# Rolling and collision

1. May 11, 2013

### Saitama

1. The problem statement, all variables and given/known data
A solid sphere rolling on a rough horizontal surface with a linear speed $v$ collides elastically with a fixed smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in backward direction.

2. Relevant equations

3. The attempt at a solution
Before collision, the velocity is $v$ and angular velocity is $v/R$. After collision the velocity is $v$ in backward direction. I am confused about what happens to the angular velocity when the collision takes place? Do the magnitude of angular velocity changes?

I think that as the particle on the edge colliding with the wall has two components of velocity, the component which is perpendicular to the wall will not change so the direction of angular velocity remains unchanged. Am I going in the right direction?

Thanks!

2. May 11, 2013

### ehild

Is there any torque during the collision due to interaction with the wall?

3. May 11, 2013

### Saitama

About CM (I think any other point would also do), there is no torque during collision.

4. May 11, 2013

### Kakashi24142

I think you have to consider the torque about the point of contact of the sphere with the surface because pure rolling can be viewed as pure rotation about the point of contact. During the collision, the normal force exerted by the wall on the sphere will produce a non-zero torque about the point of contact of the sphere with the ground.

5. May 11, 2013

### Saitama

I think ehild meant torque on the system and I think it is zero if you consider both the reaction forces during collision i.e. force due to wall on the sphere and force due to sphere on the ball.

6. May 11, 2013

### Kakashi24142

Yes, the net torque of the system is indeed zero.

I think the torque about the point of contact is non-zero for the system consisting of the sphere only, as per the following argument. Correct me if I am wrong.

Consider only the sphere at the instant of collision, the forces are:
(1) Normal force due to ground (point of application is the point of contact)
(2) Normal force due to wall (point of application is the point where the sphere touches the wall)
(3) Weight of the sphere (point of application is the COM)

Now calculate torque about the point of contact (call it P). The torque about P from (1) is zero since the point of application is on P. The torque about P from (3) is also zero because the line of action of force lies along the line connecting P and the COM. The torque about P from (2) at some instant during the collision is

$$\tau = N\cdot \sin{45^\circ} \cdot R\sqrt{2} = NR$$

where N is the magnitude of the normal force at some instant during the collision and R is the radius of the sphere.

7. May 11, 2013

### Tanya Sharma

Hello Pranav

Let the set up be like

The sphere moves from left to right towards the wall and after collision moves towards left.i.e the velocity is towards right and angular velocity clockwise .

Just after collision,the angular velocity remains same since the force from the wall will be perpendicular to the wall passing through the CM of the sphere (the wall is smooth ) ,hence no angular impulse.That is angular velocity remains clockwise with same magnitude but the velocity of the sphere is towards left and magnitude same as before (elastic collision). So the ball rolls with slipping (skids )towards left .

Here comes friction in picture,which was not playing any role till now.

The friction acts rightwards providing anticlockwise torque about CM which reduces the angular velocity as well as decelerates the sphere .

Gradually a moment comes when the sphere stops rotating but is still moving towards left .

Just after this due to the anticlockwise torque provided by the friction the sphere starts rotating anticlockwise ,gradually gaining angular speed untill the sphere rolls without slipping .

Here role of friction ends .

Now the sphere happily rolls without slipping towards left with anticlockwise angular velocity .

I hope this helps you .

Last edited: May 11, 2013
8. May 11, 2013

### sankalpmittal

Firstly you know that in elastic collision (in inelastic also !!) momentum of the sphere system is conserved.

From there you can find something, just after the small time lag when sphere collides with wall and then moves leftwards. (You know what.)

Now, consider the axis of rotation passing through the contact point of sphere with ground. Then you need not consider torque by friction, and angular momentum of sphere system IS conserved. Apply that. Find initial angular momentum when it was going due right and then the same when it was going due left. The both time lags when there is conservation of angular momentum is rolling without slipping.

9. May 11, 2013

### Saitama

Thank you Tanya, your explanation helped a lot. :)

Conserving angular momentum after collision,
$$mvR-Iv/R=mv'R+Iv'/R$$
where v' is the final velocity.

Solving I get, $v'=3v/7$. Thank you everyone!

10. May 12, 2013

### sankalpmittal

A ditto question from H.C. Verma, rotation : 86th one.

If you read my reply, (previous post) I told the same thing. You know why we are substituting vcm = Rω in both L.H.S and R.H.S respectively, right ?

And you understand that why we are conserving angular momentum about the axis passing through contact point of sphere with rough surface and not about the centre of mass itself, right ?

Last edited: May 12, 2013
11. Mar 1, 2015

### lavankohsa

@sankalpmittal: Please tell me why we are conserving angular momentum about the axis passing through contact point of sphere with rough surface and not about the centre of mass itself ??

12. Mar 2, 2015

### haruspex

As sankalpmittal posted, it allows you to ignore friction. The frictional force acts along the ground, so has no moment about the point of contact with the ground. You can do it taking moments about the mass centre, but then you have to bring in another equation for linear momentum and elminate the unknown friction force between the two equations.

13. Mar 3, 2015

### lavankohsa

If i take angular momentum w.r.t. centre of mass then accord to me the eqn will be
-I(v/r)=Iw'

since v and r both in same direction w.r.t. to centre of mass.
where w' is angular velocity after pure roling has started.
Where i am wrong in above eqn ??

14. Mar 3, 2015

### haruspex

You are to find the rotation after it has resumed rolling. Immediately after the collision it will be sliding. To get from sliding to rolling there will be a frictional torque from the ground. That is missing from your equation.

15. Mar 3, 2015

### lavankohsa

But when we take angular momentum w.r.t. point of contact than we are not taking frictional torque. Why is it so ?

16. Mar 3, 2015

### haruspex

Because the line of action of the frictional force is along the floor. That passes through the point of contact, so has no moment about it.

17. Mar 3, 2015

### lavankohsa

So that means angular momentum is conserved w.r.t. to point of contact and not conserved w.r.t. centre of mass. Is conservation of angular momentum depends on reference.

18. Mar 3, 2015

### TSny

Yes, that's right. Suppose you drop a ball and let if fall along a vertical line. Is the angular momentum of the ball conserved about a point fixed on the vertical line? About a point not on the vertical line?

For the rolling ball problem, it is best not to pick the point of contact as an origin since this point is accelerating relative to the inertial frame of the floor. But you can pick a fixed point on the floor along the line that the point of contact is moving and still get zero net torque on the ball about that point.

The center of mass is special. It accelerates relative to the inertial frame of the floor. Nevertheless, you can still apply $\tau_{net} = \frac{dL}{dt}$ relative to the the center of mass.

19. Jan 7, 2016

### Prannoy Mehta

After going through Tanya Sharma's answer. (I have a doubt, and I clearly know I am mistaken) The sphere moves towards left so the equation should be,

$$mvR-Iv/R=-mv'R+Iv'/R$$

Right? The Initial case involved it move in the forward direction and clock wise so mvR and -IvR are justified. And -mv'R should be there because it now moves towards the left. Iv'R is justified as it moves anticlockwise.

20. Jan 7, 2016

### TSny

The equation relates two different times after the collision. $v$ is the velocity just after the collision. $v'$ is the velocity after the collision when pure rolling starts. Both velocities are in the same direction.