Cylinder lying on conveyor belt

[kuruman]

Yes, I was reluctant to remain involved for the same reason, but we seem to be in different timezones and I did not want to leave the OP dangling. Anyway, it seems to me that the thread has since then stuck with the non-inertial view, so no need for you to hold back.

OK, then. We still need to hear from @JD_PM about what reference axis one ought to calculate the torque acting on the cylinder. One choice is better than others in the sense that Newton's 2nd law for rotations is sufficient to determine the acceleration of the CM in the non-inertial frame.

 

[JD_PM]

OK, then. We still need to hear from @JD_PM about what reference axis one ought to calculate the torque acting on the cylinder. One choice is better than others in the sense that Newton's 2nd law for rotations is sufficient to determine the acceleration of the CM in the non-inertial frame.

To Recap: I am still convinced that the right answer is(see #9 for more details):

$$v(t) = v_{CM} + V(t) = 2R \omega + V(t)$$

Do you agree with me?

 

[JD_PM]

To Recap: I am still convinced that the right answer is(see #9 for more details):

$$v(t) = v_{CM} + V(t) = 2R \omega + V(t)$$

Do you agree with me?

And I still have the following doubt(stated previously on #10):

Now I am confused on the following:

I know that the translational speed of the centre of mass for pure rolling motion on a non moving floor and measured from an inertial reference frame is:

$$v_{CM} = \frac{ds}{dt} = R\omega$$

Where $s$ is the arc of the cylinder

But how is it possible that I get $2R \omega$ just with the difference that in this case the floor (belt) is moving?

 

[JD_PM]

My solution is obtained from an inertial reference frame.

@kuruman you were interested in solving the problem from a non inertial reference frame. why?

 

[JD_PM]

That does not mean anything unless you specify the axis.
What about the angular acceleration?

I would say that the axis plays a role if we use:

$$\tau = I \alpha$$

The angular acceleration is obtained from the stated equation:

$$\alpha = \frac{\tau}{I}$$

 

[kuruman]

To Recap: I am still convinced that the right answer is(see #9 for more details):

$$v(t) = v_{CM} + V(t) = 2R \omega + V(t)$$

Do you agree with me?

I do not agree with you. In the special case $V(t)=0$ for all $t$, your expression predicts that the cylinder will be moving. Does that make sense? Besides, what is $\omega$? Your final answer must be in terms of the given quantities and these are $M,~k,~R$ and $V(t)$.

My solution is obtained from an inertial reference frame.

@kuruman you were interested in solving the problem from a non inertial reference frame. why?

The strategy for finding $v_{cm}(t)$ is to use dynamics to find $a_{cm}(t)$ and integrate. $\vec F_{net}= M\vec a_{cm}$ is insufficient to complete the task because there is also angular acceleration. So you need $\tau=I\alpha$. I am arguing that if you transform to the non-inertial frame, it is sufficient to use $\tau=I\alpha$ in order to find the acceleration of the cm in the non-inertial frame and hence in the inertial frame. However, for that to happen, you need to choose wisely the axis about which you the torque and the moment of inertia are to be expressed.

I would say that the axis plays a role if we use:
$$\tau = I \alpha$$
The angular acceleration is obtained from the stated equation:
$$\alpha = \frac{\tau}{I}$$

It's not "if we use", it's "because we have to use".

 

[haruspex]

I would say that the axis plays a role if we use:

$$\tau = I \alpha$$

The angular acceleration is obtained from the stated equation:

$$\alpha = \frac{\tau}{I}$$

The torque of a force is always in respect of an axis. Different axis, different torque. Likewise moment of inertia. Even the angular acceleration changes with axis; if your axis is not through the mass centre then the linear acceleration of the body contributes to its angular acceleration about the axis. You must use the same axis for all three.
E.g. consider a force F applied at the mass centre, but an axis distance s off to one side. The force has torque Fs about the axis. The body does not rotate about its own centre, but the axis sees the body as accelerating around it at angular rate a/s=F/(ms). The moment of inertia about this axis (since the body is not rotating about its own centre) is ms². Torque Fs=(ms²)(F/(ms)).

Further, it is advisable to use either a point fixed in space or the mass centre of the body as the axis. Anything else can yield the wrong answer.

 

[kuruman]

Further, it is advisable to use either a point fixed in space or the mass centre of the body as the axis. Anything else can yield the wrong answer.

Yes! In the inertial frame there is no point on the cylinder that is fixed in space, so one must use the cylinder axis for referencing torques and the moment of inertia. In the non-inertial frame, however, there is a point that is fixed in space and one can use it as reference instead of the cylinder axis.

 

[haruspex]

In the inertial frame there is no point on the cylinder that is fixed in space

True, but you do not have to use a point that is fixed to the cylinder. Any point fixed in space will do.

 

[kuruman]

True, but you do not have to use a point that is fixed to the cylinder. Any point fixed in space will do.

Yes, of course.

 

[JD_PM]

I do not agree with you. In the special case $V(t)=0$ for all $t$, your expression predicts that the cylinder will be moving. Does that make sense? Besides, what is $\omega$? Your final answer must be in terms of the given quantities and these are $M,~k,~R$ and $V(t)$.

The strategy for finding $v_{cm}(t)$ is to use dynamics to find $a_{cm}(t)$ and integrate. $\vec F_{net}= M\vec a_{cm}$ is insufficient to complete the task because there is also angular acceleration. So you need $\tau=I\alpha$. I am arguing that if you transform to the non-inertial frame, it is sufficient to use $\tau=I\alpha$ in order to find the acceleration of the cm in the non-inertial frame and hence in the inertial frame. However, for that to happen, you need to choose wisely the axis about which you the torque and the moment of inertia are to be expressed.

It's not "if we use", it's "because we have to use".

Using $\tau=I\alpha$ I got:

$$\tau=I\alpha = Rf $$

$$Mk^2 \alpha = RMa_{CM}$$

$$a_{CM} = \frac{k^2 \alpha}{R}$$

NOTE: I used an axis passing through the centre of mass of the cylinder. You will get the same result using $\tau=I\alpha$ no matter from what reference frame you analyse the rotation. This is because the fictitious force F (in non inertial reference frame case) exerts no torque on the cylinder.

Before proceeding to calculate $v_{CM}$ please tell me how you see the reasoning above.

 

[haruspex]

Using $\tau=I\alpha$ I got:

$$\tau=I\alpha = Rf $$

$$Mk^2 \alpha = RMa_{CM}$$

$$a_{CM} = \frac{k^2 \alpha}{R}$$

NOTE: I used an axis passing through the centre of mass of the cylinder. You will get the same result using $\tau=I\alpha$ no matter from what reference frame you analyse the rotation. This is because the fictitious force F (in non inertial reference frame case) exerts no torque on the cylinder.

Before proceeding to calculate $v_{CM}$ please tell me how you see the reasoning above.

From the above I deduce you are assuming $f =Ma_{CM}$. Presumably a_CM is inthe reference frame of the belt. What happened to the fictional force?

 

[JD_PM]

From the above I deduce you are assuming $f =Ma_{CM}$. Presumably a_CM is inthe reference frame of the belt. What happened to the fictional force?

Note that the fictitious force F is exerted on the centre of mass of the cylinder i.e. F produces no torque on the cylinder.

See on my diagrams how F acts on the COM.

 

[haruspex]

Note that the fictitious force F is exerted on the centre of mass of the cylinder i.e. F produces no torque on the cylinder.

See on my diagrams how F acts on the COM.

I have no issue with your torque equation. As you say, F does not have a torque about the cylinder's centre. The problem is with your second equation in post #41, where you introduce a_CM. It seems to me that to get that you must have assumed $f=Ma_{CM}$. I can't see where you define a_CM. If it is the acceleration of the mass centre in the belt frame then F should appear in the equation.
If you disagree, please post the missing steps in your working.

 

[JD_PM]

I have no issue with your torque equation. As you say, F does not have a torque about the cylinder's centre. The problem is with your second equation in post #41, where you introduce a_CM. It seems to me that to get that you must have assumed $f=Ma_{CM}$. I can't see where you define a_CM. If it is the acceleration of the mass centre in the belt frame then F should appear in the equation.
If you disagree, please post the missing steps in your working.

OK I see what you mean. Actually there is a mistake, as the second line at #41 should be:

$$Mk^2 \alpha = RMa_{o}$$

Having $f =Ma_{o}$, where $a_{o}$ is the tangential acceleration.

@kuruman suggested solving the problem being on a non inertial reference frame and using 2nd Newton's law for rotation. Well, in that scenario we have the picture #20 but with the fraction pointing to the right (I made I mistake when I posted the diagram #20). In such a scenario, using 2nd Newton's law for rotation we have:

$$\tau=I\alpha = Rf$$

From this equation there is no way you get $a_{CM}$ because:

$$F = Ma_{CM}$$

And F exerts no torque on the cylinder, therefore it does not appear on $\tau=I\alpha = Rf$

I made some research and found you can get both $v_{CM}$ and $a_{CM}$ using:

$$v_{CM} = \frac{ds}{dt} = R\frac{d\theta}{dt} = R\omega$$

But for that you need to assume constant acceleration...

And this approach does not use 2nd Newton's law... so I am basically stuck.

 

[haruspex]

OK I see what you mean. Actually there is a mistake, as the second line at #41 should be:

$$Mk^2 \alpha = RMa_{o}$$

Having $f =Ma_{o}$, where $a_{o}$ is the tangential acceleration.

@kuruman suggested solving the problem being on a non inertial reference frame and using 2nd Newton's law for rotation. Well, in that scenario we have the picture #20 but with the fraction pointing to the right (I made I mistake when I posted the diagram #20). In such a scenario, using 2nd Newton's law for rotation we have:

$$\tau=I\alpha = Rf$$

From this equation there is no way you get $a_{CM}$ because:

$$F = Ma_{CM}$$

And F exerts no torque on the cylinder, therefore it does not appear on $\tau=I\alpha = Rf$

I made some research and found you can get both $v_{CM}$ and $a_{CM}$ using:

$$v_{CM} = \frac{ds}{dt} = R\frac{d\theta}{dt} = R\omega$$

But for that you need to assume constant acceleration...

And this approach does not use 2nd Newton's law... so I am basically stuck.

I might return to this when I've more time to suggest how you could continue with that line, but for now I'd like to suggest a far simpler approach.
In the inertial frame, can you think of an axis about which there are no torques? What would that allow you to do?

 

[JD_PM]

I might return to this when I've more time to suggest how you could continue with that line, but for now I'd like to suggest a far simpler approach.
In the inertial frame, can you think of an axis about which there are no torques? What would that allow you to do?

SCENARIO ANALYSED FROM AN INERTIAL FRAME

We have to point out that in an inertial frame of reference we just have one external force: the static friction (no slipping rolling motion).

You asked for a selection of an axis which allows the system having $\tau = 0$ . That would be achievable if we were to select the point of contact cylinder-belt as the one the axis passes through perpendicularly.

With respect to that axis angular momentum is conserved. Is it that your idea? If that is the case, how could we get $v_{CM}$ from AM conservation?

 

[jbriggs444]

SCENARIO ANALYSED FROM AN INERTIAL FRAME
With respect to that axis angular momentum is conserved. Is it that your idea? If that is the case, how could we get $v_{CM}$ from AM conservation?

If I am understanding the problem correctly and seeing where @haruspex is going, we have a pretty simple known and unchanging quantity to utilize.

1. Starting angular momentum is zero
2. Angular momentum is conserved in the absence of external torques.
3. There are no external torques.

If angular momentum is zero and we know the velocity of the bottom contact surface of the cylinder with the belt, what else can we calculate?

 

[haruspex]

select the point of contact cylinder-belt as the one the axis passes through perpendicularly.

It depends exactly what you mean by that. Recall that we must choose a reference axis that is either the mass centre or is fixed in space. Since the belt is moving, the "point of contact" moves.

Your other equation should express that it is rolling contact.

 

[JD_PM]

If angular momentum is zero and we know the velocity of the bottom contact surface of the cylinder with the belt, what else can we calculate?

I guess you are driving me to the vectorial angular momentum equation:

$$L = r p$$

 

[JD_PM]

It depends exactly what you mean by that. Recall that we must choose a reference axis that is either the mass centre or is fixed in space. Since the belt is moving, the "point of contact" moves.

Your other equation should express that it is rolling contact.

I think I am going to think about the problem and see if I am able to get something out of it.

I will post whatever I get in the end

 

[jbriggs444]

I guess you are driving me to the vectorial angular momentum equation:

$$L = r p$$

What I have in mind is something else. We are picking an axis of rotation that is fixed somewhere on the plane of the belt. With respect to this axis of rotation, we have already argued that angular momentum is always zero. [I am assuming an axis fixed in space rather than an axis fixed to the belt, else the angular momentum conservation argument misses a fictitious torque].

We have the moving cylinder. We can decompose its motion into two parts. One part is its linear motion fore and aft down the belt. The other part is its rotary motion. Since the center of mass of the cylinder does not align horizontally with the chosen reference axis, the horizontal linear motion contributes to angular momentum. The rotary motion does as well. That means that we can write down a useful equation.

The no slip condition should allow us to write down another.

Without having done the algebra, it seems clear that the solution will reduce very nicely.

 

[JD_PM]

What I have in mind is something else. We are picking an axis of rotation that is fixed somewhere on the plane of the belt. With respect to this axis of rotation, we have already argued that angular momentum is always zero. [I am assuming an axis fixed in space rather than an axis fixed to the belt, else the angular momentum conservation argument misses a fictitious torque].

We have the moving cylinder. We can decompose its motion into two parts. One part is its linear motion fore and aft down the belt. The other part is its rotary motion. Since the center of mass of the cylinder does not align horizontally with the chosen reference axis, the horizontal linear motion contributes to angular momentum. The rotary motion does as well. That means that we can write down a useful equation.

The no slip condition should allow us to write down another.

Without having done the algebra, it seems clear that the solution will reduce very nicely.

OK I see, so you propose:

Linear motion contribution to the angular momentum:

$$L = RMV(t)$$

But what about rotational contribution? Well, I have been thinking how to use:

$$L = I \omega$$

But the moment of inertia gets cumbersome... I mean, we no longer have a cylinder rotating about an axis passing through the CM. What is the MoI now?

 

[haruspex]

Linear motion contribution to the angular momentum:
$$L = RMV(t)$$

No, V(t) is the belt speed. You need the cylinder's CoM speed.

But what about rotational contribution?

Well, I have been thinking how to use:
$$L = I \omega$$
But the moment of inertia gets cumbersome... I mean, we no longer have a cylinder rotating about an axis passing through the CM. What is the MoI now?

You can decompose the cylinder's motion into a linear motion (u(t), say) and a rotation about its centre. You have dealt with the linear motion's contribution to the angular momentum about the belt-level axis above. The angular momentum due to its rotation about its own axis is invariant, i.e. you can choose any axis and it is still the same, just I_CMω.

 

[JD_PM]

No, V(t) is the belt speed. You need the cylinder's CoM speed.

You can decompose the cylinder's motion into a linear motion (u(t), say) and a rotation about its centre. You have dealt with the linear motion's contribution to the angular momentum about the belt-level axis above. The angular momentum due to its rotation about its own axis is invariant, i.e. you can choose any axis and it is still the same, just I_CMω.

No, V(t) is the belt speed. You need the cylinder's CoM speed.

You can decompose the cylinder's motion into a linear motion (u(t), say) and a rotation about its centre. You have dealt with the linear motion's contribution to the angular momentum about the belt-level axis above. The angular momentum due to its rotation about its own axis is invariant, i.e. you can choose any axis and it is still the same, just I_CMω.

OK So what you mean is:

$$L = RMv_{CM} + I_{CM} \omega$$

Because of AM conservation:

$$L = 0$$

$$v_{CM} = -\frac{I_{CM} \omega}{RM}$$

Regarding vector directions: $V(t)$: points to the right. The belt moves horizontally towards right (I took that decision) and makes the cylinder spin counterclockwise, which means that the friction has to make the cylinder spin clockwise. Thus, friction points to the left.

NOTE: the negative sign makes sense because I regarded the angular velocity as positive (counterclockwise), which makes the cylinder rotate to the left (I regarded the right direction as +ive).

Let me know if you agree with me and we go to analyse the scenario on the non inertial frame.

Lately I have been thinking about the no appearance of $V(t)$ on my equation. There has to be something missing...

 

[haruspex]

$$v_{CM} = -\frac{I_{CM} \omega}{RM}$$.

Right, but you need another equation. Express the fact of rolling contact in terms of the radius, the rotation rate and the two velocities.

 

[JD_PM]

Right, but you need another equation. Express the fact of rolling contact in terms of the radius, the rotation rate and the two velocities.

Let's take left direction as positive now so as to get a positive velocity.

Why would not be correct stating the following?:

$$v_{CM} = \frac{I_{CM} \omega}{RM} + V(t)$$

It makes sense to me, as if the cylinder were to stop spinning it would continue having $v_{CM}$ because the belt moves horizontally.

 

[haruspex]

Let's take left direction as positive now so as to get a positive velocity.

Why would not be correct stating the following?:

$$v_{CM} = \frac{I_{CM} \omega}{RM} + V(t)$$

It makes sense to me, as if the cylinder were to stop spinning it would continue having $v_{CM}$ because the belt moves horizontally.

Why would it be correct? What is your reasoning for it?
It certainly does not represent the rolling condition, which is a purely kinematic relationship.

 

[JD_PM]

Why would it be correct? What is your reasoning for it?
It certainly does not represent the rolling condition, which is a purely kinematic relationship.

I just think that $V(t)$ has to contribute somehow to $v_{CM}$.

So the equation you suggest is missing is a kinematics one? Didn't we say that we were not going to assume constant acceleration?

 

[haruspex]

I just think that $V(t)$ has to contribute somehow to $v_{CM}$.

So the equation you suggest is missing is a kinematics one? Didn't we say that we were not going to assume constant acceleration?

Kinematics, not kinetics. (Anyway, kinetics also applies to non-constant acceleration; it's just that the simple SUVAT equations don't).
Kinematics is unconcerned with masses and forces. It just considers the physical constraints on relative motion, e.g. that the length of a string is constant.
In this case you want one expressing that the contact is rolling, not sliding. That relates the linear velocities, angular velocity and radius - nothing else.