JEE Advance - Physics (2011 - Paper 1 Offline - No. 20)
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 0.2 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 _________.
Explanation
f = $$\mu$$mg
The net torque about point P is
F $$\times$$ R $$-$$ fR = Ip$$\alpha$$
Where, Ip = mR2 + mR2 = 2mR2 (parallel axes theorem)
and, a = R$$\alpha$$
Also, f = $$\mu$$mg
$$\therefore$$ $$F \times R - \mu mgR = (2\,m{R^2}) \times \left( {{a \over R}} \right) = 2maR$$
$$ \Rightarrow F - \mu mg = 2\,ma$$
$$ \Rightarrow 2 - \mu \times 2 \times 10 = 2 \times 2 \times 0.3$$
which gives $$\mu = {{0.8} \over {2 \times 10}} = {{0.4} \over {10}}$$. Hence, P = 4
Comments (0)
