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/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 ((P/10)). The value of P is

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/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 ((P/10)). The value of P is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

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 = m R^{2} = (2) (0.5)^{2}

= \frac{1}{2} k g - m^{2}

N_{1} f_{2} = ma

or

N_{1} - f_{2} = (2) (0.3) = 0.6 N

.....(i)

a = R α = \frac{R τ}{I} = \frac{R ( f _{2} - f _{1} ) R}{I}

= \frac{R ^{2} ( f _{2} - f _{1} )}{I}

∴ 0.3 = \frac{( 0.5 ) ^{2} ( f _{2} - f _{1} )}{( 1/2 )}

or

f_{2} - f_{1} = 0.6 N

....(ii)

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

....(iii)
Further

f_{1} = μ N_{1} = (\frac{P}{10}) N_{1}

....(iv)
Solving above four equations we get,

P ≃ 3.6

Therefore the correct answer should be 4.