Question

A hoop of mass $m$ 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 condition will the hoop return back?

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

Answer 1

Answer: (B)

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

As the point of contact of hoop has non-zero velocity w.r.t. ground in the forward direction, frictional force will apply in the backward direction.

This would decrease both the linear velocity and the angular velocity of the hoop. Now, hoop will return back, if the linear velocity becomes zero before the angular velocity.

frictional force is $f=\mu mg$.

Linear deceleration is $a=f/m=\mu g$

time at which linear speed becomes zero is gives by,

$v_o-at=v_o-\mu gt=0⟹t=v_o/\mu g$

Angular deceleration is $\alpha =fR/m{R}^2=\mu g/R$

at time $t$, the angular velocity should be greater than zero, so that frictional force would continue to apply in the backward direction and the hoop will return back.

i.e.,

${\omega }_o-\alpha t>0⟹{\omega }_o-\left(\mu g/R\right)\times v_o/\mu g>0⟹{\omega }_o>v_o/R$