Rolling without slipping on a curved surface

[NTesla]

Homework Statement

A solid spherical ball of mass m and radius r rolls without slipping on a rough concave surface of
large radius R. Find the magnitude and direction of friction on the ball.

Relevant Equations

$$ \tau = I\alpha $$, F = ma

Kindly see the attached pdf. My attempt to solve it, is in it.

I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction.

I'm not able to figure out, why my solution is wrong, if it is wrong .

 

[haruspex]

the ball is undergoing angular simple harmonic motion

No, that is only an approximation for small displacements. But you don’t use it in what you posted, and the question as you have posted it doesn’t ask about period. I presume there are more parts to the question.

the acceleration of point of the ball which is instantaneously in contact with the surface, must be zero.

No, its instantaneous velocity is zero, but its acceleration will be nonzero and normal to the surface, so its tangential component is zero.

Your answer looks correct.

 

[kuruman]

Your last sentence is
"The minus sign here means that the direction of friction is up the incline."
Is it always up the incline? You need to specify the direction over an entire cycle of the motion.

 

[haruspex]

Your last sentence is
"The minus sign here means that the direction of friction is up the incline."
Is it always up the incline? You need to specify the direction over an entire cycle of the motion.

The OP's analysis appears to be independent of the point in the cycle (other than assuming a nonzero magnitude for the frictional force).

 

[NTesla]

No, its instantaneous velocity is zero, but its acceleration will be nonzero and normal to the surface, so its tangential component is zero.

Yes, that's right i think. Thanks for pointing that out.

Your answer looks correct.

I dont think it is correct. The reason is that when I'm asking AI the same question, it is giving me another answer where the magnitude of $f = 2mgsinθ/3$. And it shows the calculation too. I can't find fault in the calculation shown by the AI. But I also can't find problem with my calculation. However, both can't be right. Here's the calculation shown by the AI:

 

[NTesla]

Is it always up the incline? You need to specify the direction over an entire cycle of the motion.

More than the direction, I'm worried about the magnitude of the frictional force. Kindly see post#5 above, wherein I've shown the calculation done by AI, wherein the magnitude is not $2/7mgsinθ$ but $2/3mgsinθ$. Kindly let me know which calculation is wrong and why: mine or that of AI.

About the direction of friction in a whole cycle: For all those moment when the ball is not at point C1 (the bottom most point), the friction direction will be up the incline. At the bottom most point, I suppose the friction will be zero. so the direction of friction will be indeterminate at the bottom most point.

 

[haruspex]

Yes, that's right i think. Thanks for pointing that out.

I dont think it is correct. The reason is that when I'm asking AI the same question, it is giving me another answer where the magnitude of $f = 2mgsinθ/3$. And it shows the calculation too. I can't find fault in the calculation shown by the AI. But I also can't find problem with my calculation. However, both can't be right. Here's the calculation shown by the AI:

View attachment 366163

I have confirmed your result independently. Will try to locate the error in the AI solution, but rightnowI am a passenger on a very bumpy road!

 

[NTesla]

I have confirmed your result independently. Will try to locate the error in the AI solution, but rightnowI am a passenger on a very bumpy road!

Appreciate it very much, you taking time to help out. Will definitely wait for your answer.

P.S: Hope your journey is worth the destination.

 

[kuruman]

At the bottom most point, I suppose the friction will be zero. so the direction of friction will be indeterminate at the bottom most point.

"Indeterminate" is a word that a mathematician might use for a force vector that is zero. A physicist would say that a zero force is no force at all and does not exist. Something that does not exist cannot have a direction.

I will also look into this AI discrepancy and report later.

 

[Steve4Physics]

@NTesla, I (very nervously!) suggest that the AI answer is correct.

Friction here, apart from the lowest point, always acts uphill. This means that in your Post #1 pdf attachment:

a) the 2nd diagram is wrong as it shows friction acting downhill;

b) your expression for the acceleration of the ball’s centre of mass:
$a_{cm} = \frac {mg \sin \theta + f}{m}$
should be:
$a_{cm} = \frac {mg \sin \theta – f}{m}$
which, I think, will make your answer the same as the AI.

Minor edit.

 

[NTesla]

"Indeterminate" is a word that a mathematician might use for a force vector that is zero. A physicist would say that a zero force is no force at all and does not exist. Something that does not exist cannot have a direction.

Well said. Agreed. Though I'm not sure I'm a mathematician or a physicist. I'm just someone trying to figure things out one at a time.

I will also look into this AI discrepancy and report later.

Your contribution is very much appreciated. Will surely wait for your reply.

 

[NTesla]

the 2nd diagram is wrong as it shows friction acting downhill;

No, I don't think that's wrong. One can show any direction of frictional force(either up or down the incline) and the calculation will prove which direction friction will take. In my calculation, f has come out negative. And in the pic I've shown f down the incline. This only means that the friction is acting up the incline at all the points on the concave surface, except at the bottom most point, at which the frictional force is zero.

which, I think, will make your answer the same as the AI.

No, it wont. Even if we take your equation for $a_{cm}$, then we will have to take positive value for the torque due to friction. And we will again end up with the same magnitude of frictional force (= $2/7 mgsinθ$). Only, this time, the frictional force will come out positive in the calculation. So, we again reach the same value of frictional force.

 

[haruspex]

Ok, here’s what is wrong with the AI solution.

First, "the CM acceleration is $(R-r)\ddot\theta$, not $a$” is wrong in that it is both.
Indeed, for this part of the problem it is unnecessary to use the $(R-r)\ddot\theta$ form. We can just change all occurrences of that in the solution to “a” (having defined positive as up the slope).

Next, it fails to specify which sense is being taken as positive for $\alpha$. From its "Rolling" equation, we can infer that $\alpha$ is positive for rolling accelerating [correction to original post] up the slope, but in that case the upslope frictional force f exerts a negative torque: $I\alpha=-fr$.
Correcting that leads to your answer.

 

[kuruman]

I do not understand the rolling (no-slip) AI equation $$\alpha=\frac{(R-r)\ddot {\theta}}{r}.$$ Here is my reasoning.

The center of the shell, the center of the ball and the point of contact are collinear at all times.
When the ball rolls without slipping on the shell and rotates by angle $\Delta \varphi$, the point of contact travels by an arc length $\Delta s = r~ \Delta \varphi$ on the surface (see figure on the right).
This arc length is related to the angular displacement of the CM by $\Delta s =R~ \Delta \theta.$
It follows that $$ r~ \Delta \phi=R~ \Delta \theta \implies \Delta \varphi=\frac{R}{r}\Delta \theta.$$ Now if the definition of $\alpha$ is the angular acceleration about the CM, i.e. $\ddot {\varphi}~$, then it follows that $$\alpha=\ddot {\varphi}=\frac{R~\ddot{ \theta}}{r}.$$Did I miss something?
On edit
Yes. See post #19.

 

[NTesla]

I do not understand the rolling (no-slip) AI equation α=(R−r)θ¨r. Here is my reasoning.

Here's how I think that equation has been derived. $\ddot{\theta }$ is the angular acceleration about the point C (centre of curvature of the concave surface). Now, $a_{cm}$ = $\ddot{\theta }(R - r)$. And since the ball is rolling without slipping, $\Rightarrow$ $a_{cm} = \alpha _{cm}r$.
$$\therefore \alpha _{cm} = \frac{\left ( R - r \right)}{r}\ddot{\theta }$$
I can't say, there's any fault in this calculation. Seems right to me. Kindly let me know what's wrong with these steps.

However, I again cannot find fault in your calculation in post#14. That seems right too.
What is happening with this question !!! The more i try to untangle it, the more tangled it is becoming.

 

[haruspex]

I can't say, there's any fault in this calculation. Seems right to me. Kindly let me know what's wrong with these steps.

Did you see post #13?

 

[NTesla]

First, "the CM acceleration is (R−r)θ¨, not a” is wrong in that it is both.
Indeed, for this part of the problem it is unnecessary to use the (R−r)θ¨ form. We can just change all occurrences of that in the solution to “a” (having defined positive as up the slope).

Agreed.

Next, it fails to specify which sense is being taken as positive for α. From its "Rolling" equation, we can infer that α is positive for rolling up the slope, but in that case the upslope frictional force f exerts a negative torque: Iα=−fr.

You've mentioned that: we can infer that $\alpha$ is positive for rolling up the slope. I don't think I understand this line.
My argument is: Since the ball must stop at some point of time, after having climbed up the concave surface and then return back. While climbing up, it's $\alpha_{com}$, must be positive in anticlockwise direction, and $\alpha_{C}$ must be positive in clockwise direction. Isn't this right. ?

 

[NTesla]

Did you see post #13?

yes. kindly see post#17 above.

 

[haruspex]

When the ball rolls without slipping on the shell and rotates by angle $\Delta \varphi$, the point of contact travels by an arc length $\Delta s = r~ \Delta \varphi$ on the surface (see figure on the right).

No, $\Delta s$ is greater than that because of the curved surface.
Consider e.g. R only marginally greater than r.

 

[haruspex]

You've mentioned that: we can infer that $\alpha$ is positive for rolling up the slope. I don't think I understand this line.

My wording was sloppy; I should have said "for accelerating up the slope".
AI's rolling equation is $\alpha r=(R-r)\ddot\theta$.
If it is accelerating up the slope then the RHS is positive, so in that condition the LHS must be positive too.

 

[kuruman]

No, $\Delta s$ is greater than that because of the curved surface.
Consider e.g. R only marginally greater than r.

Of course. I knew I missed something.

 

[NTesla]

My wording was sloppy; I should have said "for accelerating up the slope".
AI's rolling equation is $\alpha r=(R-r)\ddot\theta$.
If it is accelerating up the slope then the RHS is positive, so in that condition the LHS must be positive too.

I still don't fully understand what you meant to say in the last line here.

My opinion is: $\alpha_{cm}$ and $\alpha_{about C}$ must have opposite signs all throughout the motion. Meaning that, $\alpha r= - (R-r)\ddot\theta$.

 

[kuruman]

I still don't fully understand what you meant to say in the last line here.

My opinion is: $\alpha_{cm}$, and $\alpha_{about C}$ must have opposite signs all throughout the motion. Meaning that, $\alpha r= - (R-r)\ddot\theta$.

Let's change the symbols a bit to make things clear. You can view the motion as a pendulum that swings in a circular arc of length $R$ (orbital motion about the center of the circle) while at the same time the bob spins about its center. Let's call the orbital angular acceleration $\ddot{\theta}$ and the spin angular acceleration $\ddot{\varphi}$. Let's also define "out of the screen" as positive and "into the screen" as negative.

Now, let's consider the ball rolling downhill as in your drawing (right). Both its orbital and spin angular speeds, $\dot {\theta}$ and $\dot {\varphi}$ are increasing. This means that the angular accelerations have the same sign as their respective angular velocities.

Using the right hand rule:
1. The orbital angular velocity is positive which makes $\ddot{\theta}$ positive.
2. The spin angular velocity is negative which makes $\ddot{\varphi}$ negative.
Therefore, $\ddot{\theta}$ and $\ddot{\varphi}$ have opposite signs.

This is true throughout the motion because the instantaneous point of contact and the center of the shell about which the directions of the spin and orbital angular velocities are respectively determined are always collinear and on opposite sides of the ball's center.

I agree with your opinion if your $\alpha$ is the same as my $\ddot {\varphi}.$

 

[NTesla]

α is the same as my

Yes, they are as you've written about.

Proceeding on the original question to find the time period, my calculation is not giving the answer that is in the book. However, the answer given in the book is correct since it can be reached by another method too. However, using my method, the answer is not what's given in the book. Here's my method to calculate the time period. Kindly take a look at the attached pdf and let me know, if I'm going wrong anywhere.

 

[haruspex]

My opinion is: $\alpha_{cm}$ and $\alpha_{about C}$ must have opposite signs all throughout the motion.

Yes, if you take the same sense as positive for both, but no law says you have to. If $\theta$ is the angle clockwise from vertical, as in your diagram, then $\ddot\theta$ must be positive clockwise, but AI could have taken anticlockwise as positive for $\alpha$. With those choices, AI's rolling equation would be correct but its torque equation wrong.
Whichever way you cut it, its rolling and torque equations are inconsistent.

 

[haruspex]

Yes, they are as you've written about.

Proceeding on the original question to find the time period, my calculation is not giving the answer that is in the book. However, the answer given in the book is correct since it can be reached by another method too. However, using my method, the answer is not what's given in the book. Here's my method to calculate the time period. Kindly take a look at the attached pdf and let me know, if I'm going wrong anywhere.

Your mistake is in applying the parallel axis theorem. That is only valid for a rigid body rotating as a unit about the axis. It would be correct if the sphere were a pendulum bob attached rigidly to a rod. In the present case, the motion of the sphere about its centre and the motion of that centre about the axis are rather different.

 

[NTesla]

That is only valid for a rigid body rotating as a unit about the axis.

Is the moment of Inertia dependent upon whether a body is actually rotating about some axis or not. Can a body not have moment of inertia about some axis, even when it is lying still ?

Also,

In the present case, the motion of the sphere about its centre and the motion of that centre about the axis are rather different.

How do i go about calculating the moment of inertia about the axis passing through the point C in the given question.

 

[haruspex]

Can a body not have moment of inertia about some axis, even when it is lying still ?

Of course, but it is only of interest in the context of rotation about some axis.

How do i go about calculating the moment of inertia about the axis passing through the point C in the given question.

It doesn’t have 'a' moment of inertia in that sense.
We can consider the motion of the sphere as the sum of the motion of its centre around C and of the sphere as a whole around its centre. Each of these has an angular momentum about C.
The former has angular momentum $m(R-r)^2\dot\theta$ clockwise around C, while the latter has angular momentum $I_{cm}\omega$ clockwise around every point, including C, where $\alpha_{cm}=\dot\omega$.
We can sum these and differentiate in order to write the torque equation.

 

[kuruman]

Your mistake is in applying the parallel axis theorem. That is only valid for a rigid body rotating as a unit about the axis. It would be correct if the sphere were a pendulum bob attached rigidly to a rod. In the present case, the motion of the sphere about its centre and the motion of that centre about the axis are rather different.

Just to clarify this point. The parallel axis theorem can still be applied here to find the moment of inertia about the instantaneous point of contact.

Is the moment of Inertia dependent upon whether a body is actually rotating about some axis or not. Can a body not have moment of inertia about some axis, even when it is lying still ?

The moment of inertia is a geometric quantity that depends only on how the mass is distributed about an axis. The body does not need to be rotating to have a moment of inertia just like a body does not need to be moving to have a mass.

 

[haruspex]

The parallel axis theorem can still be applied here to find the moment of inertia about the instantaneous point of contact.

Yes, because in that view all parts of the sphere are rotating about the same axis as a unit. But I think you do have to be careful taking angular acceleration as being about the point of contact on a curved surface.

 

[NTesla]

The parallel axis theorem can still be applied here to find the moment of inertia about the instantaneous point of contact.

And I'm applying the parallel axis theorem. But apparently, in it's application something other than what I've done in the pdf file, is to be done (considering the comment of @haruspex). Doesn't the parallel axis theorem just say to add m(distance to other axis)^2 to the moment of Inertia about the central diameter ?

And why would we need to calculate moment of Inertia about instantaneous point of contact, when we need to find angular frequency about axis passing through point C.

 

[NTesla]

angular acceleration as being about the point of contact on a curved surface.

Since we are taking torque about axis passing point C, and moment of Inertia is only the distribution of mass about axis passing through point C, considering the ball is isotropic, why would we need angular acceleration about point of contact.?

 

[haruspex]

And I'm applying the parallel axis theorem. But apparently, in it's application something other than what I've done in the pdf file, is to be done (considering the comment of @haruspex). Doesn't the parallel axis theorem just say to add m(distance to other axis)^2 to the moment of Inertia about the central diameter ?

The parallel axis theorem applies to a rigid body rotating, as a whole, about a given axis. In this problem, the sphere is not doing that. Its centre is rotating about C but the rest of it isn't.
Consider a pendulum bob suspended from a rod length R, but with the bob mounted on an axle at the bottom of the rod, so that as the bob oscillates the top of the bob is always the top. Now the moment of inertia of the bob about the pivot at the top of the rod is just $mR^2$.

 

[haruspex]

Since we are taking torque about axis passing point C, and moment of Inertia is only the distribution of mass about axis passing through point C, considering the ball is isotropic, why would we need angular acceleration about point of contact.?

Your approach was to take moments about C, but as I have shown that is awkward because you cannot apply the parallel axis theorem. @kuruman is suggesting instead considering the rotational acceleration about the point of contact. This is simpler because the sphere is rotating as a rigid body about that point.
(I am a bit nervous about that because I have erred in the past by considering acceleration about a point of contact when the contacting surface was curved, but it looks ok in this case. Wish I could remember what the problem was.)

 

[NTesla]

The parallel axis theorem applies to a rigid body rotating, as a whole, about a given axis. In this problem, the sphere is not doing that. Its centre is rotating about C but the rest of it isn't.
Consider a pendulum bob suspended from a rod length R, but with the bob mounted on an axle at the bottom of the rod, so that as the bob oscillates the top of the bob is always the top. Now the moment of inertia of the bob about the pivot at the top of the rod is just $mR^2$.

This I do understand.
But let's say, we have a solid sphere which is just lying at the bottom most point of the concave surface. Not rotating about any axis whatsoever. Now, it's moment of Inertia about an axis passing through point C(Centre of curvature of the concave surface), would be: $2/5mr^{2} +m(R-r)^2$. Is that right.?

If that's right, and the solid ball is isotropic, then, will the distribution of mass change about the axis passing through C, just because later on the sphere could be rotating about its own axis and also rotating about the axis through C ?

 

[haruspex]

But let's say, we have a solid sphere which is just lying at the bottom most point of the concave surface. Not rotating about any axis whatsoever. Now, it's moment of Inertia about an axis passing through point C(Centre of curvature of the concave surface), would be: $mr^{2} +m(R-r)^2$. Is that right.?

Yes, but when you say "about axis C" what you mean is that to accelerate it angularly about C at rate $\alpha$ as a rigid body (that is, every particle of the sphere has angular acceleration $\alpha$ about C) then the torque you would need to apply about C is : $(mr^{2} +m(R-r)^2)\alpha$.
That is not the way you have used it.

 

[NTesla]

Yes, but when you say "about axis C" what you mean is that to accelerate it angularly about C at rate $\alpha$ as a rigid body (that is, every particle of the sphere has angular acceleration $\alpha$ about C) then the torque you would need to apply about C is : $(mr^{2} +m(R-r)^2)\alpha$.
That is not the way you have used it.

Will the distribution of mass change just because the ball is rotating ?

 

[kuruman]

(I am a bit nervous about that because I have erred in the past by considering acceleration about a point of contact when the contacting surface was curved, but it looks ok in this case. Wish I could remember what the problem was.)

I think it might be the case where the ball is rolling inside a shell which is free to roll on a horizontal surface. Here, the instantaneous point of contact is an inertial frame.

 

[NTesla]

Will the distribution of mass change just because the ball is rotating ?

@haruspex, @kuruman
Your opinion on this.

 

[haruspex]

@haruspex, @kuruman
Your opinion on this.

We have to assume the sphere has uniform density, so the mass distribution does not change.

 

[NTesla]

We have to assume the sphere has uniform density, so the mass distribution does not change.

So, if the mass distribution doesn't change just because the ball is rotating, this means the moment of inertia should not change just because the ball is rotating. If so, why wouldn't the moment of inertia of a ball rotating about its own axis and about an axis passing through C, change from that calculated when it wasn't rotating at all.?

 

[kuruman]

why wouldn't the moment of inertia of a ball rotating about its own axis and about an axis passing through C,

I don't understand what you are asking here. Are these two different axes? Please provide a drawing of the ball showing the the two axes.

Look, the definition of the moment of inertia is $$I=\int dm~r^2.$$ It is a step by step procedure for finding the moment of inertia about an axis.
1. Choose an axis about which you are going to calculate the moment of inertia.
2. Reset to zero the memory of your calculator.
3. Subdivide the mass of the object into many infinitesimal mass elements $dm.$
4. Choose an element $dm.$
5. Find its distance $r$ from the axis and square it.
6. Multiply $dm$ by $r^2$ and add the result to the memory of your calculator.
7. Go back to step 4 and repeat.
8. When you run out of elements $dm$, recall what's in the memory of your calculator. The moment of inertia about your chosen axis will be displayed.

As you can see, there is no mention about rotation anywhere in this procedure.

 

[NTesla]

I don't understand what you are asking here. Are these two different axes? Please provide a drawing of the ball showing the the two axes.

my bad. I mistakenly wrote "wouldn't", when it should have been "would".

The correct statement is: If the mass distribution doesn't change just because the ball is rotating, this means the moment of inertia should not change just because the ball is rotating. If so, why would the moment of inertia of a ball rotating about its own axis and also simultaneously about an axis passing through C, change from that calculated when it wasn't rotating at all.?

This is what I mean: In this pic(Pic 1 below), when the ball is just lying at the bottom most point on the concave surface and is not rotating at all, the moment of inertia of the ball about the axis passing through point C, and perpendicular to the page, is = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$. I suppose you'll agree up to this point. In this situation, perpendicular axis theorem for calculation of moment of inertia has been correctly applied. I suppose you'll agree to this too.

However, now take a look at this pic(Pic 2 below), when the ball is rotating about an axis passing through it's own centre of mass, i.e. point B and perpendicular to the page, and also simultaneosuly rotating about an axis passing through the point C(centre of curvature of concave surface) and is perpendicular to the page.

According to @haruspex's comment in post#26:

Your mistake is in applying the parallel axis theorem. That is only valid for a rigid body rotating as a unit about the axis. It would be correct if the sphere were a pendulum bob attached rigidly to a rod. In the present case, the motion of the sphere about its centre and the motion of that centre about the axis are rather different.

and in post#33

The parallel axis theorem applies to a rigid body rotating, as a whole, about a given axis.

But in your post#42, you've clearly mentioned that:

As you can see, there is no mention about rotation anywhere in this procedure.

meaning that calculation of moment of inertia is NOT dependent on whether the body is rotating or not.

So, what I gather from these posts is that @haruspex is saying is that the 2 conditions(shown in the 2 pics above) are different even regarding the calculation of moment of Inertia. According to him, the moment of Inertia in the 2nd case(when the ball is rotating about 2 axes simultaneously), is NOT = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$. This is what I don't understand, since in post#40, he does agree that mass distribution doesn't change just because the ball is now rotating. And moment of inertia is just a measure of how the mass is distributed about an axis. So my contention is that in this 2nd case too, the moment of Inertia of the ball about the axis passing through point C and perpendicular to the page should be = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$.

Look, the definition of the moment of inertia is I=∫dm r2. It is a step by step procedure for finding the moment of inertia about an axis.

Appreciate the point-wise refresher course on how to calculate moment of inertia. However, I already know that and that's not what I was asking for.

I want to know what will be the moment of Inertia of the ball about the axis passing through the point C and perpendicular to the page, in the case when the ball is rotating about 2 axes simultaneously(as shown in Pic 2). In my opinion, it should be = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$.

 

[haruspex]

my bad. I mistakenly wrote "wouldn't", when it should have been "would".

The correct statement is: If the mass distribution doesn't change just because the ball is rotating, this means the moment of inertia should not change just because the ball is rotating. If so, why would the moment of inertia of a ball rotating about its own axis and also simultaneously about an axis passing through C, change from that calculated when it wasn't rotating at all.?

This is what I mean: In this pic(Pic 1 below), when the ball is just lying at the bottom most point on the concave surface and is not rotating at all, the moment of inertia of the ball about the axis passing through point C, and perpendicular to the page, is = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$. I suppose you'll agree up to this point. In this situation, perpendicular axis theorem for calculation of moment of inertia has been correctly applied. I suppose you'll agree to this too.

View attachment 366237

However, now take a look at this pic(Pic 2 below), when the ball is rotating about an axis passing through it's own centre of mass, i.e. point B and perpendicular to the page, and also simultaneosuly rotating about an axis passing through the point C(centre of curvature of concave surface) and is perpendicular to the page.View attachment 366238

According to @haruspex's comment in post#26:

and in post#33

But in your post#42, you've clearly mentioned that:

meaning that calculation of moment of inertia is NOT dependent on whether the body is rotating or not.

So, what I gather from these posts is that @haruspex is saying is that the 2 conditions(shown in the 2 pics above) are different even regarding the calculation of moment of Inertia. According to him, the moment of Inertia in the 2nd case(when the ball is rotating about 2 axes simultaneously), is NOT = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$. This is what I don't understand, since in post#40, he does agree that mass distribution doesn't change just because the ball is now rotating. And moment of inertia is just a measure of how the mass is distributed about an axis. So my contention is that in this 2nd case too, the moment of Inertia of the ball about the axis passing through point C and perpendicular to the page should be = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$.

Appreciate the point-wise refresher course on how to calculate moment of inertia. However, I already know that and that's not what I was asking for.

I want to know what will be the moment of Inertia of the ball about the axis passing through the point C and perpendicular to the page, in the case when the ball is rotating about 2 axes simultaneously(as shown in Pic 2). In my opinion, it should be = $\frac{2}{5}mr^{2} + m \left ( R-r \right )^{2}$.

As I wrote in post #28, the moment of inertia of a body is only of interest in the context of rotation about some axis. To put that more forcefully, it is only defined in respect of its actual or potential rotation as a rigid body about a specified axis - i.e., a motion in which every part of the body is rotating at the same angular rate about that axis.
Specifically, if an element dm is at r from the axis then its contribution to the angular momentum when the body is rotating at $\omega$ about the axis is $\omega r^2~dm$. To get the total angular momentum we can sum these $\int\omega r^2~dm=\omega\int r^2~dm$. Note that the ability to move the $\omega$ outside the integral depends on the conditions I stated.
From this result, we see it is convenient to consider $I=\int r^2~dm$ to be a property of the rigid body and the chosen axis.
For any other axis the MoI may be different, and for a motion of some other kind it is not meaningful at all.

 

[NTesla]

As I wrote in post #28, the moment of inertia of a body is only of interest in the context of rotation about some axis. To put that more forcefully, it is only defined in respect of its actual or potential rotation as a rigid body about a specified axis - i.e., a motion in which every part of the body is rotating at the same angular rate about that axis.
Specifically, if an element dm is at r from the axis then its contribution to the angular momentum when the body is rotating at $\omega$ about the axis is $\omega r^2~dm$. To get the total angular momentum we can sum these $\int\omega r^2~dm=\omega\int r^2~dm$. Note that the ability to move the $\omega$ outside the integral depends on the conditions I stated.
From this result, we see it is convenient to consider $I=\int r^2~dm$ to be a property of the rigid body and the chosen axis.
For any other axis the MoI may be different, and for a motion of some other kind it is not meaningful at all.

I'm not sure how that would help in finding out the moment of Inertia of the ball about the axis passing through point C, in the present question(in case of Pic 2 of post#43).

 

[haruspex]

I'm not sure how that would help in finding out the moment of Inertia of the ball about the axis passing through point C, in the present question(in case of Pic 2 of post#43). If you have any idea as to how do I find that, then kindly let me know.

An implication of my previous post is that you should think of moment of inertia as resistance to angular acceleration about the axis, just as inertial mass means resistance to linear acceleration: $I=\tau/\alpha$. In the set-up here, that resistance involves the linkage between rotation of the ball about its centre and its motion about C.
But going to the trouble of finding this virtual "MoI" is pointless. You just need the equation which relates the torque to the acceleration. That is the approach I suggested in post #28.

 

[NTesla]

resistance involves the linkage between rotation of the ball about its centre and its motion about C.

Agreed.

But going to the trouble of finding this virtual "MoI" is pointless.

I don't agree. There is a definite non-zero moment of Inertia about C, and finding it could possibly lead to increase in knowledge about the whole thing.

 

[kuruman]

There is a definite non-zero moment of Inertia about C, and finding it could possibly lead to increase in knowledge about the whole thing.

To find the moment of inertia about C, you use the parallel axis theorem. It is the moment of inertia about the center of mass of the ball plus the moment of inertia about C is the entire mass of the ball were concentrated at its center:$~I_C=I_{cm}+m(R-r)^2.$ Now the moment of inertia of a sphere about an axis going through its center is $I_{cm}=\frac{2}{5}mr^2$, therefore $$~I_C=\frac{2}{5}mr^2+m(R-r)^2.$$Note that the subscript $C$ specifies about the axis about which this moment of inertia is calculated and that it depends only on $m$ and the geometric quantities $r$ and $R.$ Also note that moment of inertia of the ball about its CM is different from the expression about point C.

Does this help your understanding?

 

[NTesla]

Does this help your understanding?

Not really, because I already do understand that.

In my calculation that I have attached as pdf in post#24, I have already written that $I_{about C} = \frac{2}{5}mr^2+m(R-r)^2.$ and then have proceeded to calculate the time period.

Thereafter, in my post#43, I again mentioned that:

So my contention is that in this 2nd case too, the moment of Inertia of the ball about the axis passing through point C and perpendicular to the page should be = 2/5mr2+m(R−r)2.

In that same post, at the end I reiterate that:

In my opinion, it should be = 2/5mr2+m(R−r)2.

But apparently you missed that thrice.

@haruspex's argument is that calculation of moment of Inertia about point C is pointless:

the trouble of finding this virtual "MoI" is pointless

He further implied through his comments that the moment of Inertia about point C will NOT be = $\frac{2}{5}mr^2+m(R-r)^2$. He is saying that parallel axis theorem can't be applied as you and I have done so far. But what modifications are needed, in applying the parallel axis theorem, for that he is saying that it is pointless to even find the moment of Inertia about the axis passing through point C. And this is what I don't agree with. I don't think it's pointless. I think that maybe some important part of the puzzle is missing and finding that part would lead to better understanding of the whole concept.

 

[haruspex]

He is saying that parallel axis theorem can't be applied as you and I have done so far.

I am saying that you cannot apply it to find the moment of inertia of the ball's motion about C. You can apply it, as @kuruman notes, to find its MoI about the point of contact.

To find the moment of inertia about C, you use the parallel axis theorem.

Not here. Please read posts 28, 30, 33, 36…

There is a definite non-zero moment of Inertia about C, and finding it could possibly lead to increase in knowledge about the whole thing.

The effective MoI, what I called its virtual MoI in post #46, about C results from the kinematic relationship between the motion of the sphere’s centre and the motion of the rest of the sphere about that centre. The MoI found by the parallel axis theorem assumes a particular kinematic relationship, namely, that the sphere moves as though it is fixed to a pendulum pivoted at C. Since that is not how it moves here, the PAT gives the wrong answer.

A linear analogy would be two masses connected by a pulley system such that if mass m is displaced by x then mass M is displaced by 2x. To someone pulling on mass m and observing the resulting acceleration, the effective mass is m+2M, not m+M.

Humour me: calculate the relationship between the angular velocity of the ball's centre about C and the angular momentum of the ball about C, as described in post #28. We can then see if this gives the book answer.