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.