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 and velocity of the cylinder at point A was zero. Assume that $r$ << $R$ then the reaction by the ground on the cylinder at point B is (Given acceleration due to gravity $=g$):

• A. $\frac{7mg}{3}$
• B. $\frac{4mg}{3}$
• C. $\frac{5mg}{3}$
• D. $\frac{2mg}{3}$

Answer 1

Answer: (A)

$\frac{7mg}{3}$

Using conservation of energy
$\Delta P.E=\Delta K.E$
$mgR=\frac{1}{2}I{\omega }^2+\frac{m{v}^2}{2}$
$=\frac{1}{2}(\frac{m{R}^2}{2}\frac{v^2}{R^2}+m{v}^2)$
$mgR=\frac{3}{4}{v}^{2}m$
$\Rightarrow v^2=\frac{4}{3}gR$
Now since the cylinder is rotating in a circle
Reaction on the ground $=\frac{m{v}^2}{R}+mg$
$=\frac{7m}{3}g$