# Cylinder lying on conveyor belt

• JD_PM
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
JD_PM

## Homework Statement

You buy a bottle of water in the store and place it on the conveyor belt with the longitudinal axis perpendicular to the direction of movement of the belt. Initially, both the belt and the bottle are at rest. We can approach the bottle as one cylinder with radius ##R##, mass ##M## and moment of inertia ##I = M R^2##, in which the mass is not distributed uniformly. The speed of the belt at time ##t## is ##V (t)##.

a) Find an expression for the speed ##v(t)## of the centre of mass of the bottle.

## Homework Equations

Second Newton's Law for rotation:

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

## The Attempt at a Solution

I got two expression for the speed of the centre of mass of the bottle, and I wanted to know if both are equivalent and correct.[/B]

Using the second Newton's Law for rotation you end up with the expression:

EQ1:

$$R\alpha=2a_0$$

Additionally, the total acceleration is the sum of the tangential and centripetal vector components of the acceleration. Thus:

EQ2:

$$a = a_o + R\alpha$$

With this info we can deduce that:

EQ3:

$$a_0=\frac13 a$$

EQ4:

$$R\alpha=\frac23 a$$

EQ5:

$$f=\frac13 Ma$$

Therefore, using the second Newton's Law for rotation I got the following speed:

EQ6:

$$v_{CM} = \frac{3R \omega}{2}$$

And using kinematics:

$$v_f^2 = v_o^2 + 2a(x_f - x_o)$$

I got:

$$v_{CM} = \sqrt{ \frac{6fd}{M}}$$

Note that ##d## is the distance the rolling cylinder covers.

Are these correct and equivalent?

Thank you

JD_PM said:
We can approach the bottle as one cylinder with radius ##R##, mass ##M## and moment of inertia ##I = M R^2##, in which the mass is not distributed uniformly.
This statement is problematic. If the cylinder has radius ##R## and the moment of inertia about its axis is ##I = M R^2##, then all the mass must be at radius ##R##. If in addition the distribution is non-uniform, it can only depend on the angle. The velocity of the cylinder ##v(t)## will have to depend on the angular distribution of the mass and that is not given.

Setting that aside and assuming that the moment of inertia is just ##I## and the mass distribution has no angular dependence, it seems to me that since the velocity of the belt is given as ##V(t)##, the answer to the problem must have it on the right hand side. You should not assume that the acceleration of the belt is constant.
JD_PM said:
Additionally, the total acceleration is the sum of the tangential and centripetal vector components of the acceleration.
These components are perpendicular to each other. You cannot add them as scalars. You need to rethink the problem. Imagine the belt starting from rest moving with velocity ##V(t)## and acceleration ##\frac{dV}{dt}##. What is the instantaneous velocity of the point on the cylinder that is in contact with the belt?

Last edited:
JD_PM
JD_PM said:

## Homework Statement

You buy a bottle of water in the store and place it on the conveyor belt with the longitudinal axis perpendicular to the direction of movement of the belt. Initially, both the belt and the bottle are at rest. We can approach the bottle as one cylinder with radius ##R##, mass ##M## and moment of inertia ##I = M R^2##,
First, unless the water is frozen things will get very complicated, so let's assume it is.
Second, are you sure it doesn't say I≠MR2? Or maybe I≠½MR2?
JD_PM said:
Using the second Newton's Law for rotation you end up with the expression:

EQ1:

$$R\alpha=2a_0$$

Additionally, the total acceleration is the sum of the tangential and centripetal vector components of the acceleration. Thus:

EQ2:

$$a = a_o + R\alpha$$
I assume you are defining a0 as the acceleration of the centre of the cylinder.
Your eq 1 seems to be assuming I=½MR2, which is not the case here.
You don't mean centripetal acceleration here. It looks like you are defining a as the linear acceleration of the point of the cylinder in contact with the belt, so your eqn 2 is just adding an acceleration to a relative acceleration in the same direction. But be careful with signs. Which way will the cylinder be rotating in relation to its forward movement?

JD_PM
JD_PM said:
And using kinematics:
The equation you quote is only valid for constant acceleration. As @kuruman points out, you should not assume that.

kuruman said:
This statement is problematic. If the cylinder has radius ##R## and the moment of inertia about its axis is ##I = M R^2##, then all the mass must be at radius ##R##. If in addition the distribution is non-uniform, it can only depend on the angle. The velocity of the cylinder ##v(t)## will have to depend on the angular distribution of the mass and that is not given.

Setting that aside and assuming that the moment of inertia is just ##I## and the mass distribution has no angular dependence, it seems to me that since the velocity of the belt is given as ##V(t)##, the answer to the problem must have it on the right hand side.

Yeah I have been suggested before about the fact that ##V(t)## should appear on my ##v_{CM}## equation. This is because the motion of a rolling object is the combination of the translation and rotation! My bad.

This is what I think it happens:

So the translational velocity of the centre of mass is:

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

#### Attachments

• Captura de pantalla (444).png
3.1 KB · Views: 443
• Captura de pantalla (445).png
3.2 KB · Views: 843
kuruman said:
You should not assume that the acceleration of the belt is constant.

I do not understand why both you and haruspex suggested that.

kuruman said:
What is the instantaneous velocity of the point on the cylinder that is in contact with the belt?

##V(t)##

haruspex said:
Second, are you sure it doesn't say I≠MR2? Or maybe I≠½MR2?

Literally says: '...We can approach the bottle as one cylinder with radius ##a##, mass ##M## and moment of inertia ##I = M k^2## (##k## in units of length), in which the mass is not distributed uniformly...'

haruspex said:
Your eq 1 seems to be assuming I=½MR2, which is not the case here.

Absolutely, my bad! Actually, it is:

$$R\alpha=a_0$$

NOTE: I will continue posting in a few...

JD_PM said:
Literally says: '...We can approach the bottle as one cylinder with radius ##a##, mass ##M## and moment of inertia ##I = M k^2## (##k## in units of length), in which the mass is not distributed uniformly...'
Ok, that makes sense. k is not the radius of the cylinder, just some constant. I think it is called the radius of inertia.

JD_PM said:
I do not understand why both you and haruspex suggested that.
Because you are not told it is constant. Instead, you are given that the speed is V=V(t), i.e. some unknown function of time.
The equation you quoted under "and using kinematics" is only valid for constant acceleration. (And by the way, you mean kinetics, not kinematics.)

haruspex said:
Ok, that makes sense. k is not the radius of the cylinder, just some constant. I think it is called the radius of inertia.
I've heard it called the radius of gyration.
JD_PM said:
##V(t)##
Yes, I think it is safe to assume that the points of contact are at rest with respect to each other, i.e. kinetic friction does not play a role here. Now pretend that you are sitting on a chair placed on the belt, in other words transform to moving belt's frame. The belt is at rest relative to you and so are (instantaneously) the points on the cylinder that are in contact with the belt. Would you see the cylinder move? If so how would you analyze its motion in the belt's frame of reference? Once you do that, you can transform back to the room's frame of reference.

JD_PM
JD_PM said:
Absolutely, my bad! Actually, it is:

$$R\alpha=a_0$$

NOTE: I will continue posting in a few...

Back at it.

Thanks to the haruspex comment I highlighted on #9, I realized that my calculations were wrong because of the bad selection of moment of inertia equation. Corrections:

$$R\alpha=a_0$$

$$a_0=\frac12 a$$

$$R\alpha=\frac12 a$$

$$f=\frac12 Ma$$

$$v_{CM} = 2R \omega$$

Based on this, the speed ##v(t)## of the centre of mass of the bottle is:

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

JD_PM said:
Back at it.

Thanks to the haruspex comment I highlighted on #9, I realized that my calculations were wrong because of the bad selection of moment of inertia equation. Corrections:

$$R\alpha=a_0$$

$$a_0=\frac12 a$$

$$R\alpha=\frac12 a$$

$$f=\frac12 Ma$$

$$v_{CM} = 2R \omega$$

Based on this, the speed ##v(t)## of the centre of mass of the bottle is:

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

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?

kuruman said:
Now pretend that you are sitting on a chair placed on the belt, in other words transform to moving belt's frame. The belt is at rest relative to you and so are (instantaneously) the points on the cylinder that are in contact with the belt. Would you see the cylinder move?

OK you propose I am on a non inertial reference frame now. Yes, I would observe the cylinder rolling. The cylinder rolls because of the torque exerted by the external force of static friction on the cylinder. Actually, the cylinder would move in the opposite direction to that of the friction and in the same to that of the fictitious force F (as we are on a non inertial reference frame we have to add a reaction force F; the action force is clearly f).

JD_PM said:
Actually, the cylinder would move in the opposite direction to that of the friction and in the same to that of the fictitious force F (as we are on a non inertial reference frame we have to add a reaction force F; the action force is clearly f).
Careful here. In the non-inertial frame, the belt is at rest, yet the cylinder accelerates. An observer in that frame deduces that there must be an external force acting on the cylinder. In what direction is that force, what is its magnitude and at what point on the cylinder is it applied?

kuruman said:
Careful here. In the non-inertial frame, the belt is at rest, yet the cylinder accelerates. An observer in that frame deduces that there must be an external force acting on the cylinder. In what direction is that force, what is its magnitude and at what point on the cylinder is it applied?

Let's say the cylinder accelerates backwards. The external force exerted on the cylinder is the static friction ##f## ,which acts on the only point of contact with the belt, pointing backwards:

$$f = Ma_o$$

Where ##a_o## is the tangential acceleration.

But this is not the only force exerted on the system. There is a fictitious force ##F## acting on the COM of the system.

Here it is a picture of what is going on (there was a mistake in the attached image to this post, now fixing):

#### Attachments

• Captura de pantalla (446).png
1.6 KB · Views: 360
JD_PM said:
Where ##a_o## is the tangential acceleration.
Please clarify what you mean by that. Why isn't f/M the linear acceleration of the cylinder, i.e. of its mass centre?

Last edited:
JD_PM said:
There is a fictitious force F acting on the COM of the system.
I don't know what to write here. I would not have taken you along a non-inertial path. Not sure what @kuruman has in mind.

haruspex said:
I don't know what to write here. I would not have taken you along a non-inertial path. Not sure what @kuruman has in mind.
On a non inertial frame of reference we have the following scenario:

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

Though I am not sure, as I am wondering if they should be opposite instead...

Actually, after having some thoughts, I think F should point backwards and the friction force forwards. Thus the diagram flaws.

#### Attachments

• Captura de pantalla (447).png
1.7 KB · Views: 1,055
JD_PM said:
On a non inertial frame of reference we have the following scenario:

View attachment 236869

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

Though I am not sure, as I am wondering if they should be opposite instead...
Ok, though I believe they would be inopposite directions, but that doesn't matter. If so, the sign will come out negative.
So what equations follow from that?

haruspex said:
Ok, though I believe they would be inopposite directions, but that doesn't matter. If so, the sign will come out negative.
So what equations follow from that?
OK let's carry on with opposite directions. Using Newton's second law for translation we get:

$$-F + f = ma_o$$

$$a_o = \frac{-F + f}{m}$$

JD_PM said:
OK let's carry on with opposite directions. Using Newton's second law for translation we get:

$$-F + f = ma_o$$

$$a_o = \frac{-F + f}{m}$$
I assume you are now defining ao as the linear acceleration of the mass centre of the cylinder in the belt's reference frame.
Ok, so what about rotational acceleration.

haruspex said:
I assume you are now defining ao as the linear acceleration of the mass centre of the cylinder in the belt's reference frame.
Ok, so what about rotational acceleration.

OK before starting with rotation one question about signs. I think ##a_o## would be negative as ##F>f##... Does this mean that both forces should have the same direction instead?

JD_PM said:
OK before starting with rotation one question about signs. I think ##a_o## would be negative as ##F>f##... Does this mean that both forces should have the same direction instead?
Yes F will have a greater magnitude than f, but that just means the resulting observed acceleration is in the same direction as F,

haruspex said:
Yes F will have a greater magnitude than f, but that just means the resulting observed acceleration is in the same direction as F,

OK I think that what happens is that ##a_o## has a negative sign as well, because is pointing backwards. Thus:

$$a_o = \frac{F - f}{m}$$

haruspex said:
Ok, so what about rotational acceleration.

$$\tau = Rf$$

$$f = \frac{\tau}{R}$$

JD_PM said:
$$\tau = Rf$$

$$f = \frac{\tau}{R}$$
That does not mean anything unless you specify the axis.

JD_PM said:
(as we are on a non inertial reference frame we have to add a reaction force F; the action force is clearly f).
It looks like you are confused about action and reaction forces. Only action forces act on the system, here the cylinder. Reaction forces act on entities outside the system and are not part of the system's free body diagram. For example, the horizontal contact force of static friction acting on the cylinder has a reaction counterpart that is in the opposite horizontal direction and is part of the belt free body diagram, the action force of gravity ##m\vec g## acting on the cylinder has a reaction force ##-m\vec g## acting on the Earth, etc. etc.
haruspex said:
I don't know what to write here. I would not have taken you along a non-inertial path. Not sure what @kuruman has in mind.
I do not wish to detract @JD_PM from the current path. When this problem has been solved to everybody's satisfaction, perhaps I will post my solution for consideration.

JD_PM
kuruman said:
I do not wish to detract @JD_PM from the current path.
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.

haruspex said:
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.

kuruman said:
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 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?
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?

My solution is obtained from an inertial reference frame.

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

haruspex said:
That does not mean anything unless you specify the axis.

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

$$\tau = I \alpha$$

The angular acceleration is obtained from the stated equation:

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

• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
27
Views
3K
• Introductory Physics Homework Help
Replies
97
Views
3K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
1K