Projected Block and Conveyor Belt

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1. Nov 1, 2016

decentfellow

1. The problem statement, all variables and given/known data
It is a Matching Type Problem(a single entry in a column can be matched with more than one in the other column):-

A block is projected with an initial velocity $v_{\text{Block}}$ on a long conveyor belt moving with velocity $v_{\text{Belt}}$(at that instant) having constant acceleration $a_{\text{Belt}}$. Mark the correct option regarding friction after long time (coefficient of friction between the block and belt$=\mu$). If:

2. Relevant equations

3. The attempt at a solution
In the first entry in Column-1 it is mentioned that the conveyor belt has no acceleration, I was really confused about it that whether the conveyor belt is allowed to have an acceleration greater than 0, because if it is not so then how could the friction act to accelerate the conveyor belt and deaccelerate the Block as the conveyor belt is constrained to move with zero acceleration.

In all the entries what is troubling me is that whether the conveyor belts are constrained to move with the acceleration indicated only (or in that range only) or can they be acted upon by the friction to change their acceleration for sometime in an attempt to make the block and the conveyor belt move together.

2. Nov 1, 2016

haruspex

That is how I read it.

3. Nov 1, 2016

decentfellow

So, if the conveyor belt is constrained to move with the acceleration as given in the entries, then doesn't it mean that the friction cant act as that changes the acceleration of the conveyor belt and block to stop any slipping. In that case all the entries will have zero friction acting b/w the surfaces, and that is not the answer that is given.

4. Nov 1, 2016

haruspex

Not at all. You have to assume there is a drive mechanism that forces the belt to accelerate at the given rate regardless of any other small forces acting on it.

5. Nov 1, 2016

decentfellow

Consider the first case-entry in Column-1 in which the conveyor belt doesn't have any acceleration. If there is a driving mechanism that keeps the acceleration constant whatsoever the small forces act on it, then due to $v_{\text{Block}}\gt v_{\text{Belt}}$ kinetic friction acts b/w the conveyor belt and the block to resist slipping which occurs b/w there surfaces. This leads to friction force acting in the forward direction on the conveyor belt hence changing its acceleration (increase in the magnitude) and deaccelerating the block. But, as there is a mechanism present that doesn't let the acceleration of the conveyor belt get affected by the small forces acting on it (considering the kinetic friction to be a small force) then the driving mechanism exerts an equal and opposite force on the conveyor to oppose the force of friction hence keeping the acceleration of the conveyor belt same as that before the block was projected on to the conveyor. In this scenario the block is the only one which is experiencing a change in acceleration (or deaccelration) and acquires a speed that is same as that of the conveyor belt.

Now on to the part that gives the answer that is needed in the book. As the conveyor belt doesn't move with accelerated motion hence after the initial phase (the one that is explained elaborately in the above para) there is no friction force acting b/w the surfaces of the block and the conveyor, because after the initial phase the blocks have already attained the no slipping b/w there surfaces.

After thinking a lot about the accelerated cases in column-1, the explanation that I came up with was as follows:-

When the block is projected on to the conveyor belt the conveyor belt due to having the drive mechanism doesn't experience any change in its acceleration but the block does experience retardation due to kinetic friction acting b/w the belt and the block and attains the same velocity as that of the conveyor belt. After this phase consider the following three cases:-
1. If $a_{\text{Belt}} \gt \mu g$ then after a long time the velocity of the conveyor belt would be greater than the block as the acceleration provided by the static friction is less than that which the conveyor belt has (acceleration of conveyor belt) so instead of static friction kinetic friction is acting as there is slipping b/w the surfaces.
2. If $a_{\text{Belt}} = \mu g$ then after a long time the velocity of the conveyor belt will be same as that of the velocity of the block due to the acceleration provided by the maximum static friction being the same as that provided by the conveyor belt, so as it is possible to have a no slipping scenario in this case if maximum static friction acts b/w the surfaces, so maximum static friction (or limiting friction) acts b/w the block and the conveyor belt.
3. If $a_{\text{Belt}} \lt \mu g$ then after a long time the velocity of the conveyor belt and the block can be same as the acceleration of the conveyor belt and the acceleration provided by the static friction(which is not the limiting friction in this case) is same, so the friction that acts b/w the surfaces is less than limiting friction.
Is the above explanation correct for the different cases.

The edits made have been italicized.

Last edited: Nov 1, 2016
6. Nov 1, 2016

haruspex

I agree with the answers as regards which items from the second column match, but your explanations confuse velocity and acceleration in some places.
Case C is interesting. The question statement does not discriminate kinetic from static friction, so take them to be the same. How would the answer depend on the initial velocity of the block?

7. Nov 1, 2016

decentfellow

I have made some edits in post #5 can you please read them and tell me if my explanation is more clear now.
Are you referring to the humungous explanation that I had written in the first para. I did it just to examine if something unusual was happening before the first time the no slipping motion is attained, and after doing that I understood that it doesn't really hold any meaning.

8. Nov 1, 2016

haruspex

the acceleration provided by the maximum static friction is that provided by the conveyor belt.
I think you mean that it is the same as the acceleration of the conveyor belt.

9. Nov 2, 2016

decentfellow

Yeah, exactly but your wordings are a lot clear.