Question

Under what condition will the hoop return back?

• A. ${\omega }_0>\frac{v_0}{R}$
• B. ${\omega }_0>\frac{v_0}{2R}$
• C. ${\omega }_0>\frac{{2v}_0}{R}$
• D. ${\omega }_0>\frac{v_0}{3R}$

Answer 1

The correct option is A ${\omega }_0>\frac{v_0}{R}$

Using Newton's second law,

$N\mu =-ma$ (Force acting on center of mass)

$f_{f}R=-I_c\alpha$ (Torque)

Here, $f_f,a,I_{c}and\alpha$ are the friction force, acceleration of center of mass, moment of inertia about center and angular acceleration of the hoop.

$I_c=m{R}^2$

$f_f=\mu mg$

Therefore,

$v=v_o-\mu gt$

$w=w_o-\frac{\mu g}{R}t$

For the hoop to roll back,

$w$ must be positive when $v=0$

So, when $t=\frac{v_o}{\mu g}$,

$w_o-\frac{\mu g}{R}t>0$

Therefore,

$w_o>\frac{v_o}{R}$

Answer: A