JEE Physics System Of Particles and Rotational Motion Notes Set 09

Rigid Body and Its Motion

A rigid body is a system of particles in which the distance between every pair of particles stays fixed. Because of this, the motion of any particle B relative to another particle A of the body is always perpendicular to the line AB: from A the body seems to turn about it. Every motion of a rigid body can be split into the translation of one chosen point (usually the centre of mass) and a rotation about that point, and the angular velocity of this rotation is the same whichever point is taken as reference.

Moment of Inertia

The moment of inertia measures a body's resistance to a change in its rotation about a given axis. For particles, \( I=\sum m_ir_i^2 \), where \( r_i \) is the perpendicular distance from the axis; for a continuous body \( I=\int r^2\,dm \). It depends on the mass, on how that mass is spread, and on the position and direction of the axis, but not on the angular velocity.

Two Theorems

1. Parallel axes theorem: \( I=I_{cm}+Md^2 \), where d is the distance between the two parallel axes, one of them through the centre of mass. The moment of inertia about an axis through the centre of mass is therefore the smallest of all parallel axes.
2. Perpendicular axes theorem (flat bodies only): \( I_z=I_x+I_y \), where x and y are perpendicular axes in the plane of the body meeting at the point where the z-axis passes through. For a disc this gives \( I_{diameter}=\tfrac12\cdot\tfrac12MR^2 \).
3. For a rod making an angle \( \theta \) with the axis, \( I=\tfrac13ML^2\sin^2\theta \) about an axis through one end.
4. Moments of inertia add for parts of a body that share the same axis; a body with a cut-out is treated as the full body minus the removed part (with the removed mass counted negative).

Radius of Gyration

The radius of gyration k is the distance at which a single point of the whole mass would give the same moment of inertia: \( I=Mk^2 \), so \( k=\sqrt{I/M} \). For a uniform disc about its axis, \( k=R/\sqrt2 \).

Torque

1. Torque of a force about a point: \( \vec\tau=\vec r\times\vec F \); its magnitude is \( rF\sin\theta \), the force times the perpendicular distance of its line of action from the point.
2. Torque about an axis is the component of the torque about any point on the axis, along the axis.
3. Two equal and opposite forces with different lines of action form a couple. Its torque \( F imes d \) is the same about every point, and its net force is zero.
4. The torque of gravity on a body can be taken as the weight acting at the centre of mass.

Rotation about a Fixed Axis

For rotation about a fixed axis the torque of all external forces about that axis, and the angular acceleration, are related by \( \tau=I\alpha \). The motion therefore parallels linear motion:

1. A point at distance r from the axis has speed \( v=\omega r \), tangential acceleration \( \alpha r \) and radial acceleration \( \omega^2r \).
2. Work done by a torque is \( \int\tau\,d\theta \); the work-energy theorem is \( W=\Delta(\tfrac12I\omega^2) \).
3. A rod of length L hinged at one end and released from the horizontal has \( \alpha=\dfrac{3g}{2L} \) at the start and angular speed \( \omega=\sqrt{3g/L} \) at the vertical position.
4. The hinge exerts a force on a rotating body; find it from the motion of the centre of mass: net force equals the mass times the acceleration of the centre of mass.

Equilibrium of a Rigid Body

1. For translational equilibrium the vector sum of all external forces is zero.
2. For rotational equilibrium the sum of the torques about any one point is zero. If the forces sum to zero, the torques are then zero about every point.
3. Three non-parallel forces keeping a body in equilibrium must be concurrent (their lines of action meet at one point).
4. Choose the axis for torques so that the most unknown forces pass through it, which removes them from the equation.
5. A body is in stable equilibrium if a small displacement raises its centre of mass, so that gravity restores it.

Worked Examples

Example 1. A uniform ladder of length 5 m rests with its top 4 m up a wall and its foot 3 m away. The coefficient of friction is the same at the wall and the floor. Find the minimum value of it for equilibrium.

1. Horizontal: \( N_2=\mu N_1 \); vertical: \( N_1+\mu N_2=mg \), so \( N_1=\dfrac{mg}{1+\mu^2} \).
2. Torque about the foot: \( 4N_2+3\mu N_2=1.5\,mg \Rightarrow\mu(4+3\mu)=1.5(1+\mu^2) \).
3. \( 3\mu^2+8\mu-3=0\Rightarrow\mu=\dfrac13 \).

Example 2. Find the moment of inertia of a uniform disc of mass m and radius r about a tangent in its plane.

1. About a diameter: \( \tfrac14mr^2 \).
2. By the parallel axes theorem: \( I=\tfrac14mr^2+mr^2=\tfrac54mr^2 \).

Example 3. A wheel of moment of inertia 0.5 kg m² is acted on by a constant torque of 4 N m starting from rest. Find the angular speed after 5 s and the kinetic energy.

1. \( \alpha=\tau/I=8 \) rad/s²; \( \omega=\alpha t=40 \) rad/s.
2. \( K=\tfrac12I\omega^2=\tfrac12(0.5)(1600)=400 \) J.

Example 4. A uniform rod of length 1 m and mass 2 kg is hinged at one end and released from the horizontal. Find the angular acceleration at the start and the force from the hinge then (g = 10 m/s²).

1. \( \alpha=\dfrac{3g}{2L}=15 \) rad/s².
2. Acceleration of the centre \( =\alpha\dfrac L2=7.5 \) m/s² downward: \( mg-N=ma \Rightarrow N=20-15=5 \) N upward.

Example 5. Two like parallel forces 30 N and 20 N act 1 m apart on a light rod. Where does the resultant act?

1. Resultant \( =50 \) N, acting closer to the larger force.
2. Distance from the 30 N force: \( \dfrac{20\times1}{50}=0.4 \) m.

Quick Revision Points

1. \( I=\sum mr^2 \), \( I=I_{cm}+Md^2 \), \( I_z=I_x+I_y \) for plates.
2. \( \tau=I\alpha \) about a fixed axis; kinetic energy \( \tfrac12I\omega^2 \).
3. Equilibrium needs zero net force and zero net torque.
4. A couple has the same torque about every point.