Question

A cylinder of mass $m$ and radius $r$ rolls down on a track from point A, as shown in the figure. Assume that the friction is just sufficient to support rolling. Velocity of the cylinder at point A was zero. Assume $r<<R$. The acceleration of the cylinder, when it reaches at point B is:

• A. $\frac{5}{3}g$
• B. $\frac{1}{3}g$
• C. $\frac{2}{3}g$
• D. $\frac{4}{3}g$

Answer 1

Answer: (D)

$\frac{4}{3}g$

Using conservation of energy we have
$\Delta P.E=\Delta K.E$
or
$mgR=\frac{1}{2}(I{\omega }^2+m{v}^2)$
$=\frac{1}{2}(\frac{m{R}^2}{2}\times \frac{v^2}{R^2}+m{v}^2)$
or
$mgR=\frac{3}{4}{v}^{2}m$
$v^2=\frac{4}{3}gR$
So acceleration $=\frac{v^2}{R}=\frac{4}{3}g$