JEE Physics System Of Particles and Rotational Motion Notes Set 11

Rotation About a Fixed Axis

For a rigid body turning about a fixed axis, every particle moves in a circle with the same angular displacement \( \theta \), angular velocity \( \omega=\frac{d\theta}{dt} \) and angular acceleration \( \alpha=\frac{d\omega}{dt} \). A particle at distance r from the axis has speed \( v=\omega r \), tangential acceleration \( a_t=\alpha r \) and centripetal acceleration \( a_c=\omega^2 r \).
For constant \( \alpha \) the equations mirror straight-line motion: \( \omega=\omega_0+\alpha t \), \( \theta=\omega_0 t+\tfrac12\alpha t^2 \), \( \omega^2=\omega_0^2+2\alpha\theta \). They apply only when \( \alpha \) is constant, so an expression such as \( \omega=t-3 \) describes uniformly accelerated rotation while \( \omega=5t^2-3 \) does not.
For a particle in a body that spins uniformly, the net force is purely centripetal (horizontal and pointing at the axis). If the spin is non-uniform, a tangential component is added, so the net force is horizontal but no longer meets the axis.

Torque and Rotational Dynamics

Torque about a point is \( \vec\tau=\vec r\times\vec F \), with magnitude \( rF\sin\theta=F\times(\text{moment arm}) \). The rotational form of Newton's second law is \( \tau_{ext}=I\alpha \) for a fixed axis, or more generally \( \vec\tau=\frac{d\vec L}{dt} \) in an inertial frame. A couple is a pair of equal and opposite forces with different lines of action; it produces torque but no net force, and its torque is the same about every point.
A body is in rotational equilibrium when the net torque about any point is zero. In a non-inertial frame the relation \( \vec\tau=\frac{d\vec L}{dt} \) needs an extra pseudo-torque, although \( \vec L=I\vec\omega \) still holds.

Moment of Inertia and Its Theorems

Moment of inertia measures how mass is spread around an axis: \( I=\sum m_i r_i^2 \) or \( \int r^2\,dm \). It depends on the axis chosen, so one body has infinitely many values. The radius of gyration k is defined by \( I=Mk^2 \).
Parallel-axis theorem: \( I=I_{cm}+Md^2 \), where d is the separation between the two parallel axes.
Perpendicular-axis theorem (flat plates only): \( I_z=I_x+I_y \) for axes x and y in the plane that meet at the point where the z-axis crosses.
Standard results about an axis through the centre: thin ring \( MR^2 \) (perpendicular to plane); disc \( \tfrac12MR^2 \); solid cylinder \( \tfrac12MR^2 \); solid sphere \( \tfrac25MR^2 \); thin spherical shell \( \tfrac23MR^2 \); thin rod about its centre and perpendicular to its length \( \tfrac{1}{12}ML^2 \), and about one end \( \tfrac13ML^2 \). Every section cut from a disc or ring (quarter, half) has the same I per unit mass as the full body about the central axis, so a quarter disc of mass M has \( \tfrac12MR^2 \).

Angular Momentum

For a particle, \( \vec L=\vec r\times m\vec v \), with magnitude \( mvr_\perp \) where \( r_\perp \) is the perpendicular distance from the reference point to the line of motion. A particle moving along a straight line at constant velocity has constant angular momentum about any fixed point. For a rigid body on a fixed axis, \( L=I\omega \).
When the net external torque is zero, L is conserved. If I changes, ω changes in inverse proportion; kinetic energy \( K=\frac{L^2}{2I} \) then changes inversely with I. A skater drawing in her arms, or a child on a turntable stretching out arms, shows this directly.

Rotational Kinetic Energy, Work and Impulse

Rotational kinetic energy is \( \tfrac12I\omega^2 \). The work done by a torque is \( W=\int\tau\,d\theta \), which equals \( \tau\theta \) for a constant torque, and power is \( P=\tau\omega \). The work-energy theorem gives \( W=\Delta(\tfrac12I\omega^2) \). Angular impulse \( \int\tau\,dt \) equals the change in angular momentum.

Rolling Motion

A body rolls without slipping when the contact point is momentarily at rest, which requires \( v_{cm}=\omega R \). Every point of the body then rotates about the contact point with the same ω, so the top point moves at \( 2v_{cm} \) and the contact point has zero velocity but acceleration \( \frac{v^2}{R} \) directed toward the centre.
Total kinetic energy is \( \tfrac12Mv_{cm}^2+\tfrac12I_{cm}\omega^2=\tfrac12Mv_{cm}^2\left(1+\frac{k^2}{R^2}\right) \). The fraction of the energy that is rotational is \( \frac{k^2/R^2}{1+k^2/R^2} \): one half for a ring, one third for a disc or solid cylinder, two sevenths for a solid sphere.
Rolling down an incline of angle \( \theta \) the acceleration is \( a=\frac{g\sin\theta}{1+k^2/R^2} \), with the friction \( f=\frac{Mg\sin\theta\,(k^2/R^2)}{1+k^2/R^2} \) acting up the slope. Bodies with the smaller \( k^2/R^2 \) reach the bottom first: solid sphere, then disc, then hollow sphere, then ring. The speed at the bottom is \( v=\sqrt{\frac{2gh}{1+k^2/R^2}} \), for example \( \sqrt{\frac{4gh}{3}} \) for a solid cylinder and \( \sqrt{gh} \) for a ring.
Pure rolling needs enough friction, \( \mu\ge\frac{\tan\theta}{1+R^2/k^2} \). On a smooth surface a body cannot start to roll, and its rotation stays unchanged. A body rolling up an incline and back down has friction acting up the incline during the whole journey, because the acceleration is directed down the slope both ways.