Rotational Motion and Rigid Body Dynamics

Notes

• Moment of inertia: The rotational analogue of mass, I=\sum m_i r_i^2=\int r^2\,dm. It depends on the axis. Rotational kinetic energy is K=\tfrac{1}{2}I\omega^2 and angular momentum is L=I\omega.
• Torque and angular acceleration: Torque \vec{\tau}=\vec{r}\times\vec{F} produces angular acceleration through \vec{\tau}=I\vec{\alpha}, the rotational form of Newton's second law. The rate of change of angular momentum equals the net torque, \vec{\tau}=\tfrac{d\vec{L}}{dt}.
• Parallel and perpendicular axis theorems: I=I_{cm}+Md^2 shifts the axis by distance d from the centre of mass; for a planar lamina in the xy-plane, I_z=I_x+I_y.
• Rolling without slipping: The contact point is instantaneously at rest, so v_{cm}=R\omega and a_{cm}=R\alpha. Total kinetic energy is K=\tfrac{1}{2}Mv_{cm}^2\left(1+\dfrac{k^2}{R^2}\right), where k is the radius of gyration.
• Conservation of angular momentum: If the net external torque is zero, L=I\omega is constant, so decreasing I (e.g. a skater pulling arms in) increases \omega.
• The parallel-axis theorem I = I_{cm} + Md^2 and perpendicular-axis theorem I_z = I_x + I_y (planar bodies only) let you build moments of inertia of composite bodies; always reduce to a known I_{cm} first.
• Angular momentum about a point is conserved only when the net external torque about that point is zero. Choosing the contact point of a rolling body or the pivot of a hinge often makes the (large) impulsive contact/hinge force pass through the axis, eliminating its torque.
• Rolling without slipping imposes v_{cm} = \omega R and a_{cm} = \alpha R; the friction here is static and does no net work. On an incline, the acceleration is a = g\sin\theta/(1 + I_{cm}/MR^2), so hollow bodies (larger I/MR^2) accelerate slower than solid ones.
• During a sudden collision/impulse on an extended body, linear impulse changes v_{cm} and angular impulse (torque impulse about cm) changes \omega; the point that instantaneously remains at rest defines the 'centre of percussion'. Striking at the centre of percussion produces no reaction impulse at the pivot.
• For a body that starts skidding (e.g. a struck billiard ball or a spinning wheel placed on ground), kinetic friction acts until the no-slip condition v_{cm} = \omega R is reached; solve using both F=ma and \tau = I\alpha simultaneously to find the transition time and final speed.

Formulas

• Definitions: I=\int r^2\,dm,\quad K_{rot}=\tfrac{1}{2}I\omega^2,\quad L=I\omega
• Dynamics: \tau=I\alpha=\dfrac{dL}{dt}
• Axis theorems: I=I_{cm}+Md^2,\quad I_z=I_x+I_y
• Rolling energy: K=\tfrac{1}{2}Mv_{cm}^2\left(1+\dfrac{k^2}{R^2}\right)
• Rolling down incline: a=\dfrac{g\sin\theta}{1+k^2/R^2},\quad \mu_{min}=\dfrac{\tan\theta}{1+R^2/k^2}
• Standard k^2/R^2: ring 1, disc/solid cylinder \tfrac12, solid sphere \tfrac25, shell \tfrac23
• a_{incline} = \dfrac{g\sin\theta}{1 + I_{cm}/MR^2} (rolling without slipping down an incline)
• L = I\omega + M(\vec{r}_{cm}\times\vec{v}_{cm}) (angular momentum: spin plus orbital)
• \ell_{percussion} = \dfrac{I_{pivot}}{M d} (distance of centre of percussion from pivot; d = pivot-to-cm distance)
• \tau = \dfrac{dL}{dt}, and KE_{roll} = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I_{cm}\omega^2

Exam traps & shortcuts

• Down an incline, the body with the smallest k^2/R^2 has the largest acceleration, so a solid sphere beats a disc, which beats a ring - independent of mass and radius.
• For a rolling body the instantaneous axis passes through the contact point, so the top point moves at 2v_{cm} and the contact point is momentarily at rest.
• Angular momentum about the point of collision is often conserved even when linear momentum is not (e.g. a ball hitting a hinged rod); pick the axis through the hinge.
• Pick the axis about which unknown impulsive forces have zero torque (contact point, hinge) to conserve angular momentum through a collision even when linear momentum is not conserved.
• For a ball struck horizontally at height h above centre: it rolls without slipping immediately (no skidding) when h = \tfrac{2}{5}R above centre for a solid sphere—the impulse's angular effect exactly matches the linear effect.
• Convert rolling energy problems using KE = \tfrac{1}{2}(1 + I/MR^2)Mv^2; the factor (1+I/MR^2) acts like an effective-mass multiplier and speeds up comparisons between shapes.

Reference tables