Question

A hoop of mass 'm' and radius 'R' is projected on a floor with linear velocity $v_0$ and reverse spin ${\omega }_0$. The coefficient of friction between the hoop and the ground is $\mu$. Under what of the following condition will the hoop return back?

• A. ${\omega }_0<\frac{v_0}{R}$
• B. ${\omega }_0>\frac{v_0}{R}$
• C. ${\omega }_0=\frac{\mu v_0}{R}$
• D. It will never return back

Answer 1

The correct option is B

${\omega }_0>\frac{v_0}{R}$

The acceleration of hoop = $-\mu g$

Angular acceleration of wheel =$\frac{-\mu g}{R}$

Linear velocity at time t, $v=v_0-\mu gt$

Angular velocity at time t, $\omega ={\omega }_0$-$\frac{\mu gt}{R}$

- Let linear velocity be zero of the hoop at any instant 't'.

So, 0 = $v_0-\mu gtort=\frac{v_0}{\mu g}$

Angular velocity of hoop at this instant,$\omega ={\omega }_0-\left(\frac{\mu g}{R}\right)\left(\frac{v_0}{\mu g}\right)={\omega }_0-\frac{v_0}{R}$

The hoop will return back if it has reverse spin at the instant its linear velocity is zero. so the condition of the hoop to return, $\omega >$ 0

$\Rightarrow -\frac{v_0}{R}>0or{\omega }_0>\frac{v_0}{R}$