Question

A uniform solid disc of radius R and mass m is free to rotate on a frictionless pivot through a point on its rim. The disc is released from rest in the position where the diameter through the pivot is along horizontal. The speed of its centre of mass when the diameter through the pivot is vertical is

• A. $\frac{2}{3}(gR{)}^{1/2}$
• B. $\left(g/R^{1/2}\right)$
• C. $\left(2g/R^{1/2}\right)$
• D. $2(\frac{gR}{3}{)}^{1/2}$

Answer 1

The correct option is D $2(\frac{gR}{3}{)}^{1/2}$

Radius$=R$

Mass$=m$

$\omega ={\omega }_0+at$

The speed of its centre of mass when the diameter through the pilot is vertical.

$MI=\frac{1}{3}M{R}^{2}v=\omega R=-\frac{a}{t}R=2(\frac{gR}{3}{)}^{1/2}$