Derive the period of a Ball rolling in a Bowl

[phantomvommand]

Homework Statement

A Ball of radius r rolls in a Bowl of Radius R. Friction exists to maintain pure rolling.

Relevant Equations

Simple Harmonic Equation
$Fr = I\phi$
Pure rolling condition

The following attempt gives the wrong answer, and I would like to know where it goes wrong.
Let $\theta$ be the angle of the ball with the vertical passing through the centre of the bowl, and $\phi$ be the angle the ball rolls through.
Let $m$ be the mass of the ball, $r$ be the radius of the ball and $R$ be the radius of the bowl.

Consider the torque of the ball about the centre of the bowl:
$-mg\sin\theta (R - r) - FR = (\frac 2 5 mr^2 + mR^2)\ddot \theta$

$Fr = \frac 2 5 mr^2 \ddot \phi$

Pure rolling condition:
$r\dot \phi = (R-r)\dot \theta$ , which implies $\ddot \phi = \frac {R-r} {r} \ddot \theta$.

Thus, $F = \frac 2 5 m (R-r) \ddot\theta$. Substitute this into the very 1st equation and get:

$-mg\theta (R-r) = \ddot\theta (\frac 2 5 mr^2 + mR^2 + \frac 2 5 mR(R-r))$

What is wrong with considering torque about the centre of the bowl?

 

[hutchphd]

I don't understand your pure rolling condition. Make it a condition on $\phi$ and $\theta$ instead of velocities first.

 

[phantomvommand]

I don't understand your pure rolling condition. Make it a condition on $\phi$ and $\theta$ instead of velocities first.

The idea is that the velocity of the centre of the ball is equal in magnitude to the velocity caused by the rolling, so that the base of the ball remains at rest.

 

[TSny]

Consider the torque of the ball about the centre of the bowl:
$-mg\sin\theta (R - r) - F = (\frac 2 5 mr^2 + mR^2)\ddot \theta$

Note that the left side should be the net torque, but the second term of the left side is a force, not a torque.

In relating $\theta$ and $\phi$, be careful with signs.
Are you taking counterclockwise rotation as positive for both $\theta$ and $\phi$?

 

[phantomvommand]

Note that the left side should be the net torque, but the second term of the left side is a force, not a torque.

In relating $\theta$ and $\phi$, be careful with signs.
Are you taking counterclockwise rotation as positive for both $\theta$ and $\phi$?

Thanks for pointing out this mistake. No, if theta counterclockwise is positive, phi clockwise is positive. This is still not the correct method, however. The correct method is to consider translational forces on the ball, and derive a SHM relationship. What about torque is wrong?

 

[hutchphd]

Your constraint equation is incorrect, I believe. Draw a picture of point of constraint.

 

[haruspex]

Your constraint equation is incorrect, I believe. Draw a picture of point of constraint.

It looks right to me. Both sides correctly express the velocity of the ball's centre in the lab frame.

 

[hutchphd]

Yes my mistake. That is the easier way to see it. Apologies.

 

[TSny]

if theta counterclockwise is positive, phi clockwise is positive.

Ok.

What about torque is wrong?

Your general approach is fine. But I think there is a mistake in the equation shown below:

Consider the torque of the ball about the centre of the bowl:
$-mg\sin\theta (R - r) - FR = (\frac 2 5 mr^2 + mR^2)\ddot \theta$

The net torque on the left looks good. But the right-hand side should express the rate of change of angular momentum about the center O of the bowl. This angular momentum can be expressed as the sum of two parts:

(1)The "orbital" angular momentum about O, which is found by considering the ball to be a point mass concentrated at its center of mass. What is the radius of motion of the center of mass of the ball?

(2) The "spin" angular momentum, which is the angular momentum due to the spinning of the ball about its center. The spin is determined by $\dot \phi$, not $\dot \theta$.

 

[phantomvommand]

Ok.Your general approach is fine. But I think there is a mistake in the equation shown below:

The net torque on the left looks good. But the right-hand side should express the rate of change of angular momentum about the center O of the bowl. This angular momentum can be expressed as the sum of two parts:

(1)The "orbital" angular momentum about O, which is found by considering the ball to be a point mass concentrated at its center of mass. What is the radius of motion of the center of mass of the ball?

(2) The "spin" angular momentum, which is the angular momentum due to the spinning of the ball about its center. The spin is determined by $\dot \phi$, not $\dot \theta$.

Thanks so much for this! I blindly applied $\tau = I\alpha$. I realize it is now wrong. However, I have a general question involving the usage of this formula:
when using the $\tau = I\alpha$ formula, $I$ is taken to be the moment of inertia about the axis, which means for a sphere it would be $\frac 2 5 mr^2 + mR^2$. However, when using $\tau = \frac {dL} {dt}$, $L$ is calculated by adding the orbital and spin components, and the moment of inertia used in the orbital component is simply $mR^2$. Is this right?

 

[haruspex]

Thanks so much for this! I blindly applied $\tau = I\alpha$. I realize it is now wrong. However, I have a general question involving the usage of this formula:
when using the $\tau = I\alpha$ formula, $I$ is taken to be the moment of inertia about the axis, which means for a sphere it would be $\frac 2 5 mr^2 + mR^2$. However, when using $\tau = \frac {dL} {dt}$, $L$ is calculated by adding the orbital and spin components, and the moment of inertia used in the orbital component is simply $mR^2$. Is this right?

Not sure what you mean by this in the present context.
You can only add moments of inertia like that if the body is rotating about the axis as though rigidly attached to a plane rotating about it, i.e. $\alpha$ is the same for the two components, in this case $\frac 2 5 mr^2$ and $m(R-r)^2$,
That is not the case here. We have $\ddot\theta$ for $m(R-r)^2$ and $-\ddot\phi$ for $\frac 2 5 mr^2$.

 

[phantomvommand]

Not sure what you mean by this in the present context.
You can only add moments of inertia like that if the body is rotating about the axis as though rigidly attached to a plane rotating about it, i.e. $\alpha$ is the same for the two components, in this case $\frac 2 5 mr^2$ and $m(R-r)^2$,
That is not the case here. We have $\ddot\theta$ for $m(R-r)^2$ and $-\ddot\phi$ for $\frac 2 5 mr^2$.

My question is not related to the current question. Thanks for your answer. Does it have to be rigidly attached? If I have 2 stars orbiting around each other, and assuming neither has any “spin”, would the angular momentum $L$ be $Iw$, where $I= \frac 25mr^2 + mR^2$

 

[haruspex]

My question is not related to the current question. Thanks for your answer. Does it have to be rigidly attached? If I have 2 stars orbiting around each other, and assuming neither has any “spin”, would the angular momentum $L$ be $Iw$, where $I= \frac 25mr^2 + mR^2$

No.
In the same way, if we have a pendulum consisting of a disc mass m radius r mounted on a frictionless axle at the end of a rod length L (the disc being coplanar with the rod) then the MoI is just $mL^2$.

 

[phantomvommand]

No.
In the same way, if we have a pendulum consisting of a disc mass m radius r mounted on a frictionless axle at the end of a rod length L then the MoI is just $mL^2$.

Thank you

 

[phantomvommand]

Ok.Your general approach is fine. But I think there is a mistake in the equation shown below:

The net torque on the left looks good. But the right-hand side should express the rate of change of angular momentum about the center O of the bowl. This angular momentum can be expressed as the sum of two parts:

(1)The "orbital" angular momentum about O, which is found by considering the ball to be a point mass concentrated at its center of mass. What is the radius of motion of the center of mass of the ball?

(2) The "spin" angular momentum, which is the angular momentum due to the spinning of the ball about its center. The spin is determined by $\dot \phi$, not $\dot \theta$.

Hi @TSny @hutchphd @haruspex . Thanks for all the help. I am hoping you guys can check through my following working, for the question. The actual question is that of a ball with a thin outer covering of mass m, and whose inside is solid and of mass M. The ball is of radius r, and bowl of radius R.

I am considering the motion of the ball as it rolls up the bowl. Angle between ball and bowl is $\theta$, angle the ball turned through is $\phi$. $\theta$ increases anticlockwise, so $\phi$ increases clockwise.

Taking torque about the centre of the bowl:
$-(m+M)g\sin\theta (R-r) -|F|R = \frac {dL} {dt}$
$\dot L = (m + M)(R-r)^2\ddot\theta + (\frac23mr^2 + \frac25Mr^2)\ddot\phi$

Consider torque about ball:
$Fr = (\frac23mr^2 + \frac25Mr^2)\ddot\phi$

No Slip:
$r\dot\phi = (R-r)\dot\theta <==> \ddot\phi = \frac {R-r} {r} \ddot\theta$

$F = (\frac23 m + \frac25 M)(R-r)\ddot\theta$

Substituting $F$, $\dot L$ and $\ddot\phi$ into the 1st equation, and using $\sin\theta = \theta$ and cancelling out $(R-r)$:
$-(m+M)g\theta - |(\frac23 m + \frac25M)\ddot\theta|R = (m + M)(R-r)\ddot\theta + (\frac23mr^2 + \frac25Mr^2)\frac1r\ddot\theta$

Noting that $\ddot\theta < 0$,
$-(m+M)g\theta = \ddot\theta(mR + MR - rm - rM + \frac23 mr +\frac 2 5 Mr - \frac23 mR -\frac 25 MR) = \ddot\theta (R-r)(\frac 35M + \frac13m)$

the correct answer should be:
$-(m+M)g\theta = \ddot\theta (R-r)(\frac 75M + \frac53m)$

Where did I go wrong? The correct answer can be obtained if we drop the absolute sign around friction (why?), and take the spin angular momentum of the ball to be negative, perhaps because about the axis of the bowl, anticlockwise is positive, although I am not sure about this.

 

[haruspex]

drop the absolute sign around friction (why?)

Because the sign of the torque exerted by the friction changes as the ball passes through the central position.

 

[phantomvommand]

Because the sign of the torque exerted by the friction changes as the ball passes through the central position.

Is F supposed to be a vector or magnitude?

 

[haruspex]

Is F supposed to be a vector or magnitude?

You can express it as a vector or as a scalar, but it has direction, so not a magnitude (unless the direction is provided by some other factor).
E.g. in the case of kinetic friction you know the friction opposes relative velocity but is otherwise not a function of it, so you could write $F=\mu_k|N|\frac v{|v|}$.
But here the friction force varies continuously from pointing in one direction to pointing in the other.

In your "torque about the ball" equation, the RHS can be negative, so you have F going negative. This is inconsistent with using |F| in the earlier torque equation.

 

[phantomvommand]

You can express it as a vector or as a scalar, but it has direction, so not a magnitude (unless the direction is provided by some other factor).
E.g. in the case of kinetic friction you know the friction opposes relative velocity but is otherwise not a function of it, so you could write $F=\mu_k|N|\frac v{|v|}$.
But here the friction force varies continuously from pointing in one direction to pointing in the other.

In your "torque about the ball" equation, the RHS can be negative, so you have F going negative. This is inconsistent with using |F| in the earlier torque equation.

F is negative when the ball is rolling upwards. Thus, in the 1st torque about centre of bowl equation, I have $- |F|$. Isn’t the meaning of “-“ to say that |F| reduces $\ddot\theta$?

 

[haruspex]

F is negative when the ball is rolling upwards.

According to your second torque equation, F is positive when $\ddot\phi$ is positive, i.e. clockwise. That will be true whenever the ball is in the left half of the bowl, no matter which way it’s moving.

 

[phantomvommand]

According to your second torque equation, F is positive when $\ddot\phi$ is positive, i.e. clockwise. That will be true whenever the ball is in the left half of the bowl, no matter which way it’s moving.

Isn’t $\ddot\phi$ negative, that is why F points “downwards” and decreases $\theta$?

 

[haruspex]

Isn’t $\ddot\phi$ negative, that is why F points “downwards” and decreases $\theta$?

In the left half of the bowl, friction acts up and to the left, exerting a clockwise torque on the ball. With clockwise positive for $\phi$, $\ddot\phi$ is positive.
If the ball is rolling up and left at this time, the ball has a decreasingly negative rotation; if moving down and right it has an increasingly positive rotation.

 

[phantomvommand]

In the left half of the bowl, friction acts up and to the left, exerting a clockwise torque on the ball. With clockwise positive for $\phi$, $\ddot\phi$ is positive.
If the ball is rolling up and left at this time, the ball has a decreasingly negative rotation; if moving down and right it has an increasingly positive rotation.

Sorry, I didn’t notice you said left. I thought you meant right half. Nonetheless, I suppose friction acts up and right in the right half. I did not realize this until you pointed it out :/. Thanks so much!

When the ball is in the right half and moves up, friction is negative, as $\ddot\phi<0$(anti-clockwise rotation). How do we reconcile this with the “torque about centre of bowl” equation, which suggests that the upward friction which causes anti-clockwise rotation of $\theta$ is positive?

 

[haruspex]

How do we reconcile this with the “torque about centre of bowl” equation, which suggests that the upward friction which causes anti-clockwise rotation of θ is positive?

You have -FR in that equation, so when F is negative (i.e. acting anticlockwise around the arc of the bowl) it is tending to increase $\dot\theta$. This is consistent with $\theta$ being positive anticlockwise.

 

[phantomvommand]

You have -FR in that equation, so when F is negative (i.e. acting anticlockwise around the arc of the bowl) it is tending to increase $\dot\theta$. This is consistent with $\theta$ being positive anticlockwise.

Sorry, I forgot to mention that I wrote that equation under the misconception that friction was downwards when the ball is in the right half and is rolling upwards.

now that friction points upwards in that case, shouldn’t it be +FR, given that Friction increases $\ddot\theta$?

 

[haruspex]

Sorry, I forgot to mention that I wrote that equation under the misconception that friction was downwards when the ball is in the right half and is rolling upwards.

now that friction points upwards in that case, shouldn’t it be +FR, given that Friction increases $\ddot\theta$?

I didn’t see a problem with that equation.
Please post the complete solution as you understand it now.

 

[phantomvommand]

I didn’t see a problem with that equation.
Please post the complete solution as you understand it now.

I am considering the motion of the ball as it rolls up the right half of the bowl. Angle between ball and bowl is $\theta$, angle the ball turned through is $\phi$. $\theta$ increases anticlockwise, so $\phi$ increases clockwise. Friction is upwards and rightwards. Friction increases $\theta$.

Taking torque about the centre of the bowl:
$-(m+M)g\sin\theta (R-r) + FR = \frac {dL} {dt}$
$\dot L = (m + M)(R-r)^2\ddot\theta - (\frac23mr^2 + \frac25Mr^2)\ddot\phi$

Consider torque about ball:
$Fr = (\frac23mr^2 + \frac25Mr^2)\ddot\phi$

No Slip:
$r\dot\phi = (R-r)\dot\theta <==> \ddot\phi = \frac {R-r} {r} \ddot\theta$

$F = (\frac23 m + \frac25 M)(R-r)\ddot\theta$

Substituting $F$, $\dot L$ and $\ddot\phi$ into the 1st equation, and using $\sin\theta = \theta$ and cancelling out $(R-r)$:
$-(m+M)g\theta + (\frac23 m + \frac25M)\ddot\theta R = (m + M)(R-r)\ddot\theta - (\frac23mr^2 + \frac25Mr^2)\frac1r\ddot\theta$

$-(m+M)g\theta = \ddot\theta(mR + MR - rm - rM - \frac23 mr - \frac 2 5 Mr - \frac23 mR -\frac 25 MR)$

the correct answer should be:
$-(m+M)g\theta = \ddot\theta (R-r)(\frac 75M + \frac53m)$

 

[TSny]

Consider torque about ball:
$Fr = (\frac23mr^2 + \frac25Mr^2)\ddot\phi$

With the direction of F now switched, is there a sign problem in the equation above?

 

[phantomvommand]

With the direction of F now switched, is there a sign problem in the equation above?

Thank you very much!