Question

A boy is pushing a ring of mass $2kg$ and radius $0.5m$ with a stick as shown in the figure. The stick applies a normal force of $2N$ on the ring and rolls it without slipping with an acceleration of $0.3m{s}^{-2}$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs, and the coefficient of friction between the stick and the ring is $\left(P/10\right)$ . The value of $P$ is

Answer 1

From free body diagram of ring, writing equation of motion of ring
$2-f=2\left[0.3\right]\Rightarrow f=2-0.6=1.4N$
As disc is rolling, $A=R\alpha \Rightarrow 0.3=\alpha \left[0.5\right]$
$\alpha =\frac{3}{5}rad/s$

Now torque equation,
${\tau }_c=I_c\alpha$
$fR-2\mu R=m{R}^2\alpha$
$f-2\mu =mR\alpha$
$1.4-2\mu =2\times 0.5\times 0.6$
$0.8=2\mu \Rightarrow \mu =0.4=\frac{P}{10}$
$\therefore P=4$