Rolling without slipping on a curved surface

[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?

 

[NTesla]

Which reduces to 5gsin⁡(θ)7(R−r).

I don't understand how does that reduce to this equation. Even if we neglect the small r term in the numerator, we can't neglect the term 12rR in the denominator.

 

[haruspex]

I don't understand how does that reduce to this equation. Even if we neglect the small r term in the numerator, we can't neglect the term 12rR in the denominator.

Factorise the denominator.

 

[NTesla]

Factorise the denominator.

After factorizing, in the denominator I'm getting: $(r - R)$, where $(R - r)$ should come. I've double checked my calculations/equations, but can't find why the sign is reversed.

 

[haruspex]

After factorizing, in the denominator I'm getting: $(r - R)$, where $(R - r)$ should come. I've double checked my calculations/equations, but can't find why the sign is reversed.

If you mean compared with what I posted in post #60, I only meant that your numerator/denominator term reduced to that. I dropped the minus sign that preceded it.
$\alpha_C$ should be negative, right?

 

[NTesla]

αC should be negative, right?

yes, and it is already negative, before we proceed on to do factorization. Kindly see last line in post#59. So, eventually $\alpha_{C}$ is coming out positive, after the factorization, which it shouldn't.

 

[haruspex]

yes, and it is already negative, before we proceed on to do factorization. Kindly see last line in post#59. So, eventually $\alpha_{C}$ is coming out positive, after the factorization, which it shouldn't.

Yes, you do seem to have a sign error in post #59, but where exactly it is depends which sense you are taking as positive for each variable.
Please state that for each of $\omega_C, \omega_{cm}, L, \alpha_C, \tau_C$.