Question

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

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

Answer 1

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

After 0.5 s from impulse, its velocity is $1\text{m/s}$

$v_0=2\text{m/s}$
$\mu =0.4$
From linear impulse equation,
$m{v}_0-ft=mv....\left(1\right)$
where $v$ is velocity acquired by ring when slipping stops.
From Angular impulse,
$0+fRt=I\frac{v}{R}......\left(2\right)$
Dividing (1) by (2), we get,
$\frac{m{v}_0-ft}{fRt}=\frac{mv}{m{R}^2.\left(\frac{v}{R}\right)}$
$fRt=m{v}_{0}R-fRt$
$\mu mgRt=m{v}_{0}R-\mu mgtR(\because f=\mu mg$
$2\mu gt=v_0$
$t=\frac{v_0}{2\mu g}=\frac{2}{2\times 0.4\times 10}=0.25s$
Thus, ring starts pure rolling after 0.25 s
From (1),
$m{v}_0-ft=mv$
$m{v}_0-\mu mgt=mv$
$v=v_0-\mu gt=2-\left(0.4\right)\left(10\right)\left(0.25\right)$
$v=1\text{m/s}$
When ring starts pure rolling, its velocity is $1\text{m/s}$.
As pure rolling strats at $0.25\text{s}$, there will be no change in velocity after $0.25\text{s}$

Why this question? Tip: There is no further need to find velocity after the pure rolling starts, since the velocity remains same after the pure rolling starts.