speed up, is smaller on a less steep slope. So, even though it covers the same height, it takes longer to get there.

Exam Tip: For rolling motion, remember that acceleration is affected by the angle of inclination and the object's shape (through \( K^2/R^2 \)). Time is inversely related to acceleration and sine of the angle, so a smaller angle means a longer time.

Question 19. A hoop of radius 2m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm¯¹. How much work has to be done to stop it?
Answer:
Given information:
Radius of the hoop \( r = 2 \text{ m} \)
Mass of the hoop \( m = 100 \text{ kg} \)
Speed of its center of mass \( v = 20 \text{ cm s}^{-1} = 0.20 \text{ m s}^{-1} \)

To stop the hoop, an amount of work equal to its total kinetic energy must be done. The total kinetic energy of a rolling hoop includes both translational and rotational kinetic energy.

First, calculate the angular velocity \( \omega \):
For rolling without slipping, \( \omega = \frac{v}{r} = \frac{0.20 \text{ m s}^{-1}}{2 \text{ m}} = 0.10 \text{ rad s}^{-1} \)

Next, calculate the moment of inertia \( I \) of a hoop about an axis passing through its center and perpendicular to its plane:
\( I = mr^2 = 100 \text{ kg} \times (2 \text{ m})^2 = 100 \times 4 = 400 \text{ kg m}^2 \)

Now, calculate the total kinetic energy (K.E. total):
\( \text{K.E.}_{\text{total}} = \text{K.E.}_{\text{translational}} + \text{K.E.}_{\text{rotational}} \)
\( \text{K.E.}_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
Substitute the values:
\( \text{K.E.}_{\text{total}} = \frac{1}{2} (100 \text{ kg}) (0.20 \text{ m s}^{-1})^2 + \frac{1}{2} (400 \text{ kg m}^2) (0.10 \text{ rad s}^{-1})^2 \)
\( \text{K.E.}_{\text{total}} = \frac{1}{2} (100) (0.04) + \frac{1}{2} (400) (0.01) \)
\( \text{K.E.}_{\text{total}} = 50 \times 0.04 + 200 \times 0.01 \)
\( \text{K.E.}_{\text{total}} = 2 \text{ J} + 2 \text{ J} = 4 \text{ J} \)

According to the work-energy theorem, the work done to stop the hoop is equal to its total kinetic energy.
Therefore, work done \( = 4 \text{ J} \).
In simple words: To stop a rolling hoop, you need to remove all its moving energy. This energy comes from two parts: the hoop moving forward (translational energy) and the hoop spinning (rotational energy). We calculate each part separately using its mass, speed, radius, and how it's shaped. Add these two energies together, and that's the total amount of work required to make it stop.

Exam Tip: Remember that for rolling objects, total kinetic energy is the sum of translational and rotational kinetic energies. Be careful with unit conversions (cm/s to m/s).

Question 20. The oxygen molecule has a mass of \( 5.30 \times 10^{-26} \) kg and a moment of inertia of \( 1.94 \times 10^{-46} \) kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 ms¯¹ and that is K.E. of rotation is \( \frac{2}{3} \) of its K.E. translation. Find the average angular velocity of the molecule?
Answer:
Given information for the oxygen molecule:
Mass \( m = 5.30 \times 10^{-26} \text{ kg} \)
Moment of inertia \( I = 1.94 \times 10^{-46} \text{ kg m}^2 \)
Mean translational speed \( v = 500 \text{ m s}^{-1} \)

We are given the condition:
Kinetic energy of rotation \( = \frac{2}{3} \times \) Kinetic energy of translation

The formulas for these energies are:
K.E. of rotation \( = \frac{1}{2}I\omega^2 \)
K.E. of translation \( = \frac{1}{2}mv^2 \)

Substitute these into the given condition:
\( \frac{1}{2}I\omega^2 = \frac{2}{3} \times \frac{1}{2}mv^2 \)
We can cancel \( \frac{1}{2} \) from both sides:
\( I\omega^2 = \frac{2}{3}mv^2 \)

Now, we solve for the average angular speed \( \omega \):
\( \omega^2 = \frac{2mv^2}{3I} \)
\( \omega = \sqrt{\frac{2mv^2}{3I}} \)

Substitute the given numerical values:
\( \omega = \sqrt{\frac{2 \times (5.30 \times 10^{-26} \text{ kg}) \times (500 \text{ m s}^{-1})^2}{3 \times (1.94 \times 10^{-46} \text{ kg m}^2)}} \)
\( \omega = \sqrt{\frac{2 \times 5.30 \times 10^{-26} \times 250000}{3 \times 1.94 \times 10^{-46}}} \)
\( \omega = \sqrt{\frac{2650000 \times 10^{-26}}{5.82 \times 10^{-46}}} = \sqrt{\frac{2.65 \times 10^6 \times 10^{-26}}{5.82 \times 10^{-46}}} \)
\( \omega = \sqrt{\frac{2.65 \times 10^{-20}}{5.82 \times 10^{-46}}} = \sqrt{\left(\frac{2.65}{5.82}\right) \times 10^{-20 - (-46)}} \)
\( \omega = \sqrt{0.4553 \times 10^{26}} = \sqrt{4.553 \times 10^{25}} \)
\( \omega = \sqrt{45.53 \times 10^{24}} \approx 6.747 \times 10^{12} \text{ rad s}^{-1} \)
Rounding to two significant figures as in the input:
\( \omega \approx 6.75 \times 10^{12} \text{ rad s}^{-1} \)
In simple words: We are given how much translational energy a molecule has and how its rotational energy relates to it. We use these facts, along with its mass and resistance to turning (moment of inertia), to figure out its spinning speed. By balancing the energy equation and plugging in the numbers, we can find the average angular velocity.

Exam Tip: Ensure correct application of kinetic energy formulas for both translational and rotational motion. Pay close attention to calculations involving powers of ten and be careful with unit consistency.

Question 21. A cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5 ms¯¹. (a) How far will the cylinder go up the plane?
Answer: First, we need to determine the acceleration of the cylinder as it rolls up the inclined plane.
The acceleration \( a \) for a solid cylinder rolling on an inclined plane is given by:
\( a = \frac{g\sin\theta}{1 + \frac{K^2}{R^2}} \)
For a solid cylinder, \( K^2 = \frac{1}{2}R^2 \), so \( \frac{K^2}{R^2} = \frac{1}{2} \).
Substituting these values:
\( a = \frac{g\sin30^\circ}{1 + \frac{1}{2}} = \frac{g \times \frac{1}{2}}{\frac{3}{2}} = \frac{g}{3} \)
Using \( g = 9.8 \text{ m s}^{-2} \):
\( a = \frac{9.8}{3} \text{ m s}^{-2} \)
Since the cylinder is moving up the plane, the acceleration is negative (deceleration), so \( a = -\frac{9.8}{3} \text{ m s}^{-2} \).

Now, we use the kinematic equation \( v^2 - u^2 = 2as \) to find the distance \( s \) it travels up the plane.
Initial speed \( u = 5 \text{ m s}^{-1} \)
Final speed \( v = 0 \) (since it momentarily stops at its highest point)
\( 0^2 - (5)^2 = 2 \times \left(-\frac{9.8}{3}\right) \times s \)
\( -25 = -\frac{19.6}{3}s \)
\( s = \frac{25 \times 3}{19.6} = \frac{75}{19.6} \approx 3.8265 \text{ m} \)
Rounding to two decimal places, \( s \approx 3.83 \text{ m} \).
So, the cylinder will go approximately \( 3.83 \text{ m} \) up the plane.
In simple words: To find out how far it goes up, we first calculate how quickly it slows down (acceleration) on the slope, considering it's a solid cylinder. Then, using its starting speed and the slowing-down rate, we can figure out the distance it travels before it stops and starts rolling back down.

Exam Tip: Remember to include the object's moment of inertia when calculating acceleration on an inclined plane for rolling motion. Always consider the direction of acceleration (positive for speeding up, negative for slowing down).

Question 21. (b) How long will it take to return to the bottom?
Answer: The time taken to return to the bottom will be equal to the time it took to go up to its highest point, assuming no energy losses.
Let \( t_{up} \) be the time taken to go up. Using the kinematic equation \( v = u + at \):
\( 0 = 5 + \left(-\frac{9.8}{3}\right) t_{up} \)
\( \frac{9.8}{3} t_{up} = 5 \)
\( t_{up} = \frac{5 \times 3}{9.8} = \frac{15}{9.8} \approx 1.5306 \text{ s} \)

So, the time to go up is approximately \( 1.53 \text{ s} \).
The total time to return to the bottom \( T = 2 \times t_{up} \):
\( T = 2 \times 1.5306 \approx 3.06 \text{ s} \)
Therefore, it will take approximately \( 3.06 \text{ s} \) to return to the bottom.
In simple words: Assuming no energy is lost, the time it takes to roll up will be the same as the time it takes to roll back down. So, we just need to calculate the time to go up and then double it to find the total time to return.

Exam Tip: For ideal rolling motion on an inclined plane, the time to go up is equal to the time to come down, and the acceleration value remains the same in magnitude, just opposite in direction.