Question

A gymnast throws a hoop of mass $m$ and radius $R$ horizontally on a rough horizontal surface(coefficient of friction $\mu$) to the right with speed $V$ and angular velocity $\omega =\frac{2V}{R}$ as shown.

• A. The angular velocity of hoop, when it stops translating is $\frac{V}{R}$
• B. The distance travelled by the hoop before it stops translating is $\frac{V^2}{2\mu g}$
• C. The time for which the hoop slips on the ground is $\frac{3V}{2\mu g}$
• D. Work done by friction during the process of motion is $\frac{7m{V}^2}{4}$

Answer 1

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

The correct options are
A The angular velocity of hoop, when it stops translating is $\frac{V}{R}$
B The distance travelled by the hoop before it stops translating is $\frac{V^2}{2\mu g}$
C The time for which the hoop slips on the ground is $\frac{3V}{2\mu g}$


From translatory motion.
$-f=m{a}_{cm}$
$-\mu mg=m{a}_{cm}$
$a_{cm}=-\mu g$
Equation of kinematics
$\Rightarrow V_t=V-\frac{f}{m}t.....\left(1\right)$
When translation stops $\left(t_1\right)$$V_t=0\Rightarrow t_1=\frac{Vm}{f}=\frac{V}{\mu g}$
For rotational motion
$\tau =I\alpha \Rightarrow fR=m{R}^2\alpha$
$\Rightarrow \alpha =\frac{f}{mR}=\frac{\mu g}{R}-\left(2\right)$

$a)\text{Angular velocity}\Rightarrow {\omega }_t=-\omega +\left(\frac{\mu g}{R}\right)t\text{When translation stops}{\omega }_t=-\frac{2V}{R}+\left(\frac{\mu g}{R}\right)\frac{V}{\mu g}=-\frac{V}{R}$
$b)\text{Distance travelled when translation stops}S=ut+\frac{1}{2}a{t}^{2}S=V\left(\frac{V}{\mu g}\right)+\frac{1}{2}(-\mu g){\left(\frac{V}{\mu g}\right)}^2$
$S=\frac{V^2}{2\mu g}$
c) For the body to start rolling
$V_t=R{\omega }_t$
$\Rightarrow V-\mu gt=R[-\omega +\frac{\mu g}{R}t]$
$\Rightarrow V-\mu gt=R[-\frac{2V}{R}+\frac{\mu g}{R}t]$
$\Rightarrow 3V=+2\mu gt\Rightarrow t=\frac{3V}{2\mu g}\text{At this time}\Rightarrow V_t=\frac{-V}{2}$
$d)\text{From Work energy theorem}\text{Work done by Friction}{W}_f=K{E}_i-K{E}_f$
$W_f=[\frac{1}{2}m{V}^2+\frac{1}{2}I{\omega }^2]-\frac{1}{2}m{V}_t^2[1+\frac{k^2}{r^2}]$
$W_f=[\frac{1}{2}m{V}^2+\frac{1}{2}m{R}^2\frac{4V^2}{R^2}]-\frac{1}{2}\frac{m{V}^2}{4}\left(2\right)=\frac{9}{4}m{V}^2$