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.

 

[Satvik Pandey]

Can this be solved without using rotating frame of reference because I haven't became familiar with it.

 

[mfb]

Yes. And there is no large difference between a fixed and a rotating frame, the equations look similar.
Which net force is needed to keep the ball on its circular track?

 

[voko]

Yes, this can be solved in an inertial frame, but it is slightly more difficult in my opinion. Compose the equations of motion for each "small particle" within the ball. Every such particle moves in a circle with known velocity and acceleration, so the net force acting on every particle is also known. The integral of the net force over the entire ball must be equal to the sum of weight, normal force and friction.

 

[Satvik Pandey]

Yes. And there is no large difference between a fixed and a rotating frame, the equations look similar.
Which net force is needed to keep the ball on its circular track?

Centripetal force.

 

[mfb]

Right. Now you'll need its value.

 

[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?

If you balance an object at a point, the full force of its weight applies at that point. Are you over-thinking this?

 

[voko]

If you balance

If you balance.

 

[Jilang]

I understand what you are saying, if there is acceleration you cannot balance the forces. But at the contact point nothing is accelerating, until it slips.

 

[voko]

But at the contact point nothing is accelerating, until it slips.

You cannot nullify the ball rotating at the contact point by looking only at the contact point. The ball is still there, and its rotation may, to say the least, affect the forces acting at the contact point. And you cannot just neglect that without any justification.

 

[Jilang]

I am not neglecting the forces. Since I am the only one who has given an answer to this question you should give me the respect to show the proper answer and the error or my ways or just shut up.

 

[voko]

you should give me the respect to show the proper answer and the error

Giving complete answers is against the rules of this forum.

Your error was explained to you.

just shut up

Being rude won't help you with physics, Jilang.

 

[Jilang]

Giving complete answers is against the rules of this forum.

Your error was explained to you.
Being rude won't help you with physics, Jilang.

With respect, I don't think the OP is any the wiser from your posturing. Nothing that has been said has convinced me that this is the wrong way so far approach the problem. But I am eager to learn the correct way. So get your fingers out!

 

[voko]

Nothing that has been said has convinced me that this is the wrong way so far approach the problem.

Your approach treats the system as if it were static. You need to justify that, and I do not think this is justifiable.

But I am eager to learn the correct way.

I outlined what I believe to be correct approaches earlier, in posts #29 and #33. If something is not clear there, let me know.

 

[Jilang]

Thank you, much appreciated, but having reviewed the link, I am non the wiser. Perhaps this problem is just too difficult for the average monkey?

 

[Satvik Pandey]

Right. Now you'll need its value.

So I have to calculate centripetal force acting on every "small particle'" inside the sphere and then I have to integrate it for whole sphere. right?

I have proceeded
Centripetal force=$m \omega^{2} r$

Let Density of sphere be $\rho$

So $dm =\rho dv$
Let this mass $dm$ rotate in the circle of radius of $x$

So Centripetal force =$\int{\rho dv \omega^{2} x}$

Don't know how to proceed ahead. Am I right till here?

 

[voko]

So I have to calculate centripetal force acting on every "small particle'" inside the sphere and then I have to integrate it for whole sphere. right?

Right.

I have proceeded
Centripetal force= $mω2rm \omega^{2} r$

Not exactly. The force is a vector. The direction of this vector changes as you go from one "small particle" to another.

Hint: let $\vec r$ be the vector from the contact point to the small particle. Can you express the force in terms of this vector?

 

[Satvik Pandey]

$\overrightarrow { r } =x\hat { i } +y\hat { j } +z\hat { k }$

and $dv=dxdydz$

$F_{x} = \int{\rho dxdydz \omega^{2} x}$

$F_{y} = \int{\rho dxdydz \omega^{2} y}$

$F_{z} = \int{\rho dxdydz \omega^{2} z}$

I haven't done anything like this before. I just tried this.

 

[voko]

Correct, except you must have the minus sign in front of each integral - can you explain why?

To simply things further, let's select some convenient coordinates. Let the origin of the coordinate system be at the contact point, and the Ox axis pass through the centre of the ball. Then it should be fairly obvious that $F_y = F_z = 0$ - is that obvious to you?

 

[Satvik Pandey]

Correct, except you must have the minus sign in front of each integral - can you explain why?

Centripetal force acts towards center. And in this case center(point of contact) is also the origin. So it Centripetal force acts towards origin. So it should be (-)ve?

To simply things further, let's select some convenient coordinates. Let the origin of the coordinate system be at the contact point, and the Ox axis pass through the centre of the ball. Then it should be fairly obvious that $F_y = F_z = 0$ - is that obvious to you?

Let blue and green line be the x and y axis. Suppose there is a mass which experience centripetal force along that blue line. That force can be broke in x and y components. Why is $F_y = F_z = 0$?

 

[voko]

That force can be broke in x and y components.

Correct. But this is the force at a single small particle.

Why is $Fy=Fz=0F_y = F_z = 0$?

These are the components of the integral force, not the components of the force at a single small particle, so your argument does not apply.

Consider two small particles with coordinates (x, y, z) and (x, -y, -z). What is the resultant of their centripetal forces?