A boy is pushing a ring of mass and radius with a stick as shown in the figure. The stick applies a force of 2N on the ring and rolls it without slipping with an acceleration of . The coefficient of friction between the ground and the ring is large eno

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 2N 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 [2011]

Smart Achievers · Accepted Answer

(4)

The stick applies (2N) force so, point of contact O of the ring with ground tends to slide. But the frictional force $f_{2}$ does not allow this and creates a torque about 'C' which starts rolling the ring. Between the ring & the stick a friction force $f_{1}$ also acts.

$F - f_{2} = m a$

$2 - f_{2} = 2 \times 0.3 \Rightarrow f_{2} = 1.4 N$

$Applying τ = I α about C$

$(f_{2} - f_{1}) R = I α = I \frac{a}{R}$ $[∵ a = R α]$

$\Rightarrow [1.4 - μ \times 2] \times 0.5 = 2 \times {(0.5)}^{2} \times \frac{0.3}{0.5}$ $[∵ I = M R^{2}]$

$∴$ $μ = 0.4 = \frac{4}{10} = \frac{P}{10} \Rightarrow P = 4$

Solution

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