# The final velocity of a ball rolling while slipping.

• brochesspro
Since the sign convention is not for the solver to choose, I thought that the velocity would be negative after the collision. However, it is actually the same.Since the sign convention is not for the solver to choose, I thought that the velocity would be negative after the collision. However, it is actually the same.f
Sorry, I thought I was using common parlance in addressing Impulse/Momentum for angular quantities.
You have correct quantities. It is simply that just reading an equation does not always provide insight into what the quantities represent and why the equality is justified.

In the case at hand you have a valid equation that can be justified as: "The change in angular momentum is the final angular momentum minus the initial angular momentum" and "The change in angular momentum is also equal to the integral of the net external torque over the interval in question".

The equation is saying that the wheel is spinning faster ( about its COM )post collision as far as I can tell?
No, that is not what the equation is saying. Please tell me where you read that from the equation.

What I see in that equation is a net integrated torque on the left hand side. That net integrated torque will turn out to be equal to ##-2 m v_0 R##. We can reach that conclusion independently by taking the vector cross product of the net linear impulse (##-2 m \vec{v_0}##) with the signed displacement (##\vec{R}##).

If we examine your equation, we can simplify the right hand side and reach the same conclusion.

If we are after the final rotation rate for the sphere, we do not care at all about this equation. All we care about is the total angular momentum for the sphere post-collision. That is given by:$$-mRv_0 + \frac{2}{5}mRv_0$$Clearly, this is lower in absolute value than the pre-collision angular momentum:$$mRv_0 + \frac{2}{5}mRv_0$$[I am using ##mRv_0## instead of ##m R^2 \omega## to emphasize consistency rather than clutter]
I thought the point of the equation was a reply to this:
Yes. We want to compute the post-collision angular momentum.
My issue is I don't understand how we get to the desired result formally. If angular momentum is always conserved about ##O##,
Angular momentum is not always conserved about ##O##. It is conserved post-collision. It is conserved pre-collision. But the collision involves a non-zero net external rotational impulse. As your equation makes clear. That integral is the non-zero net rotational impulse during the collision.
then why not simply evaluate the momentum pre-collision is equal to the angular momentum at the time when it begins pure rolling again?
The angular momentum post-collision is equal to the angular momentum at the time when it begins pure rolling again.

Last edited:
• erobz and malawi_glenn
Angular momentum is not always conserved about ##O##. It is conserved post-collision. It is conserved pre-collision. But the collision involves a non-zero net external rotational impulse. As your equation makes clear. That integral is the non-zero net rotational impulse during the collision.

The angular momentum post-collision is equal to the angular momentum at the time when it begins pure rolling again.
Ok, I see my confusion. I kept thinking that the integral was identically ##0##. I thought that's what @haruspex was implying in post #26.

The same equations of motion apply throughout, even if the velocity is instantaneously zero at some point.

Now I understand that to mean after the impact. Not throughout the entire scenario...

Thanks for taking the time setting me straight.

• jbriggs444
I love it when an understanding is reached following a vigorous discussion.

• erobz
I love it when an understanding is reached following a vigorous discussion.
Well, as often as I find myself in the weeds, I've learned to not give up so easily.

Well, as often as I find myself in the weeds, I've learned to not give up so easily.
Since @haruspex is essentially never wrong, if you catch him saying something that cannot be true, it is a good bet that there is a misunderstanding somewhere. Chasing down misunderstandings is often a worthwhile goal.

• erobz
Ok, I see my confusion. I kept thinking that the integral was identically ##0##. I thought that's what @haruspex was implying in post #26.

Now I understand that to mean after the impact. Not throughout the entire scenario...

Thanks for taking the time setting me straight.
Yes, I should have have been clearer. Ah well.

• erobz