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 P 10 . The value of P is

Step-by-step solution

There is no slipping between ring and ground. Hence, f₂ is not maximum. But there is slipping between ring and stick. Therefore, f₁ is maximum. Now, let us write the equations. aligned I & =m R^2=(2)(0.5)^2 & = 1 2 kgm ⁻² N₁-F₂ & =m a or N₁-F₂ & =(2)(0.3)=0.6 ~N a & =R = R I & = R (f₂-f₁ ) R I = R^2 (f₂-f₁ ) I 0.3 & = (0.5)^2 (f₂-f₁ ) (1 / 2) or f₂-f₁ & =0.6 ~N N₁^2+f₁^2 & =(2)^2=4 aligned Further F₁= N₁= ( P 10 ) N₁ Solving above four equation, we get P 3.6 Therefore, the correct answer should be 4 . Analysis of Q