Thread: Determine nature of the motion of the hoop?

A hoop with a radius $\displaystyle r$ rotating with an angular velocity $\displaystyle \omega_{0}$ is placed on a rough horizontal surface. A translational velocity $\displaystyle v_{0}$ is imparted to the hoop. Determine the nature of the motion of the hoop, assuming that the force of the sliding friction is $\displaystyle f$.

Answer:

Spoiler:

$\displaystyle \mbox{Hoop slides for t}\ =\frac{(v_0 + \omega_{0}R)M}{2f}\ \mbox{then rolls with}\ \omega = \frac{(v_0 + \omega_{0}r)M}{2f}$

I am unable to determine the answer. Any ideas?

Originally Posted by fardeen_gen

A hoop with a radius $\displaystyle r$ rotating with an angular velocity $\displaystyle \omega_{0}$ is placed on a rough horizontal surface. A translational velocity $\displaystyle v_{0}$ is imparted to the hoop. Determine the nature of the motion of the hoop, assuming that the force of the sliding friction is $\displaystyle f$.

Answer:

Spoiler:

$\displaystyle \mbox{Hoop slides for t}\ =\frac{(v_0 + \omega_{0}R)M}{2f}\ \mbox{then rolls with}\ \omega = \frac{(v_0 + \omega_{0}r)M}{2f}$

I am unable to determine the answer. Any ideas?

Suppose that at time t (during the sliding motion) the angular velocity of the hoop is $\displaystyle \omega$. The torque exerted by the frictional force F is Fr. By Newton's second law for rotational motion, $\displaystyle Fr = -mr^2\dot{\omega}$ (m is the mass of the hoop). Therefore $\displaystyle \dot{\omega} = -\frac F{mr}$. Integrate that to get $\displaystyle \omega - \omega_0 = -\frac{Ft}{mr}$.

Let v be the translational velocity of the hoop at time t. The frictional force is the only horizontal force acting on the hoop, so by Newton's second law (the ordinary second law, for linear motion) $\displaystyle F = -m\dot{v}$. Integrate that to get $\displaystyle v-v_0=-\frac{Ft}m$.

The hoop will stop sliding when $\displaystyle -r\omega=v$. That happens when $\displaystyle r\omega_0 -\frac{Ft}{m} = -v_0 + \frac{Ft}m$, which simplifies to $\displaystyle t = \frac{(r\omega_0+v_0)m}{2F}$. At that time, $\displaystyle \omega = \omega_0 - \frac{Ft}{mr} = \frac{r\omega_0 - v_0}{2r}$ (slightly different from the answer that you wanted).

Opalg, your result for $\displaystyle \omega $ doesn't match Fardeen's because he has rewritten the result for the time in its place.

Assuming you defined $\displaystyle \omega$ to be positive in the anti-clockwise sense (opposite to sense of translational velocity) I got the same result.