Uniform cylinder, conveyer belt

[daveed]

so you have a uniform solid cylinder with mass M and radius R at rest, and its placed on a conveyer belt moving at speed V. coefficients of friction are nonzero, how do you figure out the final velocity of the COM of the cylinder?

i don't know, this problem is confusing because you would think it would start moving with the conveyer belt, but then because its rolling in the other direction its screwy. any ideas to how this system works?

 

[DaveC426913]

I don't know the math, but intuitively, I can see that this is exactly equivalent to a cylinder, moving at speed V, being placed on a flat, stationary surface.

Does that make things easier to calculate?

 

[daveed]

im getting the idea that either the wheel starts to go backwards at V or starts to stay in place.

what do you think?

 

[krab]

Hint: If the cylinder had had all of its mass concentrated on its axis (this means the moment of inertia is zero), then it would have no reason for the COM to begin moving. It would just remain stationary, but spinning. This demonstrates that the moment of inertia of the cylinder figures in the calculation. OTOH, if all the mass were concentrated on the rim, the COM would surely have a non-zero speed. Since the cylinder is solid, the v of the COM is somewhere between zero and the v of the conveyor belt.

 

[arildno]

Hint:
Use conservation of angular momentum with respect to the moving contact point of the conveyor belt.

 

[daveed]

how do you do angular momentum w/o a mass associated with the conveyer belt? torque*time, but w/o a time?

O.O

 

[arildno]

I meant of course the angular momentum of the CYLINDER, computed with respect to the moving contact point on the conveyor belt.

(I read the exercise such that the conveyor belt has a constant speed V)

 

[daveed]

but the conveyer belt initially applies a torque to the cylinder, change in angular momentum is torque*time... and... i don't see it.

sorry

 

[rcgldr]

Since all you know about the coefficient of friction is that it's non-zero, then it must not matter in determining the final speed of the cylinder. The rate of acceleration must not matter either. You just end up with a final velocity and rate of rotation, for a given change in velocity at the surface of the cylinder.

The result is a change in the linear and rotational momentum and kinetic energy. At this point I'm stuck, I might get back later on this.

The moment of inertia for uniform cylinder is $$1/2\ m\ r^2$$

 

[Doc Al]

cute problem

Think of the problem this way: When the cylinder is placed on the moving belt, friction will exert a force that will produce a (1) rotational and (2) translational acceleration of the cylinder. So apply Newton's 2nd law (twice). Hint: The cylinder will keep accelerating until its surface speed just matches the speed of the belt.

 

[daveed]

Jeff:
yeah, that's the dilemna that i was in
doc al:
with the acceleration, i got that the rotation would accelerate it against the direction of the belt twice as much as the translational acceleration from the belt, but i didnt know how it would end up. i was thinking that it would head so that the rollingness would be exactly the same as the conveyer belt's speed and it would end up spinning in place

...does that sound about right?

 

[Doc Al]

I'm not sure what you are saying. It's true that the rotational acceleration (of the cylinder surface with respect to the center) is twice the translational acceleration. But, no, the cylinder doesn't end up spinning in place.

 

[daveed]

would it be moving backwards, then? this is really confusing... it can't end up slide partly because to slide would mean the conveyer belt is torque-ing the cylinder. so would it roll backwards, then?

 

[Doc Al]

As I stated in a previous post, when the cylinder is placed on the moving belt, friction will exert a force ($F = \mu mg$) that will produce a (1) rotational and (2) translational acceleration of the cylinder. So let's apply Newton's 2nd law:
(1) $\tau = \mu mg R = I \alpha$. Since $I = 1/2 mR^2$, $\alpha = 2\mu g/R$.
(2) $\mu mg = m a$, thus $a = \mu g$.
The acceleration of the cylinder surface with respect to the center is $\alpha R$, so the net acceleration of the surface is $a + \alpha R$. The cylinder will accelerate until its surface speed just matches the speed of the belt, so:
$v = (a + \alpha R)\Delta t$ ==> $\Delta t = v/(3\mu g)$.
Now find the speed of the center in that time:
$v_{cm} = a \Delta t = \mu g v/(3\mu g) = v/3$

Make sense?

 

[arildno]

An alternative is conservation of angular momentum with respect to the moving contact point on the conveyor belt.
First, justifying conservatiob of angular momentum:
1) It is easy to see that the forces acting on the cylinder (gravity and friction) cannot produce any torques with respect to the contact point.
2) In the moment-of-momentum equation, the cross-product term between the the contact point velocity and the C.M velocity is zero, because these velocities are parallell.

Hence, as stated angular momentum is conserved.
Initially, the cylinder is at rest, so the angular momentum is $$\vec{0}$$
At any other point of time, we therefore have:
$$\vec{0}=\vec{r}\times{M}\vec{v}_{C.M}+I\vec{\omega}$$
Where:
$$\vec{r},\vec{v}_{C.M}=v_{C.M}\vec{i},\vec{\omega}$$
are distance from contact point to C.M, velocity of C.M and angular velocity of cylinder, respectively.
M is the mass of the cylinder, whereas I is the moment of inertia of the cylinder with respect to the C.M.

Friction stops acting when rolling is achieved (rolling friction neglected), that is:
$$V\vec{i}=v_{C.M}\vec{i}+\omega\vec{j}\times(-R\vec{k})$$
Or:
$$\omega=\frac{v_{C.M}-V}{R}$$

We have:
$$\vec{r}\times{M}\vec{v}_{C.M}=M(R\vec{k})\times(v_{C.M}\vec{i})=MRv_{C.M}\vec{j}$$
Hence, we get:
$$\frac{I}{R}V=(MR+\frac{I}{R})v_{C.M}$$
Or, with $$I=\frac{1}{2}MR^{2}$$
$$v_{C.M}=\frac{1}{3}V$$
which agrees with Doc Al's answer.

 

[krab]

would it be moving backwards, then? this is really confusing... it can't end up slide partly because to slide would mean the conveyer belt is torque-ing the cylinder. so would it roll backwards, then?

Daveed: I think you find this confusing because you are not clear on your reference frame. For someone standing on the conveyor belt, yes, it is moving backward. For someone standing on the ground, watching, it is moving forward with the belt, but not as fast as the belt. Only a third as fast, as Doc Al's and Arildno's excellent analyses show. I especially like Arildno's since it shows the limiting case I mentioned. If I=0, the cylinder would not move but spin in place. OTOH, if I= very large (one could imagine that the cylinder has a much larger diameter part that extends on either side of the belt), then it will not spin at all and the cylinder will move at the speed of the belt.

 

[daveed]

haha yeah, i think i get it now, thanks guys =)