Physics Problems & Solutions: Classical Mechanics

A hoop of mass M and radius R is at rest at the top of an inclined plane. The hoop rolls down the plane without slipping. When the hoop reached the bottom, its angular momentum around its center of mass is

A. MR√(gh)
B. ½ MR√(gh)
C. M√(2gh)
D. Mgh
E. ½ Mgh

(GR8677 #76)

Solution:

Conservation of energy: Total energy at the top = Total energy at the bottom

$Mgh= \frac{1}{2} Mv^2_{\text{CM}}+\frac{1}{2} I_\text{CM} \omega^2$

Velocity: v_CM = ωR
Moment inertia of the hoop: I_CM = MR²

$\\Mgh= \frac{1}{2} MR^2\omega^2+\frac{1}{2} MR^2 \omega^2=MR^2 \omega^2\\\\\rightarrow \omega^2 = \frac{gh}{R^2}\rightarrow \omega= \frac{\sqrt{gh}}{R}$

Angular Momentum: $L=I\omega=MR^2 \frac{\sqrt{gh}}{R} = MR\sqrt{gh}$

Answer: A

Physics Problems & Solutions

Classical Mechanics - Rotational Motion

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