Question

A uniform ring placed on a rough horizontal surface is given a sharp impulse as shown in the figure-$5.130$. As a consequence, it acquires a linear velocity of $2 m / s$. If coefficient of friction between the ring and the horizontal surface is $0.4$ :

• A. Ring will start pure rolling after $0.25 \,s$
• B. When ring will start pure rolling its velocity is $1 \,m / s$
• C. After $0.5 s$ from impulse its velocity is $1 \,m / s$.
• D. After $0.125 s$ from impulse its velocity is $1 \,m / s$.

Answer 1

Answer: (A), (B), (C)

$v_{0}=2 m / s$
$\mu =0.4 $
$m v_{0}-f t =m v\,\,\,...(1)$
where $v$ is velocity acquired by ring when slipping stops.
$0+f R t=I \frac{v}{R}\,\,\,...(2)$
Dividing (1) by (2), we get,
$\frac{m v_{0}-f t}{f R t}=\frac{m v}{m R^{2} \cdot\left(\frac{v}{R}\right)}$
$ f R t =m v_{0} R-f R t $
$ \mu m g R t =m v_{0} R-\mu m g+R \,\,\,(\because f=\mu m g) $
$ 2 \mu g t =v_{0} $
$t=\frac{v_{0}}{2 \mu g}=\frac{2}{2 \times 0.4 \times 10}=0.25 s$
Thus, ring will start pure rolling after $0.25 s$
From (1),
$m v_{0}-f t =m v$
$m v_{0}-\mu m g t =m v $
$v =v_{0}-\mu g t=2-(0.4)(10)(0.25)$
$v =1 \,m / s$
When ring will start pure rolling, its velocity is $1 \,m / s$.
From $0.25$ s to $0.5$ s, i.e.,
$t =0.5-0.25=0.25\, s ,$
$v^{\prime} =v_{0}-\mu g t$
$v^{\prime} =2-1=1 \,m / s$