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 \(0.3 \mathrm{~m} / \mathrm{s}^{2}\). 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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Rotational Mechanics

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

Answer: 4

Solution

For the ring, let N be the normal force from stick and

f_{s}

be friction from stick.

Translational motion:

N - f_{s} = ma = 2 \times 0.3 = 0.6

N

Since the stick applies 2 N force, and considering geometry,

N = 2

N, so

f_{s} = 1.4

N.

For rotational motion about center:

(f_{s} - f_{K}) R = I α

, where

f_{K}

is friction from ground.

For a ring,

I = m R^{2}

and

a = R α

, so:

(f_{s} - f_{K}) R = m R^{2} \cdot \frac{a}{R} = m R a

f_{s} - f_{K} = ma = 0.6

f_{K} = 1.4 - 0.6 = 0.8 N

Coefficient of friction between stick and ring:

μ = \frac{f _{s}}{N} = \frac{0.8}{2} = 0.4 = \frac{P}{10}

Therefore, P = 4.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Rotational Mechanics

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