Question

A ring of mass M and radius R is rest at the top of an incident shown. The ring rolls down the plane without slipping. When the ring reaches bottom, its angular momentum about its moves is:

• A. $MR\sqrt{gh}$
• B. $MR\sqrt{\frac{gh}{2}}$
• C. $MR\sqrt{2gh}$
• D. None of these

Answer 1

The correct option is A $MR\sqrt{gh}$

Inertia of ring, $I=m{r}^2$

Velocity, $v=r\omega$

Potential Energy = Translation Kinetic Energy + Rotational Kinetic Energy

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$gh=\frac{v^2}{2}+\frac{1}{2}m{r}^2{\omega }^2$

$v=\sqrt{gh}$

Angular momentum about center, $mvR=mR\sqrt{gh}$