Question

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of $0.3 m/s^2$. 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 $\bigg(\frac{P}{10}\bigg).$ The value of P is

Answer 1

There is no slipping between ring and ground. Hence $f_2$ is not maximum. But there is slipping between ring and stick. Therefore, $f_1$ is maximum. Now let us write the equations.
$I=mR^2=(2)(0.5)^2$
$=\frac{1}{2}kg-m^2$
$N_1f_2=ma$
or $N_1-f_2=(2)(0.3)=0.6N$ .....(i)
$a=R\alpha=\frac{R\tau}{I}=\frac{R(f_2-f_1)R}{I}$
$=\frac{R^2(f_2-f_1)}{I}$
$\therefore 0.3=\frac{(0.5)^2(f_2-f_1)}{(1/2)}$
or $f_2-f_1=0.6N$ ....(ii)
$N^2_1+f^2_1=(2)^2=4$ ....(iii)
Further $f_1=\mu N_1=\bigg(\frac{P}{10}\bigg)N_1$ ....(iv)
Solving above four equations we get, $P\simeq3.6$
Therefore the correct answer should be 4.