Question

In the given Figure, a boy is pushing a ring of mass $2\text{kg}$ and radius $0.5\text{m}$ with a stick. The stick applies a normal reaction of $2\text{N}$ on the ring and rolls it without slipping with an acceleration of $0.3{\text{ms}}^{-2}$. The coefficient of friction between ground and ring is large enough to avoid any slipping.
If the coefficient of friction between the stick and the ring is $\frac{P}{10}$, then the value of P is .

Answer 1

Let $\mu$ be coefficient of friction between ring and stick.
Using Newton's 2nd law along horizontal direction,
$2-f=2\left(0.3\right)$
$\Rightarrow f=1.4\text{N}$ Since the ring rolls on the surface,
$a=R\alpha$ (condition for no-slipping)

$\Rightarrow 0.3=\left(0.5\right)\alpha$

$\Rightarrow$ $\alpha =\frac{3}{5}\frac{\text{rad}}{{\text{s}}^2}$

Now, ${\tau }_{net}=I\alpha$
$\Rightarrow fR-f_{L}R=m{R}^2\alpha$
$\Rightarrow fR-2\mu =m{R}^2\alpha$
[Here, we have used the limiting friction at the point of contact of stick and ring $f_L=\mu N=2\mu$]

$\Rightarrow$ $f-2\mu =mR\alpha$
$\Rightarrow$ $1.4-2\mu =2\times 0.5\times 3/5$
$\Rightarrow 2\mu =0.8$
$\Rightarrow \mu =\frac{4}{10}=\frac{P}{10}$
$\Rightarrow P=4$