Rotational Motion — Inertia, Torque & Rolling (Medium)

Quick Recap — Moment of Inertia & Rolling

• Standard $I$ (about the symmetry axis): ring/hoop $MR^2$; disc/solid cylinder $\tfrac12 MR^2$; solid sphere $\tfrac25 MR^2$; hollow sphere $\tfrac23 MR^2$; rod (centre) $\tfrac{ML^2}{12}$, (end) $\tfrac{ML^2}{3}$.
• Theorems: parallel axis $I=I_{cm}+Md^2$; perpendicular axis (planar) $I_z=I_x+I_y$. Radius of gyration $k=\sqrt{I/M}$.
• Dynamics: $\tau=I\alpha$, $L=I\omega$, KE$_{rot}=\tfrac12 I\omega^2=\dfrac{L^2}{2I}$, power $P=\tau\omega$.
• Rolling: $v=R\omega$; total KE $=\tfrac12 mv^2\!\left(1+\dfrac{I}{mR^2}\right)$; the rotational fraction is $\dfrac{I/mR^2}{1+I/mR^2}$ (so $\tfrac27$ for a solid sphere).
• On an incline: $a=\dfrac{g\sin\theta}{1+I/mR^2}$; from a height $h$, $v=\sqrt{\dfrac{2gh}{1+I/mR^2}}$.

Worked mini-example. A solid sphere ($I/mR^2=\tfrac25$) rolling down a $30^\circ$ incline has $a=\dfrac{g\sin30^\circ}{1+2/5}=\dfrac{5}{1.4}=3.6$ m/s$^2$ — less than the $5$ m/s$^2$ it would have if it simply slid.