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Cylinder lying on conveyor belt

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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 $a$, mass $M$ and moment of inertia $I = M k^2$ ($k$ in units of length), 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.
b) Explain why a bottle tends to start spinning on one moving band.

What I have tried:

a) Having both longitudinal axis and belt's velocity vector perpendicular to each other on the same plane means that there will be rolling motion if we assume that static friction will be overcome. Let's disregard slippery effects as well.

Here I used an approach based on the fact that the bottle-belt-Earth system is nonisolated in terms of energy because the belt exerts an external force on the bottle, which means that it does work on the bottle. Therefore:

$$W = \Delta K_{tr} + \Delta K_{rot}$$

$$(F-f_f)d =v_{CM}^2 (\frac{I_{CM}}{a^2} + M)$$

$$v_{CM} = \sqrt{ \frac{2(F-f_f)d}{\frac{I_{CM}}{a^2} + M}} = \sqrt{ \frac{(F-f_f)d}{ M}}$$

The result I got does not seem to be incorrect; I checked dimensions and got $\frac{L}{T}$ .My question here is if I am right including the force F. I consider this force as a fictitious one, which is triggered by the non inertial frame in which the bottle is located: the conveyor belt. Note I am analysing the scenario from an inertial reference frame: The Earth (we regard it as inertial for known reasons).

b) If the static friction force was equal to the fictitious force F the cylinder would not spin. It would just move transitionally along the belt. Actually, if we were located on the belt, we would not see the cylinder move at all.

In order to spin, the magnitude of F must exceed the magnitude of the maximum force of static friction. This friction force is recalled as force of kinetic friction.

EDIT

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:

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

This makes sense to me. However I am confused because I was suggested that I should not assume that the belt moves with constant acceleration (please see kuruman's #2 comment https://www.physicsforums.com/threads/cylinder-lying-on-conveyor-belt.963471/ ); Then how could I solve this problem?

asked Dec 22, 2018 in Physics Problems by Jorge Daniel (606 points)
edited Jan 31 by Jorge Daniel
(a) If you are using an inertial reference frame, ie one in which Newton's Laws apply, then you should not use fictitious forces. Fictitious forces only appear in **non-inertial frames** which are accelerating in some way.

In an inertial frame the only horizontal force on the cylinder is the force of friction.


(b) Not true.

The fictitious force and the friction force do not act in the same straight line. The fictitious force acts at the COM of the cylinder, the friction force acts at the point of contact with the belt. If these forces are equal and in the opposite direction they form a couple which does rotate the cylinder. There would only be no rotation if they acted in the same straight line.

If there is no slipping at the contact point then the friction force is **static** not kinetic.
a) OK. Since we are observing the system from an inertial reference frame we get:

$$v_{CM} = \sqrt{ \frac{f_f d}{ M}}$$

Letting the negative sign be absorbed by $f_f$.

b) OK I understand your explanation. However, should the fictitious force F be included in the analysis of spinning if it is being analysed from an inertial frame of reference (as it is the case)?
(a) $F$, $f$ and $d$ are all unknown, so this is not a useful answer to the question. Your answer does not include $V(t)$ which affects the motion. $I/a^2+M=\frac32M$ not $2M$ so your calculation is incorrect anyway. There is a factor of 2 missing from your middle line.

(b) Read my first comment again. Fictitious forces do not appear in inertial frames of reference.  Inertial frames are constant velocity frames. Non-inertial frames are accelerating frames. If the belt is accelerating then a frame in which the belt is stationary is a non-inertial frame. (Confusingly, fictitious forces are also called "inertial forces".)
a) From an inertial reference frame:

$$v_{CM} = \sqrt{ \frac{2fd}{\frac{I_{CM}}{a^2} + M}} = 2\sqrt{ \frac{fd}{3M}}$$

If this is not a useful answer, should I solve  $v_{CM}$ by kinematics?

b) What I meant is that your explanation is based on the fictitious force F, which means that you are analysing the scenario located on the belt.

In order to justify the spinning of the bottle form an inertial reference frame I would say that the friction force is triggered by the acceleration that the belt gives to the bottle.
You might find my answer to the following question helpful : [rotation of wheels and axle in accelerating truck](http://physics.qandaexchange.com/?qa=470/rotation-in-truck).
Thank you Professor, I think I got it!

a) No work is done by the static force of friction because the point of application of the force moves through no displacement (there is no slipping at the contact point). Therefore, **work-energy theorem is useless here**.
I am not asked for computing neither the linear acceleration nor the angular one, but for the sake of curiosity:

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

$$\alpha = \frac{\tau}{I} = \frac{2f}{ma}$$

Where $a_o$ is the linear acceleration and $a$ is the radius (sorry for the confusing notation).
Here comes my conclusion for a):

Due to the fact that there is no external force doing work on the system, the cylinder **will not have translational  motion**, which means:

$$v_{CM} = 0$$

I hope I got it right this time.

1 Answer

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Best answer

In essence this question has been answered already in rotation of wheels and axle in accelerating truck. There I show that the linear acceleration $a_0$ (relative to the ground) and the rotational acceleration $\alpha$ of a solid disk or cylinder which rolls without slipping on a lorry (or conveyor belt) which has translational acceleration $a$ are given by $$a_0=\frac13 a$$$$R\alpha=2a_0=\frac23 a$$ where $R$ is the radius of the cylinder.

The friction force is $$f=\frac13 Ma \le \mu Mg$$ so to avoid slipping the acceleration of the belt must be limited to $a \le 3\mu g$.

answered Jan 2 by sammy gerbil (28,448 points)
selected Jan 2 by Jorge Daniel
I was wrong stating that:

$$v_{CM} = 0$$

Because the cylinder has translational motion due to friction.

I got two equivalent expressions for the $v_{CM}$:

1) From 2nd Newton's Law:

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

2) From kinematics:

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

This result is obtained based on the fact that the cylinder has no initial velocity and starts its translational motion from the origin.

Note that *d* is the distance the rolling cylinder covers.
OK I am still confused on how can I get the speed of the centre of mass of the bottle. I understand what you did for getting the acceleration, so I guess that analogously we can derive the the speed of the centre of mass as:

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

Do you agree?

*EDIT*

Please see my original question again; I posted a new approach.
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