Cylinder lying on conveyor belt

In summary: I was thinking. I was trying to apply that conservation of energy equation and it wasn't working, and I was getting confused by the fact that I was adding scalar values, and now I realize I was adding components of a vector, and it just cascaded from there. I'll try to get back to this when I have a clearer mind. Thanks.In summary, the problem involves a bottle of water being placed on a moving conveyor belt and can be approached as a single cylinder with radius ##R##, mass ##M##, and moment of inertia ##I = M R^2##. The velocity of the belt is given as ##V(t)## and it is assumed that the cylinder has a uniform mass distribution
  • #36
JD_PM said:
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)##.
JD_PM said:
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.
JD_PM said:
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".
 
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  • #37
JD_PM said:
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.
 
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  • #38
haruspex said:
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.
 
  • #39
kuruman said:
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.
 
  • #40
haruspex said:
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.
 
  • #41
kuruman said:
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.
 
  • #42
JD_PM said:
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 aCM is inthe reference frame of the belt. What happened to the fictional force?
 
  • #43
haruspex said:
From the above I deduce you are assuming ##f =Ma_{CM}##. Presumably aCM 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.
 
  • #44
JD_PM said:
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 aCM. It seems to me that to get that you must have assumed ##f=Ma_{CM}##. I can't see where you define aCM. 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.
 
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  • #45
haruspex said:
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 aCM. It seems to me that to get that you must have assumed ##f=Ma_{CM}##. I can't see where you define aCM. 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.
 
  • #46
JD_PM said:
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?
 
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  • #47
haruspex said:
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?
 
  • #48
JD_PM said:
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?
 
  • #49
JD_PM said:
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.
 
  • #50
jbriggs444 said:
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:

$$L = r p$$
 
  • #51
haruspex said:
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
 
  • #52
JD_PM said:
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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  • #53
jbriggs444 said:
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?
 
  • #54
JD_PM said:
Linear motion contribution to the angular momentum:
$$L = RMV(t)$$
No, V(t) is the belt speed. You need the cylinder's CoM speed.
JD_PM said:
But what about rotational contribution?
JD_PM said:
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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  • #55
haruspex said:
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.
haruspex said:
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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  • #56
JD_PM said:
$$v_{CM} = -\frac{I_{CM} \omega}{RM}$$.
Right, but you need another equation. Express the fact of rolling contact in terms of the radius, the rotation rate and the two velocities.
 
  • #57
haruspex said:
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.
 
  • #58
JD_PM said:
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.
 
  • #59
haruspex said:
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?
 
  • #60
JD_PM said:
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.
 
  • #61
haruspex said:
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:

$$v_{CM} + V(t) = R \omega$$
 
  • #62
JD_PM said:
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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  • #63
haruspex said:
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:

$$v_{CM} = \frac{1}{2}[(\frac{I_{CM}}{R^2M} + 1) R\omega + V(t)]$$
 
  • #64
JD_PM said:
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.
 
  • #65
haruspex said:
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).
 
  • #66
JD_PM said:
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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  • #67
haruspex said:
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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  • #68
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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  • #69
kuruman said:
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##
 
  • #70
JD_PM said:
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.
 
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<h2>What is a cylinder lying on a conveyor belt?</h2><p>A cylinder lying on a conveyor belt refers to a cylindrical object that is placed or positioned on a moving conveyor belt. The conveyor belt is a mechanical system used to transport objects from one location to another.</p><h2>How does a cylinder lying on a conveyor belt move?</h2><p>A cylinder lying on a conveyor belt moves due to the rotation of the conveyor belt. The belt is powered by a motor, which causes it to move at a constant speed. As the belt moves, it carries the cylinder with it.</p><h2>What factors affect the movement of a cylinder on a conveyor belt?</h2><p>The movement of a cylinder on a conveyor belt can be affected by several factors, such as the speed of the conveyor belt, the weight and size of the cylinder, and the friction between the cylinder and the belt. Other factors may include the angle of the conveyor belt and any obstacles or inclines on the belt.</p><h2>How is the speed of a cylinder on a conveyor belt calculated?</h2><p>The speed of a cylinder on a conveyor belt can be calculated by dividing the distance traveled by the time it takes to travel that distance. This is known as the average speed. However, the actual speed of the cylinder may vary due to changes in the conveyor belt's speed or other external factors.</p><h2>What are some common uses of cylinders on conveyor belts?</h2><p>Cylinders on conveyor belts are commonly used in manufacturing and production industries to transport and sort objects. They can also be used in warehouses and distribution centers to move products from one location to another. In addition, cylinders on conveyor belts are also used in packaging and shipping processes.</p>

What is a cylinder lying on a conveyor belt?

A cylinder lying on a conveyor belt refers to a cylindrical object that is placed or positioned on a moving conveyor belt. The conveyor belt is a mechanical system used to transport objects from one location to another.

How does a cylinder lying on a conveyor belt move?

A cylinder lying on a conveyor belt moves due to the rotation of the conveyor belt. The belt is powered by a motor, which causes it to move at a constant speed. As the belt moves, it carries the cylinder with it.

What factors affect the movement of a cylinder on a conveyor belt?

The movement of a cylinder on a conveyor belt can be affected by several factors, such as the speed of the conveyor belt, the weight and size of the cylinder, and the friction between the cylinder and the belt. Other factors may include the angle of the conveyor belt and any obstacles or inclines on the belt.

How is the speed of a cylinder on a conveyor belt calculated?

The speed of a cylinder on a conveyor belt can be calculated by dividing the distance traveled by the time it takes to travel that distance. This is known as the average speed. However, the actual speed of the cylinder may vary due to changes in the conveyor belt's speed or other external factors.

What are some common uses of cylinders on conveyor belts?

Cylinders on conveyor belts are commonly used in manufacturing and production industries to transport and sort objects. They can also be used in warehouses and distribution centers to move products from one location to another. In addition, cylinders on conveyor belts are also used in packaging and shipping processes.

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