0
$\begingroup$

Suppose you have a ball with moment of inertia $I$, rolling on a surface with kinetic friction coefficient $\mu$. Initially, it has a linear velocity of $v_0$ and an angular velocity of $w_0$. Also, for notation let $v_{\ell}(t) = v(t) - R\omega(t)$ be the linear velocity (of the center of mass) excluding that from rolling, and $v_{r}(t) = R\omega(t)$ be the linear velocity of the center of mass (excluding the slipping of the ball).

Now, as it rolls with slipping, clearly $\omega(t)$ decreases. In fact, suppose there is a force $F$ applied from the friction, then

$$v_{\ell}'(t) = -\frac{F}{M}$$

However, $v_{r}$ increases by more:

$$\omega'(t) = \frac{FR}{I} = \frac{5}{2R}\frac{F}{M}$$ $$\implies$$ $$v_{r}'(t) = \frac{5}{2}\frac{F}{M} > |v_{\ell}'(t)|$$ Therefore, $$v'(t) = v_{\ell}'(t) + v_{r}'(t) = \frac{3}{2}\frac{F}{M}$$ so the ball speeds up as it moves.

But now assume that $w_0$ is negative, and in fact is large enough so that the ball will be going backwards once it starts rolling without slipping. But this means that the ball's velocity decreased overall - a contradiction to what we just proved!

So what have I done wrong - why doesn't the ball's velocity speed up?

Thought experiment provoked by Problems 2a and 5 (MIT OCW).

$\endgroup$
  • $\begingroup$ The ball HAS to decrease in BOTH rotational and translational velocity because the friction force does work, hence the total kinetic energy MUST decrease. One can also understand it as follows: the friction force doesn't only reduce the angular velocity because of torque about the central axis, it also acts against the translation, thereby slowing it down. This is not an endorsement of your calcs, I've not checked those. $\endgroup$ – Gert Sep 16 '19 at 18:04
  • $\begingroup$ Yes, but in this case the angular velocity is working against the linear velocity - so when you decrease the angular velocity, you increase the linear velocity. $\endgroup$ – programjames Sep 16 '19 at 18:11
  • $\begingroup$ Also, for notation let $vℓ(t)=v(t)−Rω(t)$ be the linear velocity (of the center of mass) excluding that from rolling, and $vr(t)=Rω(t)$ be the linear velocity of the center of mass (excluding the slipping of the ball). I don't think this is a useful, or even meaningful way of dealing with the problem. All you need is $v$ (transl. velocity of the ball) and $\omega$ (angular velocity of the ball). Both are functions of time, in your case. $\endgroup$ – Gert Sep 16 '19 at 18:20
  • $\begingroup$ If you look at the contact point between the ground and the ball, then it is moving at speed $v_{\ell}(t)$. The latter ($v_r(t)$) was just notation so that the speed of the center of mass would be $v_\ell(t) + v_r(t)$. Also, even if the linear velocity is greater than $0$, the force from friction only depends on $v_\ell(t)$. I don't see a meaningful way to determine $v(t)$ without breaking it into these components. $\endgroup$ – programjames Sep 16 '19 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.