Discussion Forum

1)

A boy is pushing a ring of mass 2 kg and a radius of 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/ s2. 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 $\frac{P}{10}$. The value of P is 

Answer:

Option C

Explanation:

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.

 $l=mR^{2}=(2)(0.5)^{2}$

 =$\frac{1}{2} kgm^{-2}$

 $N_{1}-F_{2}=ma$

 or $N_{1}=F_{2}=(2)(0.3)=0.6 N$....(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)

 $N_{1}^{2}+f_{1}^{2}=(2)^{2}=4$....(iii)

Further $F_{1}=\mu N_{1}= \left(\frac{P}{10}\right)N_{1}$  ...(iv)

 Solving above four equation , we get  P$\approx$ 3.6

 Therefore , the correct answer should be 4