Rolling without slipping 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.

 

[NTesla]

I am saying that you cannot apply it to find the moment of inertia of the ball's motion about C.

Yes, i agree that that is what you were saying. I'm not sure why you thought I was disagreeing with you on this point. In the statement: "He is saying that parallel axis theorem can't be applied as you and I have done so far." "as" is the operative word. Meaning that you are saying that PAT can't be applied "as" I and kuruman had applied to calculate the MoI about point C.

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.

I agree and have understood this part already. But when I'm urging on to try to figure out a way to find the MoI about point C, which takes into account the unusual circumstance of the sphere's movement in the original question, you said that it's pointless. My disagreement is on that point. I don't think it's pointless.

I've reversed engineered the moment of Inertia of the sphere about point C. It's coming to be = $\frac{2}{5}mr^{2} + m\left (R-r\right)^{2} - \frac{2}{5}mrR$. The first two terms are expected and are easy to understand. However, the calculative origin of the 3rd term needs to be understood.

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.

Maybe it will give the book's answer, maybe it won't. There could be and are more than 1 way to reach the answer. My need is to understand how to reach answer using my own approach, and not to abandon it, in favour of other methods. I'll give time to solve it using your suggested approach also, once I understand and solve the question using the approach that I've been on.

 

[haruspex]

My need is to understand how to reach answer using my own approach

But I am arguing that you cannot arrive at it using your approach rather than mine because the steps you have to take to find the effective MoI involve my approach.

 

[haruspex]

I've reversed engineered the moment of Inertia of the sphere about point C. It's coming to be = $\frac{2}{5}mr^{2} + m\left (R-r\right)^{2} - \frac{2}{5}mrR$.

I get the same. Curiously, that gives 0 for $5R=7r$.

 

[haruspex]

I would argue strongly against thinking of this "effective MoI" as really being an MoI.
MoIs arise usually in three contexts:

• Angular momentum, $I\omega$
• Torque, $\tau=I\alpha$
• Rotational KE, $E=\frac 12I\omega^2$.

Since the second of those is merely the derivative wrt time of the first, the effective MoI we found for the one is valid for the other, but it won’t give the right result for energy.

 

[kuruman]

It is a fact that the parallel axis theorem correctly calculates the MoI about any point when the ball is at rest at the equilibrium position. The question is how one uses correctly the MoI thus calculated for whatever one wants to do. Let's say we want to calculate the total angular momentum about point C.

According to parallel axis theorem the moment of inertia about point C is $I_C=\frac{2}{5}mr^2+m(R-r)^2.$

Case I
The ball is at the end of a light rod and is going around in a circle with angular speed $\dot{\theta}$. The angular momentum of the ball about point C is the sum of two terms.

The first term is orbital angular momentum of the ball's CM about C with angular speed $\dot{\theta}$: $~L_{\text{orb.}}=m(R-r)^2\dot{\theta}.$
The second term is spin of the ball about its CM with angular speed $\dot{\varphi}$: $~L_{\text{spin}}=\frac{2}{5}mr^2\dot{\varphi}.$

Now the spin angular velocity $\dot{\varphi}$ is the same as $\dot{\theta}$ because the ball goes once around its axis as it goes once around point C. It's the same kind of motion that the Moon undergoes as it orbits the Earth. The orbital period of the Moon is equal to its spin period hence it presents the same face to the Earth at all times. Thus, $~L_{\text{spin}}=\frac{2}{5}mr^2\dot{\theta}$

The total angular momentum of the ball in this case is $$L_{\text{tot.}}=L_{\text{orb}}+L_{\text{spin}}=m(R-r)^2\dot{\theta}+\frac{2}{5}mr^2\dot{\varphi}=I_{C~}\dot{\theta}.$$ Case II
The ball is rolling inside the spherical shell as in this problem.

The relation $L_{\text{tot.}}=L_{\text{orb}}+L_{\text{spin}}$ is still valid but the relation between orbital and spin angular velocities is not. The rolling without slipping constraint, corrected from post #14, can be found using the diagram on the right, $$s=R\theta=r(\theta+\varphi)\implies \varphi=\frac{(R-r)}{r} \theta.$$The angular momentum about C is $$\begin {align}
L_{\text{tot.}} & = L_{\text{orb}}+L_{\text{spin}} =m(R-r)^2\dot{\theta}+\frac{2}{5}mr^2\dot{\varphi} \nonumber \\
& = m(R-r)^2\dot{\theta}+\frac{2}{5}mr^2\frac{(R-r)}{r} \dot{\theta} \nonumber \\
& = m(R-r)\left(R-\frac{3}{5}r\right)\dot{\theta} \neq I_{C~}\dot{\theta}. \nonumber

\end{align}$$

 

[haruspex]

It is a fact that the parallel axis theorem correctly calculates the MoI about any point when the ball is at rest at the equilibrium position. The question is how one uses correctly the MoI thus calculated for whatever one wants to do. Let's say we want to calculate the total angular momentum about point C.

According to parallel axis theorem the moment of inertia about point C is $I_C=\frac{2}{5}mr^2+m(R-r)^2.$

Case I
The ball is at the end of a light rod and is going around in a circle with angular speed $\dot{\theta}$. The angular momentum of the ball about point C is the sum of two terms.

The first term is orbital angular momentum of the ball's CM about C with angular speed $\dot{\theta}$: $~L_{\text{orb.}}=m(R-r)^2\dot{\theta}.$
The second term is spin of the ball about its CM with angular speed $\dot{\varphi}$: $~L_{\text{spin}}=\frac{2}{5}mr^2\dot{\varphi}.$

Now the spin angular velocity $\dot{\varphi}$ is the same as $\dot{\theta}$ because the ball goes once around its axis as it goes once around point C. It's the same kind of motion that the Moon undergoes as it orbits the Earth. The orbital period of the Moon is equal to its spin period hence it presents the same face to the Earth at all times. Thus, $~L_{\text{spin}}=\frac{2}{5}mr^2\dot{\theta}$

The total angular momentum of the ball in this case is $$L_{\text{tot.}}=L_{\text{orb}}+L_{\text{spin}}=m(R-r)^2\dot{\theta}+\frac{2}{5}mr^2\dot{\varphi}=I_{C~}\dot{\theta}.$$ Case II
The ball is rolling inside the spherical shell as in this problem.

View attachment 366299The relation $L_{\text{tot.}}=L_{\text{orb}}+L_{\text{spin}}$ is still valid but the relation between orbital and spin angular velocities is not. The rolling without slipping constraint, corrected from post #14, can be found using the diagram on the right, $$s=R\theta=r(\theta+\varphi)\implies \varphi=\frac{(R-r)}{r} \theta.$$The angular momentum about C is $$\begin {align}
L_{\text{tot.}} & = L_{\text{orb}}+L_{\text{spin}} =m(R-r)^2\dot{\theta}+\frac{2}{5}mr^2\dot{\varphi} \nonumber \\
& = m(R-r)^2\dot{\theta}+\frac{2}{5}mr^2\frac{(R-r)}{r} \dot{\theta} \nonumber \\
& = m(R-r)\left(R-\frac{3}{5}r\right)\dot{\theta} \neq I_{C~}\dot{\theta}. \nonumber

\end{align}$$

https://en.wikipedia.org/wiki/Moment_of_inertia disagrees.

"The moment of inertia, otherwise known as the mass moment of inertia, angular/rotational mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is defined relatively to a rotational axis. It is the ratio between the torque applied and the resulting angular acceleration about that axis."​

 

[kuruman]

https://en.wikipedia.org/wiki/Moment_of_inertia disagrees.

"The moment of inertia, otherwise known as the mass moment of inertia, angular/rotational mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is defined relatively to a rotational axis. It is the ratio between the torque applied and the resulting angular acceleration about that axis."​

OK, can you help me figure out how to apply this definition in the case of the rolling ball? There are two torques acting on the ball, one from gravity acting on the CM and one from static friction acting on the point of contact. I assume that "by the torque applied" the wikipedia definition means the net applied torque. That is easy to figure out as $$\boldsymbol {\tau}_{net}=-R~f_s(\mathbf {\hat r}\times\mathbf{\hat{\theta}})+(R-r)mg\sin\theta~(\mathbf {\hat r}\times \mathbf{\hat{\theta}}).$$ What about "the resulting angular acceleration about that axis?" that is needed for the ratio? Is it simply the vector sum of the orbital and spin terms?

 

[haruspex]

Is it simply the vector sum of the orbital and spin terms?

Yes.

 

[NTesla]

Here's the calculation taking the angular momentum about point C.
$L_{aboutC} = -mw_{C}(R-r)^2 + \frac{2}{5}mr^2w_{cm}$

and using the equation: $w_{cm} = \frac{R-r}{r}w_{c}$, we get:
$L_{aboutC} = \left ( -R^2 - \frac{7}{5}r^2 + \frac{12}{5}rR\right )mw_{c}$

Differentiation both sides wrt t, we get:
$\tau _{C} = \frac{\mathrm{d} L_{about C}}{\mathrm{d} t}$ = $\left ( -R^2 -\frac{7}{5}r^2 + \frac{12}{5}rR\right )m\alpha _{C}$

and $\tau _{C} = -\left ( mgsin\theta (R-r) - \frac{2}{7}mgRsin\theta \right )$

This gives, $\alpha _{C} = - \frac{\left ( \frac{5}{7}R - r \right )gsin\theta}{\frac{12}{5}rR- R^2 - \frac{7}{5}r^2}$.

But this is not correct value of $\alpha _{C}$ which could give the correct value of time period. I don't know what am i missing or doing wrong.

 

[haruspex]

This gives, $\alpha _{C} = - \frac{\left ( \frac{5}{7}R - r \right )gsin\theta}{\frac{12}{5}rR- R^2 - \frac{7}{5}r^2}$.

Which reduces to $\frac {5g\sin(\theta)}{7(R-r)}$.
Do you know what the answer is supposed to be?