# 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

#### brochesspro

Homework Statement
A solid sphere rolling on a rough horizontal surface with a linear speed ##v_0## collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.
Relevant Equations
Conservation of angular momentum.
The moment of inertia of a solid sphere of mass ##m## and radius ##R## is ##\frac 2 5mR^2##.
Here is the problem statement along with the figure.  Here, I take the right-ward and anti-clockwise directions to be positive.
After the ball collides with the wall, its angular velocity remains the same and its velocity changes direction while remaining the same in magnitude.
Using the conservation of angular momentum about the bottom-most point, I get:
##-mv_0R -\frac 2 5mR^2\omega_0 = -mv_fR + \frac 2 5 mR^2\omega_f##
##\Rightarrow -mv_0R -\frac 2 5mv_0R = -mv_fR + \frac 2 5 mv_fR##
since ##v_0 = R\omega_0## and ##v_f = R\omega_f##
##\Rightarrow -\frac 7 5 mv_0R = -\frac 3 5 mv_fR##
##\Rightarrow 7v_0 = 3v_f##
##\Rightarrow v_f = \frac 7 3 v_0##
This is clearly false as the sphere slows down due to friction acting in the positive direction. May I know what exactly I have done wrong?
Thanks.

When can you use conservation of angular momentum? Can you do that when there is slipping?

When can you use conservation of angular momentum? Can you do that when there is slipping?
You can use the conservation of angular momentum about a point about which the net external torque on the system is zero. Whether slipping occurs or not does not matter, I think.

You can use the conservation of angular momentum about a point about which the net external torque on the system is zero. Whether slipping occurs or not does not matter, I think.
It will be conserved for the entire system yes but not for individual objects. Consider the spinning disc wich you place another disc from above. Torque due to friction will be same but opposite in direction on those two discs, leding to zero net external torque on the system of two discs. But each individual disc will have its angular momentum altered

It will be conserved for the entire system yes but not for individual objects. Consider the spinning disc wich you place another disc from above. Torque due to friction will be same but opposite in direction on those two discs, leding to zero net external torque on the system of two discs. But each individual disc will have its angular momentum altered
What does this situation have to do with mine? And the system is taken as the rolling sphere, so I do not think it makes a difference.

What does this situation have to do with mine?
It seems to me that you use that angular momentum of that disc is conserved.

It seems to me that you use that angular momentum of that disc is conserved.
Yes, you are right.

Yes, you are right.
I meant sphere, not disc...

Well its not conserved because spehere is not the entire system here.

I meant sphere, not disc...

Well its not conserved because spehere is not the entire system here.
What do you mean? And is the system not up to the solver to choose?

What do you mean? And is the system not up to the solver to choose?
You can choose system yes but if your system is the sphere then you have a non zero external torque due to sliding and friction... simple as that

You can choose system yes but if your system is the sphere then you have a non zero external torque due to sliding and friction... simple as that
How is the net external torque non-zero about the bottom-most point? All three forces on the sphere, i.e. its weight and the contact force(comprising of the frictional force and the normal reaction from the floor) from the floor pass through the bottom-most point of the sphere, which is also its point of contact with the ground.

To be honest, I think there is a problem with the sign convention I choose, it is just a feeling though.

To be honest, I think there is a problem with the sign convention I choose, it is just a feeling though.
should be this I think
##-mv_0R +\frac 2 5mR^2\omega_0 = -mv_fR - \frac 2 5 mR^2\omega_f##
i.e. change signs of omega

should be this I think
##-mv_0R +\frac 2 5mR^2\omega_0 = -mv_fR - \frac 2 5 mR^2\omega_f##
i.e. change signs of omega
However, I chose the anticlockwise direction to be the positive direction, so the signs should stay the same according to me.

if v is positive to the right, then clockwise should be positive angular direction. As in the case when the sphere is approaching the wall.

if v is positive to the right, then clockwise should be positive angular direction. As in the case when the sphere is approaching the wall.
Why so? Is the sign convention not for the solver to choose?

Here, I take the right-ward and anti-clockwise directions to be positive.

##-mv_0R -\frac 2 5mR^2\omega_0 = -mv_fR + \frac 2 5 mR^2\omega_f##
Why ##-mv_0R##? I thought you were taking anticlockwise rotation as positive. After the collision, the velocity is ##-v_0##.

• brochesspro
Why ##-mv_0R##? I thought you were taking anticlockwise rotation as positive. After the collision, the velocity is ##-v_0##.
I see what mistake I did, thanks. I have corrected my mistake. I have gotten the correct answer of ##v_f = -\frac 3 7v_0##. Thanks for your help.

• haruspex
Why so? Is the sign convention not for the solver to choose?
Then you need to adjust the condition for pure rolling too.

You can choose system yes but if your system is the sphere then you have a non zero external torque due to sliding and friction... simple as that
We are evaluating angular momentum and torque about the lab bench -- the point O. There is no torque about this point because the only relevant force (friction) has a line of action that passes through point O.

So angular momentum is conserved during the post-collision spin-up. Angular momentum about this axis is not conserved during the collision.

There is an angular momentum balance that can be written down. However, it is not the one you are thinking of.

• malawi_glenn
We are evaluating angular momentum and torque about the lab bench -- the point O. There is no torque about this point because the only relevant force (friction) has a line of action that passes through point O.

So angular momentum is conserved during the post-collision spin-up. Angular momentum about this axis is not conserved during the collision.

There is an angular momentum balance that can be written down. However, it is not the one you are thinking of.

yeah I could not see the figure properly I thought O was the point of impact on the wall :)

Trying to wrap my head around this. Is the entire initial rotational energy ##\frac{1}{2}I \omega_o^2## required to be converted to heat?

On a frictionless table I would expect it bounces off the wall at ##-v_o##, but maintains its initial rotational kinetic energy throughout the duration of motion.

So I'm thinking if its going to change its direction of rotation on a rough surface, its angular velocity must pass through zero (at the end of its slipping motion phase) before pure rolling may resume?

EDIT:

I see that what I'm thinking can't be correct. The change in energy is ## \frac{4}{7}mv_o^2##, and the initial rotational KE is ##\frac{1}{5}mv_o^2##.

So its losing all its initial rotational KE and some translational KE on top of that in the process between the final state and the collision with the wall. That seems to make sense.

Last edited:
So I'm thinking if its going to change its direction of rotation on a rough surface, its angular velocity must pass through zero (at the end of its slipping motion phase) before pure rolling may resume?
Right. That is the picture in my mind's eye.

We have an elastic collision. But the ball is still spinning in its original direction. This is wrong for its new velocity. So it continues to spin one way while moving the other. It spins down through zero angular velocity (we assume!) and then picks up angular velocity in its new direction of forward motion until, eventually, the spin rate and the movement rate match up for rolling without slipping.

It may be worth checking the results of our algebra to make sure that spinning actually stops before rolling without slipping is achieved. Having practiced rolling a hula-hoop with backspin on it for fun, I know for a fact that such will not be achieved for all combinations of forward velocity, backward spin and moment of inertia.

It may be worth checking the results of our algebra to make sure that spinning actually stops before rolling without slipping is achieved. Having practiced rolling a hula-hoop with backspin on it for fun, I know for a fact that such will not be achieved for all combinations of forward velocity, backward spin and moment of inertia.
Yeah, I figure I'd might as well just check after posting that. From what I'm getting it loses all its rotational KE and a portion of its translational KE as well in the process between the wall and its final state.

Yeah, I figure I'd might as well just check after posting that. From what I'm getting it loses all its rotational KE and a portion of its translational KE as well.
I've not done the calculation. My intuition is that for a solid sphere, it will pick some rotational KE back up. It is certain that the final translational KE and final rotational KE will be in direct proportion. The constant of proportionality will depend on the moment of inertia of the shape.

• erobz
It may be worth checking the results of our algebra to make sure that spinning actually stops before rolling without slipping is achieved.

Do you not trust the algebra in post #1, as corrected in post #18? It should produce the right sign for the velocity. The same equations of motion apply throughout, even if the velocity is instantaneously zero at some point.

Do you not trust the algebra in post #1, as corrected in post #18? It should produce the right sign for the velocity. The same equations of motion apply throughout, even if the velocity is instantaneously zero at some point.
I had not reviewed the algebra in post #1 or #18. The result in #18 after correcting for the sign error seems plausible. However, I want to attack the problem in a slightly different way. Rather than verifying algebra, I want to verify an intuition. Always try to test with a different approach from the one used to generate the result.

Pre impact we have angular momentum of ##v_0 R## from the linear motion of the ball and ##\frac{2}{5}v_0 R## from its rotation. Total ##\frac{7}{5}v_0 R##. Clockwise. That part of the algebra was solid.

Post impact the linear motion is reversed but the rotational motion is not. So we should have a total of ##\frac{3}{5}v_0 R##. We are subtracting now rather than adding. Clearly the linear contribution is dominating. So the total should be counter-clockwise.

Final angular momentum is ##\frac{3}{5}v_0 R## counterclockwise. Initial angular momentum was ##\frac{7}{5}v_0 R## clockwise. Angular momentum (when rolling without slipping) is proportional to linear velocity. So the result that ##v_f = - \frac{3}{7} v_0## is upheld.

Since the final velocity is in the direction away from the barrier, we are not facing the case where the backspin dominates and a second collision with the barrier ensues.

If I am not mistaken, one would need a thin hoop for backspin to cancel with linear momentum. One would need a rather more exotic mass distribution to go past ##I=mR^2## and get backspin to dominate.

Last edited:
I still don't understand how the sign convention works...

If I choose ##\rightarrow^+##, ##\circlearrowright^+##, I keep getting:

$$mRv_o + \frac{2}{5}mRv_o = -mRv_f -\frac{2}{5}mR v_f$$

Obviously that's not the result...but I can't see why for some reason?

I still don't understand how the sign convention works...

If I choose ##\rightarrow^+##, ##\circlearrowright^+##, I keep getting:

$$mRv_o + \frac{2}{5}mRv_o = -mRv_f -\frac{2}{5}mR v_f$$

Obviously that's not the result...but I can't see why for some reason?
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.

One would need a rather more exotic mass distribution to go past I=mR2 and get backspin to dominate.
E.g. a yoyo structure, where the narrow radius in the middle is what contacts the frictional surface.

• jbriggs444
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.

Last edited:
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:

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

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

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

?

Last edited:
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:$$- m R v_0 + \frac{2}{5}m R v_0$$

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?

Last edited: