Question

A rod of mass $m$ and length $2R$ is attached to a circular ring of mass $mandradiusR$ along its vertical diameter as shown. Horizontal force $F$ is applied at top. Friction is sufficient for pure rolling, then friction force acting on the ring, at the instant shown is

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

Answer 1

The correct option is B $\frac{F}{5}$


$F+f=2ma$

$F-f)r=l\alpha ,a=\alpha r$