# Help with a solid sphere sliding and then rolling

Initially at time $t=0$, a solid sphere slides with velocity $v$ along a horizontal surface. The coefficient of friction is u. Find the required time for the sphere to stop sliding (the sphere will then be rolling).

Ive tried using Newtons equations of motion but I'm not sure what the final velocity will be? At first I thought it may be zero because it will be rolling but obviously it will still have some translational velocity. I've also found the work done by the friction on the sphere and taking this figure away from the initial kinetic energy leaving (what I thought) would be the rotational kinetic energy that the ball has. However I can't get to the answer. I'm almost certain I'm just missing a small trick but I have no idea where to go with it.

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You can relate the change in angular momentum of the sphere to the change in linear momentum - they are related by the radius. That is one equation. The second equation tells you the relationship between angular velocity and linear velocity when the sphere stops sliding. Again there is a simple factor of $r$.

Finally you have the coefficient of friction so you know how large the force is. With the first two equations you get $F \cdot t$, and with the third one you find $t$ from $F$ .

Does that help?

Those two should be sufficient to solve this.

UPDATE

Here are the equations, and how to combine them:

Momentum equations:

\begin{align} \Delta p &= m \Delta v = F \Delta t\\ \Delta L &= I\Delta \omega = F r \Delta t\\ I &=\frac{2}{5}mr^2\\ \end{align}

Initial condition:

\begin{align} v &= v_0\\ \omega &= 0 \end{align}

Final condition (for rolling sphere):

$$v = \omega r$$

Combining:

\begin{align} m\Delta v &= F \Delta t\\ m(v_0-v_f)&= F \Delta t &(1)\\ \\ I\Delta \omega &= F r \Delta t\\ I \omega _f&= F r \Delta t &(2)\\ \\ r \omega_f &= v_f \end{align} Eliminating $v_f$ from (1) using (2):

\begin{align} m(v_0 - r \omega _f) &= F \Delta t \\ m(v_0-\frac{Fr^2}{I}\Delta t)&=F\Delta t\\ m v_0 &= F \Delta t(1 + \frac{mr^2}{I})\\ &=F \Delta t(1+\frac52) \\ \Delta t &= \frac27\frac{mv_0}{m g u}\\ &= \frac27\frac{v_0}{gu} \end{align}

Sanity check:

• If initial velocity is greater, skidding time increases
• If friction coefficient is greater, skidding time decreases
• If gravity is greater, skidding time decreases

All of these make sense.

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You're right, the fact that the sphere will still be rolling means that its velocity won't necessarily be zero. But you don't have to know its actual velocity; you just have to know something about its velocity. In mathematical terms, remember that you're solving a system of equations, (at least) one of which involves the final velocity, and you only need as many equations as you have variables. So it's not necessary that the equation that involves final velocity is actually of the form $v = \text{const.}$

Here's a big hint: how can you relate the translational velocity of the sphere to its rotational velocity once it stops sliding?

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