Dynamics Question/Circular Motion

[Oblivion77]

Homework Statement

A constant speed conveyor belt delivers stones and drops them in a bin
as shown. The conveyor ends with a pulley with a 100 mm radius. The
coefficients of friction on the belt are μs = 0.20 and μk = 0. If the velocity
of the pulley is 0.1 m/s, what is the angle,theta, at which the stones fall off the
belt?

The Attempt at a Solution

I am kinda lost on this question, it says "when the stone falls off the belt" so I am guessing at this moment the normal force is 0? I also don't think there is any tangential acceleration, but I think there would be normal acceleration. If there is no normal force when the stone falls off the rock, would there then be no friction? This is what I got using this method, which doesn't seem possible.

I drew a FBD and got the equation mgCos(theta) = m(v^2/r), the m's will cancel and your left with gCos(theta) = v^2/r, solving for theta you get 89.42 degrees, which seems much too large.

 

[tiny-tim]

Hi Oblivion77!

… it says "when the stone falls off the belt" so I am guessing at this moment the normal force is 0?

No, you're thinking of loop-the-loops, where the stone would be under the belt.

Here, the stone will fall when the slope is so high that it just beats the friction.

I also don't think there is any tangential acceleration, but I think there would be normal acceleration.

Yes, that's right … the speed is constant, so the tangential acceleration is zero.

The normal acceleration is a matter of geometry, not physics … for speed v, it's … ?

(and then use N µN mg and a)

 

[Oblivion77]

So would I be using static coefficient of friction? Would the kinetic coefficient of friction come into play?

 

[tiny-tim]

No, µ_k won't be relevant … the stone will not slip until µ_s is "reached".

 

[Jaz]

Im doing the same assignment. No, Tim is wrong.
-yes the normal force will be zero when it falls off, that's what allows it to break free from the surface. This is true for any example.

-If you draw a free body diagram at the point where it falls off, you'll see that all you have is the weight MG and that as a result you have resultant forces in the normal AND TANGENTIAL direction. You can't assume that the velocity is constant. And it isnt.
WHY? Because when the block starts slipping the velocity of the block starts to increase, otherwise it wouldn't be slipping... slipping would mean that there exists a relative velocity between the belt and the block, if the velocity is constant (same as belt) this won't happen.

-Uk of 0 simply tells you that when the block is slipping the only thing that is changing its velocity is the force of gravity acting on it, because there is no friction. It makes calculations easier and allows you to use energy, which normally cannot be used with friction (at least with what we have learned) because this is a thermal energy.

 

[tiny-tim]

Welcome to PF!

Hi Jaz! Welcome to PF!

(have a mu: µ )

Yes, I think you're right … I was answering when the stone would slip, not when it would slip off.

Once it reaches the critical angle θ_k, the tangential component of g will be greater than µ_sN, and its tangential speed will become greater than v, with the motion being governed by µ_k, which in this case is given as zero.

So we find θ_k, and then treat it as if the stone were falling down a frictionless stationary cylinder, starting at θ_k with speed v (using energy, as you suggest).