The final velocity of a ball rolling while slipping.

[erobz]

You are making the same mistake as in post #1. The velocity after the collision is $-v_0$.
You seem to be taking $v_f$ as positive left, which is a bit confusing.

I’m not following…I’m I supposed to be looking at immediately before and after the collision? Is that where I’m confused, because when the sphere has started to roll again it has both negative angular momentum terms as far as I can tell…or is that incorrect. I’m all tangled up.

 

[jbriggs444]

I’m not following…I’m I supposed to be looking at immediately before and after the collision? Is that where I’m confused, because when the sphere has started to roll again it has both negative angular momentum terms as far as I can tell…or is that incorrect. I’m all tangled up.

Immediately after a collision with a smooth wall, linear momentum of the sphere is reversed. So that contribution to total angular momentum is reversed. However, rotation of the sphere about its own center of mass is not reversed.

Immediately before collision, the two contributions are both in the same direction. They add.
Immediately after collision, the two contributions are in opposite directions. They subtract.

Long after the collision, the two contributions will return to the 5:2 proportion characteristic of a rolling sphere. But we need to take one step at a time. First determine the total angular momentum immediately after the collision.

There will be time afterward decide what that means for the final linear velocity after things settle down.

 

[erobz]

Immediately after a collision with a smooth wall, linear momentum of the sphere is reversed. So that contribution to total angular momentum is reversed. However, rotation of the sphere about its own center of mass is not reversed.

Immediately before collision, the two contributions are both in the same direction. They add.
Immediately after collision, the two contributions are in opposite directions. They subtract.

Long after the collision, the two contributions will return to the 5:2 proportion characteristic of a rolling sphere. But we need to take one step at a time. First determine the total angular momentum immediately after the collision.

There will be time afterward decide what that means for the final linear velocity after things settle down.

Well, the only parameter free to change is the angular velocity of the wheel immediately after impact $\omega$. The linear velocity is given as $-v_o$. Thus, taking $\rightarrow^+$, $\circlearrowright^+$:

$$ \int\tau_{ext}~dt = \left( -mRv_o + \frac{2}{5}mR^2 \omega \right) - \left( mRv_o + \frac{2}{5}m R v_o \right) $$

Then I believe what is being said is the integral is $0$ since the frictional force from the floor causes no torque about its point of contact(our reference for angular momentum), hence:

$$\cancel{ \int\tau_{ext}~dt }^0 = \left( -mRv_o + \frac{2}{5}mR^2 \omega \right) - \left( mRv_o + \frac{2}{5}m R v_o \right) $$

$$ \implies 0 = -mRv_o + \frac{2}{5}mR^2 \omega - mRv_o - \frac{2}{5}m R v_o$$

$$ \implies \omega = \frac{6 v_o}{R} = 6 \omega_o$$

?

 

[jbriggs444]

Well, the only parameter free to change is the angular velocity of the wheel immediately after impact $\omega$. The linear velocity is given as $-v_o$. Thus, taking $\rightarrow^+$, $\circlearrowright^+$:

$$ \int\tau_{ext}~dt = \left( -mRv_o + \frac{2}{5}mR^2 \omega \right) - \left( mRv_o + \frac{2}{5}m R v_o \right) $$

Going through term by term...

You have that integral on the left hand side. But it is an indefinite integral. We do not know what interval it is supposed to be evaluated over. It would be helpful if you had explained what you are doing rather than writing down an equation. I am forced to decode what you are doing by reverse engineering the equation.

But yes, we likely have some sort of angular momentum balance. The change in angular momentum from some initial state to some final state will be equal to the external torque applied to the system in the interval between the initial and final states.

We have that first term on the right hand side. This looks like the pre- post-collision angular momentum. You are writing down $L_f - L_i$. We have $mRv_0$ where $m$, $R$ and $v_0$ are all positive (clockwise angular momentum) but you put a minus sign on it so that the result will be counter-clockwise. So you must be thinking of the post-collision contribution from linear momentum. You have $\frac{2}{5}mr^2$ which is positive (clockwise) so that is the post-collision contribution from rotation. That all checks out. So that first term on the right hand side is the post-collision angular momentum.

The other term on the right hand side is then obvious. It is the pre-collision angular momentum. Both contributions are positive (clockwise). Perfect.

Which means that the integral on the left hand side is to be evaluated over the brief interval during which the collision takes place. It is the angular impulse delivered by the external force that is responsible for reversing the linear momentum of the sphere. Plainly, that impulse is the linear impulse: $-2 mv_0$ that is delivered by the wall multiplied by the offset from the selected reference axis, $R$.

So the equation that you have provided is correct. It simplifies to $-2 mv_0 R = -2 mv_0 R$. It is consistent with the post-collision angular momentum being given by:$$- m R v_0 + \frac{2}{5}m R v_0$$

 

[erobz]

You have that integral on the left hand side. But it is an indefinite integral. We do not know what interval it is supposed to be evaluated over. It would be helpful if you had explained what you are doing rather than writing down an equation. I am forced to decode what you are doing by reverse engineering the equation.

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

So the equation that you have provided is correct. It simplifies to $-2 mv_0 R = -2 mv_0 R$. It is consistent with the post-collision angular momentum being given by:$$- m R v_0 + \frac{2}{5}m R v_0$$

The equation is saying that the wheel is spinning faster ( about its COM )post collision as far as I can tell? I thought the point of the equation was a reply to this:

But we need to take one step at a time. First determine the total angular momentum immediately after the collision.

My issue is I don't understand how we get to the desired result formally. If angular momentum is always conserved about $O$, then why not simply evaluate the momentum pre-collision is equal to the angular momentum at the time when it begins pure rolling again?

 

[jbriggs444]

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.

 

[erobz]

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.

 

[jbriggs444]

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.

 

[haruspex]

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.