Question

A uniform disc of mass m and radius R is released gently on a horizontal rough surface. Such that centre of the disc has velocity $V_0$ towards right and angular velocity ${\omega }_0$(anti clockwise) as shown. Disc will certainly come back to its initial position if ____.

Answer 1

Given: a uniform disc of mass m and radius R is released gently on a horizontal rough surface. Such that centre of the disc has velocity $V_0$ towards right and angular velocity ${\omega }_0$(anti clockwise) as shown.

To find the condition when disc will certainly come back to its initial position

Solution:

Let v be the velocity of center of mass

$\omega$ be the angular velocity

and $f$ be the frictional force.

We know, $f=\mu mg........\left(i\right)$

Now

$v=u+at⟹v=at⟹t=\frac{v}{a}..........\left(ii\right)$

and,

$\tau =I\alpha ⟹\alpha =\frac{\tau }{I}⟹\alpha =\frac{FR}{\frac{m{R}^2}{2}}⟹\alpha =\frac{2F}{mR}$

$⟹\alpha =\frac{2\mu mg}{mR}$ [from eqn(i)]

$⟹\alpha =\frac{2\mu g}{R}............\left(iii\right)$

Now

${\omega }_f={\omega }_0-\alpha t$

$⟹{\omega }_0=\alpha t$ [as ${\omega }_f=0$ wen disc completely stops]

Now substituting the values from eqn(ii) and eqn(iii), we get

${\omega }_0=\frac{2\mu g}{R}\times \frac{v}{a}⟹{\omega }_0=\frac{2\mu g}{R}\times \frac{v}{\mu g}⟹{\omega }_0=2v$

So disc will certainly come back to its initial position if ${\omega }_0>2v$