# Homework Help: Angular momentum conservation.

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1. Jan 6, 2017

### cheapstrike

1. The problem statement, all variables and given/known data

A uniform solid sphere of radius R, rolling without sliding on a horizontal surface with an angular velocity ωo, meets a rough inclined plane of inclination θ=60°. The sphere starts pure rolling up the plane with an angular velocity ω. Find the value of ω.

2. Relevant equations

Conservation of angular momentum.
3. The attempt at a solution

If we conserve angular momentum about the point of contact with inclined plane, what will be the equation that we will get. I am confused as to what velocity will we take as the initial velocity of the centre of mass about the point of contact with the incline.
How do i upload photo on mobile? I am having little trouble. Also, I couldn't explain my problem in detail because I am unable to type properly on mobile

2. Jan 6, 2017

### PeroK

Until forces have had time to act, the motion of the sphere won't change. Are you asked to calculate $\omega$ as a function of time?

3. Jan 6, 2017

### haruspex

That's the right approach.
What is the angular momentum of the sphere about the point of impact immdiately before impact? You can consider the rolling motion as the sum of a linear velocity of the mass centre plus a rotation about the mass centre. Each contributes to the angukar momentum about the point of impact.
In terms of ω, what is the angular momentum about that point just after impact?
I'm guessing this is an impact, not a smooth transition.

4. Jan 6, 2017

### cheapstrike

Equation for conservation of angular momentum is given in my solution as:
2/5mR2ωo+mvoRcos60°=7/5mR2ω.
I understand that the first term on the lhs is the angular momentum due to rotation where is the second term in the lhs is that's due to the centre of mass but I am not understanding why we have taken it mvoRcos60°.

5. Jan 6, 2017

### haruspex

How far does the original line of motion of the mass centre pass from the point you are taking moments about?

6. Jan 6, 2017

### cheapstrike

In this picture, are the two points (COM and point of contact of incline and sphere) correctly marked? If so, then the distance should be R, right? And if we calculate angular momentum of the mass center about the point of contact, then it will be R(vector)×mVcm(vector), which will be mRVsinα, where α is the angle between R and Vcm vectors, which should be 120°?

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Last edited: Jan 6, 2017
7. Jan 6, 2017

### cheapstrike

So sorry, the angle between R and vcm vectors will be 150° and therefore angular momentum is mvrcos60°.

8. Jan 6, 2017

### PeroK

What do you plan to do next? The way I looked at it, the sphere initially had components of velocity normal to and tangential to the slope. As it didn't bounce, the normal component was lost in a totally inelastic collision, leaving just the tangential component, which is $v \cos \theta$.

Now, I initially assumed that the ball would take some time to speed up linearly and slow its rotation and stabilise into rolling, but this brings gravity into play. So, @haruspex, as always, is correct and we have to assume that the stabilisation happens instantaneously.

In any case, how are you going to generate equations relating the initial and final states?

Last edited: Jan 6, 2017
9. Jan 6, 2017

### cheapstrike

Why will the normal component of initial velocity of the sphere make it leave the surface of incline when it is acting downwards wrt the incline?
Btw I used conservation of angular momentum and got the equation as:
Iwo+mvo Rcos60 = Iw+mRv.
w and v are final angular and linear velocities respectively(just after it starts moving on the incline). v=Rw. And therefore w can be calculated in terms of wo.

Last edited: Jan 6, 2017
10. Jan 6, 2017

### haruspex

We have a rotating ball striking a surface at an angle. In reality, we would expect it to bounce a little, but for the purposes of the question we take it as completely inelastic.
Anyway, you now have the two terms making up the initial angular momentum about the impact point, matching the left hand side of your equation in post #4. Are you ok with the term on the right hand side?

11. Jan 6, 2017

### cheapstrike

Yeah, I just edited my last post. Thanks

12. Jan 6, 2017

Good.