Finding Angle of Sphere Falling From Table Edge

[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?

 

[voko]

What should I do?

Think about #53 :)

 

[haruspex]

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.

As has been explained to you, you are neglecting the internal forces of the ball. Yes, the gravitational force acts equally throughout the ball, but intermolecular forces do not. Since the ball undergoes a rotational acceleration, there are internal torques. By your argument, you could replace the rigid ball by a ball of water (except for the bit in contact).

 

[ehild]

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?

The centripetal force is not a real force. It does not "act". It has no sense to "add the centripetal forces " . You have to add the real forces and see if they give the centripetal force needed to a circular motion. But the real forces acting on a particle of the sphere include the internal forces between the particles. You do not know them.

In case of a system of material points, you can write Newton's second equation for each particle of mass m_i, and add the equations together. Each particle of the sphere is subject of internal forces between the particles and there are also external forces: m_ig for each particle and the normal force and friction where the sphere touches the edge.

The internal forces cancel in the sum, and you get that ∑m_ia_i=∑F(external).
But ∑m_ia_i=Ma_com where M is the total mass.
The centre of mass moves as a material point when the resultant of the external forces act on it.

The whole motion of a rigid body can be described as a translational motion of the CoM and a rotation. The angular acceleration of the rotation about an axis is equal to the torque divided by the moment of inertia about the axis.

So you need three equations: two for the translational motion of the CoM (actually, it is an accelerating circular motion about the edge, so you need an expression both for the centripetal acceleration and tangential acceleration), and you need an equation for the rotation.
Keep in mind that the angular velocity of the circular motion is the same as that of the rotation.

 

[Satvik Pandey]

The centripetal force is not a real force. It does not "act". It has no sense to "add the centripetal forces " . You have to add the real forces and see if they give the centripetal force needed to a circular motion. But the real forces acting on a particle of the sphere include the internal forces between the particles. You do not know them.

In case of a system of material points, you can write Newton's second equation for each particle of mass m_i, and add the equations together. Each particle of the sphere is subject of internal forces between the particles and there are also external forces: m_ig for each particle and the normal force and friction where the sphere touches the edge.

The internal forces cancel in the sum, and you get that ∑m_ia_i=∑F(external).
But ∑m_ia_i=Ma_com where M is the total mass.
The centre of mass moves as a material point when the resultant of the external forces act on it.

The whole motion of a rigid body can be described as a translational motion of the CoM and a rotation. The angular acceleration of the rotation about an axis is equal to the torque divided by the moment of inertia about the axis.

So you need three equations: two for the translational motion of the CoM (actually, it is an accelerating circular motion about the edge, so you need an expression both for the centripetal acceleration and tangential acceleration), and you need an equation for the rotation.
Keep in mind that the angular velocity of the circular motion is the same as that of the rotation.

I have made those equations in #post 1. Could you please tell me if they are right or not?

 

[ehild]

I have made those equations in #post 1. Could you please tell me if they are right or not?

They look correct. Go ahead.

 

[voko]

The centripetal force is not a real force. It does not "act".

I find this statement mistaken. There is nothing mystic about centripetal forces, and they are very real. If a particle in a system of particles rotates about a fixed centre, then the resultant of forces acting on it has a centripetal component. It is also valid to sum such forces, exactly like it is valid to sum unknown internal forces, which is the foundation of your argument.

 

[ehild]

The centripetal(radial) component of a force in a polar system of coordinates is not the same concept as the centripetal force.

 

[haruspex]

The centripetal(radial) component of a force in a polar system of coordinates is not the same concept as the centripetal force.

True, but I agree with voko that
(a) the centripetal force required by an object to achieve a motion is a real force, but, as you say, it is a resultant force, not an applied force; and
(b) it makes sense to sum the centripetal forces required by the elements of a rigid body to obtain the centripetal force required by the body as a whole.

 

[voko]

The centripetal(radial) component of a force in a polar system of coordinates is not the same concept as the centripetal force.

This is trivially false, because a component of a force is a force. This is also irrelevant, because the polar system of coordinates was not employed above.

 

[ehild]

True, but I agree with voko that
(a) the centripetal force required by an object to achieve a motion is a real force, but, as you say, it is a resultant force, not an applied force; and
(b) it makes sense to sum the centripetal forces required by the elements of a rigid body to obtain the centripetal force required by the body as a whole.

Ok, you can call it real in the sense that it is not imaginary or mistic, but it is not an applied force.

The motion of a rigid body (at least, in a plane) can be considered as a translational motion of the CM and a rotation about the CM.

The translational motion of the CM is under the effect of the resultant of the external forces. As it is taught in high-schools:

1. The net (external) force on a system of particles equals the mass of the system times the acceleration of the system's center of mass. (Newton's Second Law applied to the center of mass).
2. If the net (external) force on a system of particles is zero, the center of mass of the system will not accelerate (Newton's First Law applied to the center of mass.
3. The system behaves as if all of its mass were located at the system's center of mass.