Finding Angle of Sphere Falling From Table Edge

[Satvik Pandey]

Homework Statement

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.

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

I don't know if these equations are right. Please help.

 

[haruspex]

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

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

 

[Satvik Pandey]

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

Could please explain, why? How to proceed?

 

[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.

 

[Satvik Pandey]

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.

Could you please show how did you find that.

 

[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.

 

[voko]

Can this be solved by just considering the forces at the point of contact tangent to the sphere?

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?

 

[Satvik Pandey]

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?

voko ,what do you think about my equations?

 

[voko]

voko ,what do you think about my equations?

I agree with haruspex.

 

[mfb]

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.

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

See haruspex.

 

[Satvik Pandey]

I agree with haruspex.

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

 

[voko]

Could you please explain why that is not correct?

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

 

[Jilang]

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?

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

 

[Satvik Pandey]

Then how should I proceed?

 

[voko]

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

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

 

[Jilang]

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

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

 

[voko]

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

Why?

 

[Jilang]

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

 

[voko]

Then how should I proceed?

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.

 

[Jilang]

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.

Does that give a different answer?

 

[voko]

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

If that is a question to me, the answer is no.

Regardless, I asked you to justify your statements. Please do so.

 

[Satvik Pandey]

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.

What is 'co-rotating frame'?
Are my other equations correct?

 

[Satvik Pandey]

No, it's ok - just didn't seem right, and I didn't have time to check it before.

voko has just said that it's not correct. What should I do?:(

 

[haruspex]

Could please explain, why? How to proceed?

Calculate it from first principles by integration.

 

[haruspex]

voko has just said that it's not correct. What should I do?:(

No, I mean I hadn't checked ... I have now and it is wrong.

 

[Satvik Pandey]

Consider a uniform bar with its centre as the point of contact. It would require a centripetal force to get around the corner even though its mass centre does not.
Calculate it from first principles by integration.

I don't understand what you want me to calculate. Please explain more.

 

[voko]

Hm, interesting. Unless I made a mistake in integration, total centrifugal force is $m \omega^2 r$, that of a point mass, which means Satvik's disputed normal force equation was correct :)

 

[Satvik Pandey]

Hm, interesting. Unless I made a mistake in integration, total centrifugal force is $m \omega^2 r$, that of a point mass, which means Satvik's disputed normal force equation was correct :)

But was the way in which I found it incorrect?

 

[voko]

But was the way in which I found it incorrect?

As I said earlier, you assumed you could treat the ball as a point mass, and that was not justified.

Have a look here:

http://en.wikipedia.org/wiki/Rotating_reference_frame#Newton.27s_second_law_in_the_two_frames

The sum of the centrifugal force, Euler force, normal force, weight and friction must be zero (the Coriolis force is zero because the ball is stationary with respect to itself). The centrifugal and the Euler force must be integrated to obtain the force balance.

 

[haruspex]

Hm, interesting. Unless I made a mistake in integration, total centrifugal force is $m \omega^2 r$, that of a point mass, which means Satvik's disputed normal force equation was correct :)

I managed to convince myself it could not be right, but now I see a flaw. I'll trust your integration.