# Find angle of the sphere losing contact with the surface

1. Dec 1, 2016

1. The problem statement, all variables and given/known data
A solid sphere rolls down a hemisphere from rest. Find the angle at which the sphere loses contact with the surface.
2. Relevant equations
ΣFr = Macm,r
N-mgcosθ = -mVcm2/(R+a)
N = mgcosθ - mvcm2/(R+a) eq. (1)

Conservation in mechanical energy
mg(R+a) = 1/2mvcm2 + 1/2Icmω2 + mg(R+a)cosθ
mg(R+a) = 1/2mvcm2 + 1/2(2/5ma2)(Vcm/a)2 + mg(R+a)cosθ
mg(R+a) = 7/10mVcm2 + mg(R+a)cosθ
Simplified
7/10mVcm2 + mg(R+a)(1-cosθ)
Vcm2 = 10g/7 (R+a)(1-cosθ) eq. (2)

eq. (2) into eq. (1)
N = mgcosθ - m10g(R+a)/(R+a) (1-cosθ)
N= 17mgcos/7 - 10mg/7
N goes to 0 since the sphere loses contact with the surface, thus:
0 = 17cosθ-10
θ = cos-1(10/17)
θ = 54°

3. The attempt at a solution

My attempt solution has been trying to prove this problem experimentally. By rolling a solid sphere on a hemisphere, the angle I found was much smaller than 54°. Any thoughts?

2. Dec 1, 2016

### PeroK

I found some notes on this problem and that looks like the right answer. Although, that is for a small ball where $a << R$.

PS I think it's the same answer for any $a$.

Also, as rolling without slipping requires a minimum amount of friction to accelerate, the sphere is bound to slip before it comes off. In practice, therefore, it will start to slip and gain higher speed earlier, so come off earlier.

Last edited: Dec 1, 2016
3. Dec 1, 2016

Thus, the sphere would start slipping right before it comes off? I can't seem to notice that. Also, would it be possible to do a calculation to find a better and more approximate angle?

4. Dec 1, 2016

### haruspex

You could redo the calculation assuming no friction, so no rotation. That at least would give you a range for the angle.
In between, you would have to work in terms of forces and torques rather than energy.

What angles are you observing? How are you measuring them?

5. Dec 1, 2016

I'm observing an angle approximately half of the angle I measured per my calculations. I'm just measuring it by observation. I'm still very confused in why the experiment is not working as desired and how I could get a better approximation.

6. Dec 1, 2016

### haruspex

So you are seeing it become airborne when only about a third of the way around the arc?

7. Dec 1, 2016

Yes. I was expecting it to come off at around 45 degrees.

8. Dec 1, 2016

### haruspex

But you calculated 54 from vertical, no?
What are you using for the hemisphere? How accurately shaped?

9. Dec 1, 2016