# Ball Falling

1. Nov 9, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data
A solid spherical ball is placed carefully on the edge of a table in the position shown in the figure. The coefficient of static friction between the ball and the edge of the table is 0.5 . It is then given a very slight push. It begins to fall off the table.

Find the angle (in degrees)(with vertical) turned by the ball before it slips.

2. Relevant equations

3. The attempt at question

I have came up with some equations. Let $\theta$ be the angle(with vertical) at which the sphere begins to slip.

By conservation of energy
$mgr-mgrcos\theta =\frac { 1 }{ 2 } { I }_{ 0 }{ \omega }^{ 2 }$

As $v=r \omega$

So $g(1-cos\theta )=\frac { 7 }{ 10 } \frac { { v }^{ 2 } }{ r }$

By finding torque about the contact point

$mgsin\theta r=\frac { 7 }{ 10 } m{ r }^{ 2 }\alpha$

As $a=r \alpha$

So $\frac { 5gsin\theta }{ 7 } =a$

Also from FBD of the block

$mgcos\theta -N=m\frac { { v }^{ 2 } }{ r }$

and $\\ mgsin\theta -\mu N=ma$

Last edited: Nov 9, 2014
2. Nov 9, 2014

### haruspex

I doubt that's right for the centripetal force on a sphere.

3. Nov 9, 2014

### Satvik Pandey

Could please explain, why? How to proceed?

4. Nov 9, 2014

### Jilang

Can this be solved by just considering the forces at the point of contact tangent to the sphere? That would give:
sin theta = 1/2 cos theta at the time of slipping.

5. Nov 9, 2014

### Satvik Pandey

Could you please show how did you find that.

6. Nov 9, 2014

### Jilang

The frictional force at the point of contact would be mg cos theta x 1/2 at a tangent to the sphere and the gravitational force would be mg sine theta.

7. Nov 9, 2014

### voko

If you can find the normal force at the contact, then it is simple. Your result follows from equating the normal force with the radial component of the weight, but how do you justify this?

8. Nov 9, 2014

### Satvik Pandey

voko ,what do you think about my equations?

9. Nov 9, 2014

### voko

I agree with haruspex.

10. Nov 9, 2014

### Staff: Mentor

Not in the way you write, as the ball is not moving in a straight line.

See haruspex.

11. Nov 9, 2014

### Satvik Pandey

Could you please explain why that is not correct?:p

12. Nov 9, 2014

### voko

Because that assumes that the entire mass of the ball is at its center.

13. Nov 9, 2014

### Jilang

Consider the element of the sphere just touching the corner, it is stationary.

14. Nov 9, 2014

### Satvik Pandey

Then how should I proceed?

15. Nov 9, 2014

### voko

But the rest of the ball is not. How do you justify the insignificance of that?

16. Nov 9, 2014

### Jilang

The sphere is rigid, you need to assume that the force on every single atom is equal.

17. Nov 9, 2014

### voko

Why?

18. Nov 9, 2014

### Jilang

Do you think that the force of gravity only acts on the centre of mass?

19. Nov 9, 2014

### voko

Personally, I would use a co-rotating frame. In that frame, the ball is in equilibrium. The sum of real and fictitious forces is zero.

20. Nov 9, 2014

### Jilang

Does that give a different answer?