Finding Angle of Sphere Falling From Table Edge

[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?

 

[Satvik Pandey]

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

The resultant of y and z component of the forces is zero. right?
Thank you. I got that.
What to do next?

 

[Satvik Pandey]

$\rho \omega ^{ 2 }\iiint { { x }dxdydz }$
or $\rho \omega ^{ 2 }\frac { 4{ r }^{ 2 } }{ 2 } yz$
Is it right?

 

[voko]

Next you need to compute $F_x$. It can be done using more or less the same consideration we used for $F_y$ and $F_z$, except we cannot find a point with coordinates $(-x, -y, -z)$ for some $(x, y, z)$, because $x$ varies for zero to the diameter of the sphere. But, if you have $(x, y, z)$, can you find another point $(x', -y, -z)$ in the ball so that $(x, y, z) + (x', -y, -z) = \mathrm{constant}$?

 

[Satvik Pandey]

Next you need to compute $F_x$. It can be done using more or less the same consideration we used for $F_y$ and $F_z$, except we cannot find a point with coordinates $(-x, -y, -z)$ for some $(x, y, z)$, because $x$ varies for zero to the diameter of the sphere. But, if you have $(x, y, z)$, can you find another point $(x', -y, -z)$ in the ball so that $(x, y, z) + (x', -y, -z) = \mathrm{constant}$?

What do you think about my #post 52?

 

[voko]

What do you think about my #post 52?

That is wrong. The integral is a definite integral taken over the volume of the ball, so there cannot be any variables in the result.

 

[Satvik Pandey]

That is wrong. The integral is a definite integral taken over the volume of the ball, so there cannot be any variables in the result.

Evaluating multiple integral is done by integrating first integral keeping parameters of 2nd and 3rd Integral Constant and then integrate 2nd Integral by keeping parameters of 3rd Integral Constant and then solving 3rd integral. Am I Right ??

 

[voko]

Evaluating multiple integral is done by integrating first integral keeping parameters of 2nd and 3rd Integral Constant and then integrate 2nd Integral by keeping parameters of 3rd Integral Constant and then solving 3rd integral. Am I Right ??

Before you start talking about multiple integrals, you need to convert the volume integral to multiple integrals. That has not been done, so you have no valid argument.

My advice here is to avoid having to evaluate the integral explicitly. We could deduce the value of the y and z components without evaluating their respective integrals, and it is possible for the x component in a similar way. Think about #53.

 

[Satvik Pandey]

Before you start talking about multiple integrals, you need to convert the volume integral to multiple integrals. That has not been done, so you have no valid argument.

My advice here is to avoid having to evaluate the integral explicitly. We could deduce the value of the y and z components without evaluating their respective integrals, and it is possible for the x component in a similar way. Think about #53.

dv=dxdydz.

Is this is not the way to convert the volume integral to multiple integrals?:s

 

[voko]

The volume integral is taken over a particular 3D domain, a ball in this case. Each integral in a multiple integral is taken over a 1D line segment, which may parametrically depend on the other integration variables. There is more than one way to convert a 3D domain into parametrized 1D segments and so convert a volume integral into a triple integral - have you studied this?

 

[Satvik Pandey]

The volume integral is taken over a particular 3D domain, a ball in this case. Each integral in a multiple integral is taken over a 1D line segment, which may parametrically depend on the other integration variables. There is more than one way to convert a 3D domain into parametrized 1D segments and so convert a volume integral into a triple integral - have you studied this?

Sorry but I haven't studied that yet. What should I do?