Find the velocity when the ball rolls without slipping

[vbrasic]

Homework Statement

A thin spherical shell is sliding with velocity $v_0$ on a table initial until friction eventually causes it to roll without slipping. Find its translational velocity when the it rolls without slipping as a fraction of $v_0$.

Homework Equations

$$I=\frac{2}{3}MR^2$$
$$I\dot{\omega}=RF$$

The Attempt at a Solution

We have for a thin spherical shell that $$\frac{2}{3}MR^2\dot{\omega}=RM\dot{v}.$$ Using this we can get a relation between $\dot{\omega}$ and $\dot{v}$. We have that, $$\frac{2}{3}R\dot{\omega}=\dot{v}.$$ Integrating both sides with respect to time gives, $$\frac{2}{3}R\omega=v-v_0.$$ When the ball rolls without slipping we have that $\omega=\frac{v}{R}$. So we have that $$\frac{2}{3}v=v-v_0,$$ such that $$v_0=1/3v\to 3v_0=v.$$ This doesn't make logical sense to me however, as $v_0$ should be the max translational velocity. According to this the ball's velocity is increasing which shouldn't be the case.

 

[kuruman]

We have for a thin spherical shell that
$$\frac{2}{3}MR^2\dot{\omega}=RM\dot{v}.$$

Why is this true? Angular momentum is conserved about the point of contact because there is no net torque about that point. Start from there.

 

[Dr Dr news]

This is just like a bowling ball. The ball slides initially and then as friction takes over it rolls. Depending on your release you can even impart a side spin resulting in a curved path with the intent of striking at a point to knock over all ten pins.

 

[haruspex]

This is just like a bowling ball. The ball slides initially and then as friction takes over it rolls. Depending on your release you can even impart a side spin resulting in a curved path with the intent of striking at a point to knock over all ten pins.

How is that relevant to the question?

 

[haruspex]

This doesn't make logical sense

Your mistake is that you have been inconsistent regarding signs.
In $\frac{2}{3}MR^2\dot{\omega}=RM\dot{v}$, think about how the positive sense of ω relates to the positive direction of v.
Compare that with their relationship in $\omega=\frac{v}{R}$.

Why is this true?

By considering how the frictional force affects linear and angular acceleration.

Angular momentum is conserved about the point of contact because there is no net torque about that point.

That is certainly the easiest method in this problem, but there is potential for ambiguity regarding "point of contact" as a reference axis. It should be a point fixed in an inertial frame which happens to be on the locus of the points of contact. This is different from point of contact as a moving (and accelerating) point; that would likely lead to the wrong answer.

 

[kuruman]

That is certainly the easiest method in this problem, but there is potential for ambiguity regarding "point of contact" as a reference axis. It should be a point fixed in an inertial frame which happens to be on the locus of the points of contact. This is different from point of contact as a moving (and accelerating) point; that would likely lead to the wrong answer.

In this particular problem, one could pick as reference the starting point on the flat surface at which the ball has velocity v₀ or any other fixed point on the surface. It was not my intention to introduce ambiguity by implying that one may calculate the angular momentum about a moving point of contact which in this case accelerates until the sphere transitions to rolling without slipping. "Fixed point on the surface" is less ambiguous.

 

[Dr Dr news]

How is that relevant to the question?

Engineering students like to have concrete examples they can visualizw.

 

[haruspex]

Engineering students like to have concrete examples they can visualizw.

This student seems to be pretty much on top of the conceptual aspect, just made a mistake over signs.