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Cylinder lying on conveyor belt

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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

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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.
 
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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

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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 ICMω.
 
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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 ICMω.
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 ICMω.
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...
 
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haruspex

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$$v_{CM} = -\frac{I_{CM} \omega}{RM}$$.
Right, but you need another equation. Express the fact of rolling contact in terms of the radius, the rotation rate and the two velocities.
 
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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

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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.
 
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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

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

$$v_{CM} + V(t) = R \omega$$
 

haruspex

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You mean:

$$v_{CM} + V(t) = R \omega$$
Since both those linear velocities are in the lab frame, and I assume are positive in the same direction, adding them looks wrong.
 
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Since both those linear velocities are in the lab frame, and I assume are positive in the same direction, adding them looks wrong.
My bad, regarding left as positive, we would have:

$$v_{CM} - V(t) = R \omega$$

Together with the equation obtained by conservation of AM:

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

Getting as the final answer:

$$v_{CM} = \frac{1}{2}[(\frac{I_{CM}}{R^2M} + 1) R\omega + V(t)]$$
 

haruspex

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My bad, regarding left as positive, we would have:

$$v_{CM} - V(t) = R \omega$$

Together with the equation obtained by conservation of AM:

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

Getting as the final answer:

$$v_{CM} = \frac{1}{2}[(\frac{I_{CM}}{R^2M} + 1) R\omega + V(t)]$$
Two problems there.
Your first two equations are not handling signs consistently. If velocities are positive to the left then for your first equation to be right anticlockwise must be positive for the rotation. The second equation was obtained using a different pair of definitions.

Secondly, you are asked to express vCM in terms of k (where I=Mk2), R, M and V(t). ω should not feature.
 
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Two problems there.
Your first two equations are not handling signs consistently. If velocities are positive to the left then for your first equation to be right anticlockwise must be positive for the rotation. The second equation was obtained using a different pair of definitions.

Secondly, you are asked to express vCM in terms of k (where I=Mk2), R, M and V(t). ω should not feature.
OK I redid the problem from scratch:

From second (translation) Newton's law:

$$a_o= \frac{f}{M}$$

From second (rotation) Newton's law:

$$\tau = I \alpha = fa$$

$$k^2 \alpha= a_o a$$

$$\alpha = \frac{a_o a}{k^2}$$

Where ##a_o## is the acceleration of the cylinder measured from the ground.

Assuming that the acceleration of the belt is constant (##a_b##):

$$ \frac{V}{t} = a_b$$

The acceleration of the cylinder with respect to the ground accounts for the acceleration of the belt and the angular acceleration of the cylinder (which has a negative sign because I considered the cylinder spinning clockwise i.e. the belt accelerating to the left and regarding that direction as the positive one). Therefore:

$$a_o = a_b - a\alpha$$

$$a_o = \frac{V}{t} -a\frac{a_o a}{k^2}$$

$$a_o = \frac{Vk^2}{(k^2 + a^2)t}$$

We know by kinematics that:

$$v_{cm} = \frac{Vk^2}{k^2 + a^2}$$

The issue here is that I had to assume the acceleration of the belt being constant so as to solve the problem. The answer makes sense to me but I would like to know how to solve it without assuming that ##a_b = const##. (Recall that @kuruman suggested on #2 comment that we should not assume it).
 

haruspex

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OK I redid the problem from scratch:

From second (translation) Newton's law:

$$a_o= \frac{f}{M}$$

From second (rotation) Newton's law:

$$\tau = I \alpha = fa$$

$$k^2 \alpha= a_o a$$

$$\alpha = \frac{a_o a}{k^2}$$

Where ##a_o## is the acceleration of the cylinder measured from the ground.

Assuming that the acceleration of the belt is constant (##a_b##):

$$ \frac{V}{t} = a_b$$

The acceleration of the cylinder with respect to the ground accounts for the acceleration of the belt and the angular acceleration of the cylinder (which has a negative sign because I considered the cylinder spinning clockwise i.e. the belt accelerating to the left and regarding that direction as the positive one). Therefore:

$$a_o = a_b - a\alpha$$

$$a_o = \frac{V}{t} -a\frac{a_o a}{k^2}$$

$$a_o = \frac{Vk^2}{(k^2 + a^2)t}$$

We know by kinematics that:

$$v_{cm} = \frac{Vk^2}{k^2 + a^2}$$

The issue here is that I had to assume the acceleration of the belt being constant so as to solve the problem. The answer makes sense to me but I would like to know how to solve it without assuming that ##a_b = const##. (Recall that @kuruman suggested on #2 comment that we should not assume it).
To avoid assuming constant acceleration you will need to avoid involving acceleration at all.
You were nearly there with post #63. You just need to get your use of signs consistent and express the answer in terms of the variables given in the problem statement.
 
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To avoid assuming constant acceleration you will need to avoid involving acceleration at all.
You were nearly there with post #63. You just need to get your use of signs consistent and express the answer in terms of the variables given in the problem statement.
OK, Fixing signs and ##k## issues one gets:

$$v_{CM} = V(t) - R \omega$$

$$v_{CM} = - \frac{k^2 \omega}{R}$$

$$v_{CM} = \frac{1}{2}[-(\frac{k^2}{R^2} + 1) R\omega + V(t)]$$

The only thing still disturbs me is that ##\omega## is still there. I am thinking now

OK It is just about using ##v_{CM} = \omega R## again. Actually doing so I get a pretty nice solution for ##v_{CM}##:

$$v_{CM} = \frac{V(t)}{\frac{k^2}{R^2} + 3}$$

In terms of dimensions makes sense because k has dimensions of length, then the right hand side of the equation has dimensions of velocity.

WoW it has been a long journey but I enjoyed it a lot! :)
 
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kuruman

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What's counter-intuitive about this problem is that if the belt brakes suddenly, the cylinder will stop moving just as suddenly.
 
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What's counter-intuitive about this problem is that if the belt brakes suddenly, the cylinder will stop moving just as suddenly.
Mmm true but I think that is due to not considering an initial velocity on the cylinder. If we were to consider one different from zero, we would get an extra term that would make ##v_{CM}## different from zero even when ##V(t) = 0##
 

haruspex

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it has been a long journey
And not quite over. The answer you got in post #65 was correct, but it assumed constant acceleration. You need to get the same answer without that assumption.

In post #63, the signs on ω in the first two equations were inconsistent. In post #67 you have changed the sign on both occurrences of ω, so they are still inconsistent.

Once you have those two equations right, you don't need the third equation in those two posts. In fact, I am not sure how you arrived at it. Just eliminate ω between the first two.
 

kuruman

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Mmm true but I think that is due to not considering an initial velocity on the cylinder. If we were to consider one different from zero, we would get an extra term that would make ##v_{CM}## different from zero even when ##V(t) = 0##
The problem specifies that the bottle (cylinder) and the belt are initially at rest.
 
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And not quite over. The answer you got in post #65 was correct, but it assumed constant acceleration. You need to get the same answer without that assumption.

In post #63, the signs on ω in the first two equations were inconsistent. In post #67 you have changed the sign on both occurrences of ω, so they are still inconsistent.

Once you have those two equations right, you don't need the third equation in those two posts. In fact, I am not sure how you arrived at it. Just eliminate ω between the first two.
Yeah the following equation is wrong (algebraic mistake):

$$v_{CM} = \frac{1}{2}[-(\frac{k^2}{R^2} + 1) R\omega + V(t)]$$

Then, combining the following two equations so as to solve for ##v_{CM}## one gets the same result than #65:

$$v_{CM} = V(t) - R \omega$$

$$v_{CM} = - \frac{k^2 \omega}{R}$$

$$v_{cm} = \frac{Vk^2}{k^2 + R^2}$$
 

haruspex

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Yeah the following equation is wrong (algebraic mistake):

$$v_{CM} = \frac{1}{2}[-(\frac{k^2}{R^2} + 1) R\omega + V(t)]$$

Then, combining the following two equations so as to solve for ##v_{CM}## one gets the same result than #65:

$$v_{CM} = V(t) - R \omega$$

$$v_{CM} = - \frac{k^2 \omega}{R}$$

$$v_{cm} = \frac{Vk^2}{k^2 + R^2}$$
Now you have arrived.
 
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Now you have arrived.
The problem has been solved.

However I am still interested in the approach @kuruman suggested on 36#; solving the problem from a non-inertial reference frame. I guess this approach is based on the second Newton's Law for rotation:

$$f - F = I \alpha$$

Where ##f## is the friction and ##F## is a fictitious force.
 

kuruman

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It's not much different from what you did. The acceleration of the belt is ##\frac{dV}{dt}##. In the non-inertial frame it manifests itself as a fictitious force ##ma## acting on the CM in the opposite direction as the acceleration of the belt. The torque equation about the point of contact on the belt gives ##MaR=(Mk^2+MR^2)\alpha## (note the use of he parallel axis theorem). Let ##a'_{cm}## be the acceleration of the CM in the non-inertial frame. Then ##a'_{cm}=\alpha/R##. Put this back in the torque equation, cancel the masses and get $$a'_{cm}=\frac{R^2}{k^2+R^2}a=\frac{R^2}{k^2+R^2}\left(-\frac{dV}{dt}\right).$$ The negative sign is introduced because ##a'_{cm}## and ##\frac{dV}{dt}## are in opposite directions. In the inertial frame, the acceleration of the CM is $$a_{cm}=a+a'_{cm}=\frac{dV}{dt}-\frac{R^2}{k^2+R^2}\frac{dV}{dt} =\frac{k^2}{k^2+R^2}\frac{dV}{dt} $$The velocity of the CM is$$v_{cm}(t)=\int a_{cm}dt=\frac{k^2}{k^2+R^2}\int \frac{dV}{dt}dt=\frac{k^2}{k^2+R^2}V(t).$$I am not convinced that this is a better way to approach the problem however.
 

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