Question 755: Centre of Mass and Rotational Motion

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 0.2 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 _________.

Accepted Answer

f = $$\mu$$mg
The net torque about point P is
F $$	imes$$ R $$-$$ fR = I
p
$$\alpha$$
Where, I
p
= mR
2
+ mR
2
= 2mR
2
(parallel axes theorem)
and, a = R$$\alpha$$
Also, f = $$\mu$$mg
$$	herefore$$ $$F 	imes R - \mu mgR = (2\,m{R^2}) 	imes \left( {{a \over R}} ight) = 2maR$$
$$ \Rightarrow F - \mu mg = 2\,ma$$
$$ \Rightarrow 2 - \mu  	imes 2 	imes 10 = 2 	imes 2 	imes 0.3$$
which gives $$\mu  = {{0.8} \over {2 	imes 10}} = {{0.4} \over {10}}$$. Hence, P = 4

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type