# Ball roll down from the top of a rough spherical dome

1. Apr 12, 2015

### Yoonique

1. The problem statement, all variables and given/known data
A ball is rolling down from the top of a rough spherical dome with negligible initial velocity and angular velocity. Show that the ball must slide before losing the contact with the dome.

2. Relevant equations
ΣF=ma
Στ = Fr = Iα
fs = μsN
vcm = rω
Δmgh = 0.5mvcm2 + 0.5Icmω2
I = 2(mr2)/5

3. The attempt at a solution
Along the sphere:
ΣF = ma
mgsinθ - fs = matan

Στ = Fr = Iα
fsr = Iα

Let θ1 be where the ball lose contact. R be the radius of the sphere, r be the radius of the ball.
mgh = 0.5mvcm2 + 0.5Icmω2
mg(R+r)(1-cosθ1) = 0.7mvcm2

When N=0,
gcosθ1 = vcm2/(R+r)
vcm2 = (R+r)gcosθ1
mg(R+r)(1-cosθ1) = 0.7(R+r)mgcosθ1
cosθ1 = 1/1.7
θ1 = 54.0°

Let θ2 be where the ball starts to slide.
To show that the ball slides before it lose contact, θ2 < θ1
When is starts to slide, fs = μsN
fsr = Iα
μsN = 2mrα/5
N = 2mrα/5μs

And if I sub it in to the equations, it gets pretty complicated. I can't solve for θ. So how do I continue from here?

Last edited: Apr 12, 2015
2. Apr 12, 2015

### Merlin3189

Assuming you've worked out when the ball loses contact, when the required centripetal force is greater than the normal (radial) component of weight, the angle where it slides is simply where θ2 excedes the angle of friction,
tan θ2 > μs, or in other words,
the radial component of weight x μs < the tangential component of weight.

3. Apr 12, 2015

### Yoonique

I have already worked out when the ball loses contact which is 54.0°
mgsinθ2 > μsN
sinθ2 > 2rα/5g
But I got a variable α and I can't solve the value for θ.

4. Apr 12, 2015

### Yoonique

Can anyone give me some hints to solving this question?

5. Apr 14, 2015

### Yoonique

BUMP, went to the second page. Anyone?

6. Apr 14, 2015

### PeroK

Just an idea. If you have $\theta_1$ where the ball would fall off, then plug that value back into the equation for rolling without slipping and see whether it balances. Or, whether the frictional force would have to be too large by that stage.

7. Apr 14, 2015

### ehild

When N =0 fs must be also zero, so fs=μN holds. That is slipping!

From energy considerations you get the linear speed of the CM. From that, you get its linear and angular acceleration, and you can obtain expressions for N and for fs as function of the angle theta. Find the condition for theta when fs=μN, and check if it is greater or less than the angle of separation if pure rolling would happen.

8. Apr 14, 2015

### PeroK

Actually, with this insight, you may not have to do any calculations at all!

9. Apr 14, 2015

### Yoonique

If I just plug in the value of θ1 into the equation, I would get N=0 and fs = μ(0) = μN (condition for slipping). But this only shows that the moment it loses contact it slips. The question wants me to show that it is already slipping BEFORE it loses contact. So I need to find a value for θ2 and show that it is less than θ1.

Okay I will try ehild's method.

10. Apr 14, 2015

### PeroK

Yes, using ehild's insight is the way to go.

For future reference, if you show that $q_1 < q_2$ at some point, then $q_1$ must have been less than $q_2$ before that point. Where $q_1$ and $q_2$ are any continuously changing quantities.

11. Apr 14, 2015

### Yoonique

I tried but my equation simplified to θ2 and μs. μs is an unknown, thus I can't solve for θ2. Is there an equation I missed out?

12. Apr 14, 2015

### PeroK

I would try to solve this problem without any calculations. Sometimes, when the equations get complicated, it's good to solve the problem using logical arguments to minimise the calculations you need.

Here are the two key things you know:

1) The ball starts to roll and gets faster (its energy must keep increasing).

2) The normal force must reduce to 0 by the time it leaves the sphere.

Try to solve it logically using those facts and what you know about the conditions for rolling without slipping.

13. Apr 14, 2015

### ehild

Why don't you show your work in detail?

14. Apr 14, 2015

### Yoonique

Ok, here it is:

Let θ2 be where the ball starts to slide.
To show that the ball slides before it lose contact, θ2 < θ1

mg(R+r)(1-cosθ1) = 0.7mvcm2
vcm2 = (10/7)g(1-cosθ)(R+r)

mgcosθ2 - N = 10/7 mg(1-cosθ)
N = mg(10/7 cosθ2 - 10/7)

fs = 2/3mr2(a/r2) = 2/3 ma

mgsinθ2 - fs = ma
mgsinθ2 = 3.75fs

When is starts to slide, fs = μsN
mgsinθ2 = 3.75μsN = 3.75μsmg(10/7 cosθ2 - 10/7)
sinθ2 = 3.75μs(10/7 cosθ2 - 10/7)

I can't solve for θ2 with μs being unknown.

15. Apr 14, 2015

### Yoonique

Edited my post after thinking through.

As the ball speeds up, N decreases so fs decreases thus the tangential acceleration of the ball increases. Since fs = 2/3ma, fs increases? But i already said fs decrease at the start.. I'm confused..

Last edited: Apr 14, 2015
16. Apr 14, 2015

### ehild

You obtained that in case the ball rolled without slipping all the way along the dome, it would leave it at the angle where 10/7 cosθ2 - 10/7=0. As the left-hand side must be positive, the right-hand side is also positive. This expression decreases with increasing angle. Is the angle where slipping occurs less than the one it would leave the dome if it rolled all the way ?

17. Apr 14, 2015

### PeroK

Suppose the ball rolls without slipping. The tangential force (gravity) is increasing, so the friction force must be increasing. It can't go on increasing if, eventualy, it must get to 0.

18. Apr 14, 2015

### Yoonique

Ops I found a mistake in my working, it should be N = mg(17/7 cosθ2 - 10/7).
So sinθ2 = 3.75μs(17/7 cosθ2 - 10/7)
Okay I start to understand a little. In short for sinθ2 = 3.75μs(17/7 cosθ2 - 10/7) to be defined (which means sliding occurs), θ2 must be less than 54.0° = θ1.
Since sinθ2 > 0,
17/7 cosθ2 - 10/7 > 0 so cosθ2 > 10/17 > cosθ1.
So θ2 < θ1.

Okay I think I got it.

Last edited: Apr 14, 2015
19. Apr 15, 2015

### PeroK

Here's an alternative. If $F(\theta)$ is the tangential force and $f(\theta)$ is the frictional force, you can show that, for rolling without slipping:

$F(\theta) = kf(\theta)$ (for some positive constant $k$)

$F$ is strictly increasing, so $f$ must be strictly increasing. Therefore, slipping must occur when $f$ can no longer increase, which must be before the ball loses contact.

It's worth thinking about how powerful mathematics can be when used in this way.

Last edited: Apr 15, 2015
20. Apr 15, 2015

### Yoonique

Why is fs strictly increasing? I thought fs = μsN. As θ increasing, N actually decreases so fs decreases.