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jbriggs444

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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.erobz said:Sorry, I thought I was using common parlance in addressing Impulse/Momentum for angular quantities.

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".

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

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]

Yes. We want to compute the post-collision angular momentum.erobz said:I thought the point of the equation was a reply to this:

Angular momentum is noterobz said:My issue is I don't understand how we get to the desired result formally. If angular momentum isalwaysconserved about ##O##,

**conserved about ##O##. It is conserved post-collision. It is conserved pre-collision.**

*always***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 momentumerobz said:then why not simply evaluate the momentum pre-collision is equal to the angular momentum at the time when it begins pure rolling again?

**post-**collision is equal to the angular momentum at the time when it begins pure rolling again.

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