Question

A cylinder of mass m and radius r rolls down a circular track from point A as shown in the figure. Assume $r<<R$ and the friction is just sufficient to support the rolling. Velocity of the cylinder at point A is zero. The reaction by the track on the cylinder at point B is

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

Answer 1

The correct option is A $\frac{7}{3}mg$

Using the law of conservation of energy:
$mgR=\frac{1}{2}m{v}^2(1+\frac{k^2}{r^2})$ where $\frac{k^2}{r^2}=\frac{1}{2}$. or $v^2=\frac{4gR}{3}$ ----(1)
Also at the lowest point: $N-mg=\frac{m{v}^2}{R}$ or $N=mg+\frac{m}{R}\frac{4gR}{3}=\frac{7}{3}mg$