Question

A boy is pushing a ring of mass $2$ kg and radius $0.5$ m with a stick shown in the figure. The stick applies a force of $2$ N on the ring and rolls it without slipping with an acceleration of $0.3m/s^2$. The coefficient of friction between the ground and ring is large enough that rolling always occurs and coefficient of friction between the stick and the ring is ($P/10$). The value of $P$ is ?

Answer 1

Given,

$m=2kg$

$r=0.5m$

$N=2N$

$a=0.3m/s^2$

Let $f_s$ is friction between surface and sphere.

Let $f_a$ is friction between surface and stick

force acting on ring

$N-f_s=ma2-f_s=2\times 0.3f_s=2-0.6=1.4N$

Torque acting on the center of ring is

$(f_s-f_a)r=I\alpha$

$a=r\alpha$ and $I=m{r}^2$

Now,

$(f_s-f_a)r=m{r}^2\times \frac{a}{r}(1.4-2\mu )0.5=2\times {0.5}^2\times \frac{0.3}{0.5}1.4-2\mu =\frac{0.3}{0.5}2\mu =0.8\mu =0.4$

In question

$\mu =P/10$

$0.4=p/10$

$p=4$