Question

A uniform disc of mass $m$ and radius $R$ is placed on a smooth horizontal surface. An impulse $I$ is given horizontally to the sphere at height $h=\frac{3R}{5}$ above the centre line. Then which of the following option(s) is correct?

• A. The minimum time after which the highest point touches the ground is $\frac{\pi Rm}{2I}$
• B. The minimum time after which the highest point touches the ground is $\frac{5\pi Rm}{6I}$
• C. The displacement of the COM during this interval is $\frac{5\pi R}{6}$
• D. The displacement of the COM during this interval is $\frac{\pi R}{2}$

Answer 1

Answer: (B), (C)

The correct options are
B The minimum time after which the highest point touches the ground is $\frac{5\pi Rm}{6I}$
C The displacement of the COM during this interval is $\frac{5\pi R}{6}$

Assume just after impulse given to the disc, linear velocity of COM of the disc is $v$ and angular velocity is $\omega$.


Impulse $I=\Delta P=P_f-P_i$
$(\because \text{disc is at rest},P_i=0)$
$I=mv-0$
or $v=\frac{I}{m}$
Angular impulse $J=\Delta L=L_f-L_i$ $(\because \text{disc is at rest initially},L_i=0)$
$I\times h=L_f$
$I\times \left(\frac{3R}{5}\right)=I_0\omega$
Here, $I_0=\frac{m{R}^2}{2}$
$I\times \left(\frac{3R}{5}\right)=\left(\frac{m{R}^2}{2}\right).\omega$
or $\omega =\frac{6I}{5Rm}$

When the highest point reaches the ground, it rotates by an angle $\pi$.
Hence $\theta =\omega t$
$\pi =\left(\frac{6I}{5Rm}\right)t$
$t=\frac{5\pi Rm}{6I}$
And the distance travelled by the COM in that time
$d=v.t$
$d=\frac{I}{m}.\left(\frac{5\pi Rm}{6I}\right)$
$d=\frac{5\pi R}{6}$