JEE Physics System Of Particles and Rotational Motion Notes Set 08

Rotational Kinematics and Torque

A rigid body keeps the distances between its particles fixed. In pure rotation about a fixed axis every particle moves in a circle about the axis and all particles share the same angular displacement, angular velocity \( \omega \) and angular acceleration \( \alpha \). Linear quantities of a particle at distance r are \( v=\omega r \), tangential acceleration \( a_t=\alpha r \) and centripetal acceleration \( a_c=\omega^2r \).
For constant angular acceleration the equations mirror linear motion: \( \omega=\omega_0+\alpha t \), \( \theta=\omega_0t+\tfrac12\alpha t^2 \), \( \omega^2=\omega_0^2+2\alpha\theta \).
Torque about a point is \( \vec\tau=\vec r\times\vec F \), with magnitude \( \tau=rF\sin\theta=F\times(\text{moment arm}) \). A pair of equal and opposite forces with different lines of action is a couple; its torque \( F\times d \) is the same about every point. Rotational Newton's law: \( \tau_{net}=I\alpha \) for a fixed axis.

Moment of Inertia

The moment of inertia \( I=\sum m_ir_i^2 \) measures how hard it is to change the rotation of a body about a given axis. It depends on the mass, on how that mass is distributed and on the position of the axis, but not on the angular velocity.
Radius of gyration k is defined by \( I=Mk^2 \).
Standard results (M mass, R radius, L length):

Parallel-axis theorem: \( I=I_{cm}+Md^2 \) for an axis at distance d from a parallel axis through the centre of mass. Perpendicular-axis theorem (flat bodies only): \( I_z=I_x+I_y \) for two perpendicular axes in the plane meeting at the point where the z-axis passes.
To handle a body with a hole, take the moment of inertia of the full body and subtract that of the removed piece (about the same axis, using the parallel-axis theorem for the piece). Mass is proportional to area for a uniform plate.

Example 1. A uniform disc of mass 9m and radius R has a hole of radius R/3 cut out, the centre of the hole being at distance 2R/3 from the centre of the disc. Find the moment of inertia of the remaining part about the axis through the centre of the disc, perpendicular to its plane.
Answer:
The hole has 1/9 of the area, so its mass is m.
Full disc: \( \frac12(9m)R^2=4.5\,mR^2 \).
Hole about the same axis: \( \frac12m\left(\frac R3\right)^2+m\left(\frac{2R}{3}\right)^2=0.5\,mR^2 \).
Remaining part: \( 4mR^2 \).

Angular Momentum and Its Conservation

For a particle, \( \vec L=\vec r\times\vec p \), magnitude \( mv\times(\text{perpendicular distance of the line of motion from the point}) \). For a rigid body rotating about a fixed axis, \( L=I\omega \). A particle moving in a straight line with constant velocity has constant angular momentum about any fixed point.
The rotational form of Newton's second law is \( \vec\tau_{ext}=\frac{d\vec L}{dt} \) (valid in an inertial frame, or about the centre of mass). When the net external torque is zero, L is conserved. Then \( I_1\omega_1=I_2\omega_2 \): if I decreases, ω increases, and since \( K=\frac{L^2}{2I} \), the kinetic energy rises as well (the extra energy comes from the work done by internal forces).

Example 2. A ring of mass M and radius R rotates about its axis with angular velocity ω. Two small masses m are attached gently at the opposite ends of a diameter. Find the new angular velocity.
Answer:
No external torque acts, so \( MR^2\omega=(M+2m)R^2\omega' \).
\( \omega'=\frac{M\omega}{M+2m} \)

Rotational Energy, Work and Angular Impulse

Kinetic energy of rotation about a fixed axis: \( K=\tfrac12I\omega^2 \). Work done by a torque: \( W=\int\tau\,d\theta \) (equal to \( \tau\theta \) for constant torque); power \( P=\tau\omega \). The work-energy theorem holds: \( W=\Delta K \).
Angular impulse \( \int\tau\,dt=\Delta L \). When an impulse J acts on a free body at a distance d from its centre of mass, the centre of mass gets \( v=J/m \) and the body gets \( \omega=\frac{Jd}{I_{cm}} \). For a body hinged at a point, take angular impulse about the hinge, as the hinge exerts an impulsive force too.
For a rod of mass m and length l hinged at one end and hanging at rest, an impulse J at the free end gives \( \omega=\frac{Jl}{ml^2/3} \).

Rolling Motion

In pure rolling the point of contact is instantaneously at rest, so \( v_{cm}=\omega R \). The velocity of the top point is \( 2v_{cm} \), and a point at angle θ from the top has speed \( 2v_{cm}\cos(\theta/2) \). The acceleration of the contact point is \( \omega^2R \) towards the centre (when the motion is uniform); static friction does no work in pure rolling.
Total kinetic energy: \( K=\tfrac12Mv^2\left(1+\frac{I_{cm}}{MR^2}\right) \), so the rotational share is \( \frac{I_{cm}/MR^2}{1+I_{cm}/MR^2} \) (1/2 for a ring, 1/3 for a disc, 2/5 for a solid sphere).
Rolling down an incline of angle θ: \( a=\frac{g\sin\theta}{1+I_{cm}/MR^2} \), with friction \( f=\frac{Mg\sin\theta}{1+MR^2/I_{cm}} \). The minimum coefficient of friction needed is \( \mu\ge\frac{\tan\theta}{1+MR^2/I_{cm}} \): \( \tfrac13\tan\theta \) for a solid cylinder, \( \tfrac27\tan\theta \) for a solid sphere and \( \tfrac12\tan\theta \) for a ring. Speed at the bottom from height h: \( v=\sqrt{\frac{2gh}{1+I_{cm}/MR^2}} \). A body with smaller \( I_{cm}/MR^2 \) rolls faster: solid sphere, then disc, then hollow sphere, then ring.
When a body slips initially, kinetic friction acts at the contact point and angular momentum about the contact point (or about a line on the surface) is conserved; use this to find the final rolling speed. For example, a solid sphere thrown with speed \( v_0 \) and no spin begins rolling with \( v=\frac57v_0 \).

Example 3. A solid cylinder rolls down an incline of angle 30 degrees without slipping. Find the acceleration of its centre of mass and the minimum coefficient of friction required.
Answer:
For a solid cylinder \( I_{cm}=\frac12MR^2 \), so \( a=\frac{g\sin\theta}{1.5}=\frac23g\sin30^\circ=\frac g3 \).
Friction \( f=\frac13Mg\sin\theta \) must not exceed \( \mu Mg\cos\theta \).
\( a=\frac g3\approx3.3\ \text{m/s}^2 \) and \( \mu_{min}=\frac{\tan30^\circ}{3}\approx0.19 \).

Example 4. A solid sphere rolls along a horizontal surface with speed v and then climbs a rough incline without slipping. Find the maximum height reached.
Answer:
Total kinetic energy \( \frac12mv^2+\frac12\cdot\frac25mR^2\frac{v^2}{R^2}=\frac7{10}mv^2 \).
At the top it is all potential energy: \( mgh=\frac7{10}mv^2 \).
\( h=\frac{7v^2}{10g} \), independent of mass, radius and angle.