Finding Angle of Sphere Falling From Table Edge

[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.

 

[Satvik Pandey]

They look correct. Go ahead.

Do you think my method is correct too?

 

[ehild]

Do you think my method is correct too?

Which method, where? You have the necessary equations as far as I saw in the first post, only replace v with rω. Find theta when Fs=μFn.
You know quite much, just go ahead with the solution. Ask if the result is wrong or you have a special question. I would like to see your result.
You do not need to derive Physical theorems or to solve volume integrals. And no co-rotating frame of reference is needed. Use what you have learnt.

 

[Satvik Pandey]

Which method, where? You have the necessary equations as far as I saw in the first post, only replace v with rω. Find theta when Fs=μFn.
You know quite much, just go ahead with the solution. Ask if the result is wrong or you have a special question. I would like to see your result.
You do not need to derive Physical theorems or to solve volume integrals. And no co-rotating frame of reference is needed. Use what you have learnt.

Well I got this
$17cos\theta -4sin\theta =10$
or ${ (17cos\theta -10) }^{ 2 }=16(1-{ cos }^{ 2 }\theta )$
Am I right till here?

 

[voko]

As it is taught in high-schools

I do not think you or anyone else can really say what and how is taught in high schools the world over. I certainly have no idea about India's high school curriculum. Even if that is how it is taught, that was not stated by the original poster, not even when I explicitly said that a justification was required. To me, that is a strong indicator that the equation was obtained by sheer luck, without really understanding the physics behind it, which is a shaky foundation for "going ahead".

That said, you might be right in that the volume integration is not the easiest path to the solution, even if it is valid in principle.

 

[Tanya Sharma]

You do not need to derive Physical theorems or to solve volume integrals. And no co-rotating frame of reference is needed.

+1 on this .

ehild's response has been the most sensible and appropriate one . Considering the fact that OP had got it right in post#1 , it is quite puzzling to me why a simple problem was needlessly made to look like a toughie by others .

 

[Tanya Sharma]

Well I got this
$17cos\theta -4sin\theta =10$

Correct .

 

[Satvik Pandey]

Correct .

I got $\theta ={ cos }^{ -1 }(0.74)=41.82$
and $\theta ={ cos }^{ -1 }(0.36)=68.89$
Is it correct?

 

[ehild]

I got $\theta ={ cos }^{ -1 }(0.74)=41.82$
and $\theta ={ cos }^{ -1 }(0.36)=68.89$
Is it correct?

One is certainly correct. But check if the normal force is positive for both of them.

 

[Tanya Sharma]

I got $\theta ={ cos }^{ -1 }(0.74)=41.82$
and $\theta ={ cos }^{ -1 }(0.36)=68.89$
Is it correct?

One of them is correct . The one which satisfies the equation i have quoted in post#76 is the correct solution.

 

[ehild]

I do not think you or anyone else can really say what and how is taught in high schools the world over.

I am sure that you too, learned it at high school or during the first Physics course at the college:

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.

It is worth remembering.

 

[Satvik Pandey]

One is certainly correct. But check if the normal force is positive for both of them.

I got that positive for 41.82 only.

One of them is correct . The one which satisfies the equation i have quoted in post#76 is the correct solution.

41.82 satisfies that equation too.:)

 

[Satvik Pandey]

That is the correct answer:). Thank you ehild and Tanya for helping me.:D
Thank you voko, haruspex and mfb.

 

[ehild]

Well done! But a solution is really good if you start with explaining the applied concepts. "The relevant equations" is not enough.
You should have started with

" The centre of the sphere performs circular motion about the edge, and at the same time, the sphere rotates .
The in-plane motion of a rigid body is equivalent with the motion of the CoM and rotation about the CoM.
In case of no-slip the angular frequency of the circular motion of the CoM is the same as that of the rotational motion. That ensures that the point of contact has zero velocity.
For the motion of the CoM we can write ma_CoM=∑F(external).
The external forces are gravity, normal force from the edge and force of friction"

 

[voko]

a simple problem was needlessly made to look like a toughie by others

Really? As I said in #29, in the co-rotating frame, the problem is reduced to an equilibrium problem. The force balance is $$ \vec F + \vec N + \vec P + \vec C + \vec E = 0, $$ where the terms are the force of friction, normal force, weight, total centrifugal force, and total Euler force. In the coordinate system suggested in #48, $\vec F$ and $\vec E$ have only the y-components, both directed upward, $\vec N$ and $\vec C$ have only the x-components, both directed rightward, and $\vec P$ has both components, which are $mg \cos \theta$ and $mg \sin \theta$, directed leftward and downward, correspondingly, thus yielding $$ F + E = mg \sin \theta \\ N + C = mg \cos \theta . $$ The centrifugal and Euler forces are trivially computed, which gives $$ fN + m \dot \omega r = mg \sin \theta \\ N + m \omega^2 r = mg \cos \theta . $$ How is that tough, exactly?

 

[Satvik Pandey]

Well done! But a solution is really good if you start with explaining the applied concepts. "The relevant equations" is not enough.
You should have started with

" The centre of the sphere performs circular motion about the edge, and at the same time, the sphere rotates .
The in-plane motion of a rigid body is equivalent with the motion of the CoM and rotation about the CoM.
In case of no-slip the angular frequency of the circular motion of the CoM is the same as that of the rotational motion. That ensures that the point of contact has zero velocity.
For the motion of the CoM we can write ma_CoM=∑F(external).
The external forces are gravity, normal force from the edge and force of friction"

Thank you ehild :). I will keep that in mind before starting new thread.

 

[Satvik Pandey]

So the ball first rotate about the contact point then it slips and after it looses contact with the cliff. right?
I was wondering at what angle the ball will loose contact with the cliff.
No normal force will act on the ball at the moment it lost contact with the cliff.

So $\frac { m{ v }^{ 2 } }{ r } =mgcos\theta$
I think we can not use energy conservation to find other equation because when the sphere begins to slip on the cliff the force of friction begins to do work on it. right?
So, what should I do?

 

[ehild]

You have the equations for the motion of the CoM, but with kinetic friction this time. The coefficient of the kinetic friction is not known, but It can be the same as the coefficient of static friction. Eliminate the normal force and solve the equation for v as function of theta with the initial condition for theta and v at the instant when the slipping starts.

 

[Satvik Pandey]

So I have 2 equations-

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

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

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

or and $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=ma$

Here we can not use $a=r \alpha$

So $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{dt}$

or $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{d \theta} \omega$

But $\omega=\frac { v }{ r }$

So $mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r } )=m\frac { vdv }{ rd\theta }$

Let ${ v }^{ 2 }=a\\ So 2vdv=da$ ( here $a$ is not acceleration)

$gsin\theta -\mu (gcos\theta -\frac { a }{ r } )=\frac { da }{ 2rd\theta }$

$\frac { da }{ 2rd\theta } =gsin\theta -\mu gcos\theta +\mu \frac { a }{ r }$

$\frac { da }{ d\theta } -2\mu a+2rg(\mu cos\theta -sin\theta )=0$

I think this is linear differential equation. Should I solve this with integrating factor method?
Am I correct till here?

 

[ehild]

Yes, you can apply the integrating factor method at the end.
Explain your method first, and define the variables. What are v, a, and alpha? What is omega? What do you mean that you can not use a=rα? And do not use the symbol a for two different things (acceleration of the CoM and v².

 

[Satvik Pandey]

Yes, you can apply the integrating factor method at the end.
Explain your method first, and define the variables. What are v, a, and alpha? What is omega? What do you mean that you can not use a=rα? And do not use the symbol a for two different things (acceleration of the CoM and v².

Sorry ehild, I am getting late for the school:p. I will do that after coming from the school.:)

 

[voko]

But $\omega=\frac { v }{ r }$

Is that true when the ball is slipping?

 

[Satvik Pandey]

Yes, you can apply the integrating factor method at the end.
Explain your method first, and define the variables. What are v, a, and alpha? What is omega? What do you mean that you can not use a=rα? And do not use the symbol a for two different things (acceleration of the CoM and v².

Let $\theta$ be the angle at which the ball is slipping and rotating together.
The CoM of the sphere performs circular motion about the edge. right?
Let the tangential velocity and the acceleration of the CoM at that particular angle $\theta$ be $v$ and $a$ respectively.
Let $\omega$ be the angular velocity of the CoM of the ball about the edge.

What do you mean that you can not use a=rα? And do not use the symbol a for two different things (acceleration of the CoM and v².

I meant $\alpha$ to be the angular acceleration of the that ball about the CoM. We can not use this formula as the ball is slipping.

Sorry for using a for two different things. I am going to change it.

Is that true when the ball is slipping?

Here v is the tangential velocity of the CoM at angle $\theta$ and $\omega$ is the angular velocity of the CoM of the ball and the ball is doing circular motion about the edge. So can we not use this equation $v=r \omega$?

 

[Satvik Pandey]

So I have 2 equations-

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

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

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

or and $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=ma$

Here we can not use $a=r \alpha$

So $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{dt}$

or $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{d \theta} \omega$

But $\omega=\frac { v }{ r }$

So $mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r } )=m\frac { vdv }{ rd\theta }$

Let ${ v }^{ 2 }=x\\ So 2vdv=dx$

$gsin\theta -\mu (gcos\theta -\frac { x }{ r } )=\frac { dx }{ 2rd\theta }$

$\frac { dx }{ 2rd\theta } =gsin\theta -\mu gcos\theta +\mu \frac { x }{ r }$

$\frac { dx }{ d\theta } -2\mu x+2rg(\mu cos\theta -sin\theta )=0$

I think this is linear differential equation. Should I solve this with integrating factor method?
Am I correct till here?

I have now edited it. Please see the quoted text.

 

[voko]

Ah, so $\omega$ is the angular velocity of the CoM, not of the ball itself. Disregard #91 in that case.

 

[Satvik Pandey]

Ah, so $\omega$ is the angular velocity of the CoM, not of the ball itself. Disregard #91 in that case.

Is differential equation that I have derived correct?

 

[voko]

I do not see any obvious mistake.

 

[ehild]

Let $\theta$ be the angle at which the ball is slipping and rotating together.

Theta is the angle of the radius, connecting the centre of the sphere with the edge, with respect to the vertical.
As the sphere touches the edge, the centre moves along a circle, but that circular motion is decoupled from the rotation of the sphere.

The CoM of the sphere performs circular motion about the edge. right?
Let the tangential velocity and the acceleration of the CoM at that particular angle $\theta$ be $v$ and $a$ respectively.
Let $\omega$ be the angular velocity of the CoM of the ball about the edge.

that is correct.

I meant $\alpha$ to be the angular acceleration of the that ball about the CoM. We can not use this formula as the ball is slipping.

You can totally disregard the rotation of the ball about its centre because of the slipping. The slipping ensures that the force of friction is kinetic, wirth defined magnitude Fk=μN.
But you can define angular variables for the circular motion of the centre. The angular velocity of the circular motion is ω=v/r, and the angular acceleration is α=a/r.

 

[ehild]

I have now edited it. Please see the quoted text.

The editing looks terrible, :Dbut the equation is OK.

 

[Satvik Pandey]

The editing looks terrible, :Dbut the equation is OK.

It is very lengthy to solve that equation with Integrating Factor method.
I hope that I solve it soon.

 

[ehild]

There is an other method to solve the linear inhomogeneous differential equation x'-2μx=f(θ).
Take the general solution of the homogeneous part: it is x=Ce^2μθ.
Add a particular solution X_p of the inhomogeneous equation: the general solution will be x=Ce^2μθ+X_p.
In case the inhomogeneous part contains sine and cosine functions, you can seek the solution in the form Xp=Acos(θ)+Bsin(θ). Replacing X_p =Acos(θ)+Bsin(θ) into the equation, you can find the constants A and B.