Momentum and energy conservation in a vertical loop

[Eitan Levy]

Homework Statement

Two beads with masses of M and m are threaded on a vertical loop with radius of R.
M is released without velocity from a height of 1.5R from the bottom of the loop.
The collision between the beads is completely elastic.
What is the smallest mass M that will make the second bead reach the top of the loop?
What is the smallest mass M that will cause the second bead to not apply force on the loop when it reaches the top of it?

Homework Equations

Conservation of momentum: m₁v₁+m₂v₂=m₁u₁+m₂u₂
Conservation of energy: m₁v₁²+m₂v₂²=m₁u₁²+m₂u₂²

The Attempt at a Solution

Basically I can't understand how the first case is possible. I understand that in order for the second bead to be able to complete a full loop it will have to have a velocity of √(gR). However I can't understand how it is possible that the bead will reach the top without any pace (According to the answers I am supposed to just make sure that the second bead will have energy of 2mgR, without any velocity).
Any explanation would be appreciated.

 

[Doc Al]

However I can't understand how it is possible that the bead will reach the top without any pace

I think you mean that the second bead will just barely reach the top of the loop, with zero velocity at that point. Why do you think that is not possible? (I presume that second bead is resting at the bottom, prior to the collision.)

 

[Andrew Mason]

Homework Statement

Two beads with masses of M and m are threaded on a vertical loop with radius of R.
M is released without velocity from a height of 1.5R from the bottom of the loop.
The collision between the beads is completely elastic.
What is the smallest mass M that will make the second bead reach the top of the loop?
What is the smallest mass M that will cause the second bead to not apply force on the loop when it reaches the top of it?

Homework Equations

Conservation of momentum: m₁v₁+m₂v₂=m₁u₁+m₂u₂
Conservation of energy: m₁v₁²+m₂v₂²=m₁u₁²+m₂u₂²

The Attempt at a Solution

Basically I can't understand how the first case is possible. I understand that in order for the second bead to be able to complete a full loop it will have to have a velocity of √(gR). However I can't understand how it is possible that the bead will reach the top without any pace (According to the answers I am supposed to just make sure that the second bead will have energy of 2mgR, without any velocity).
Any explanation would be appreciated.

I assume that m is at rest at the bottom of the loop.

This is essentially a 2 dimensional elastic collision problem. The energy of the system is given by the initial potential energy of mass M. That energy is then distributed between m and M in the collision. You want it distributed in such a way that m has enough kinetic energy from the collision to get it to the top of the loop. You should be able to express what v_m (m's speed immediately after the collision) has to be in order to achieve that height. Use two dimensional collision analysis to express v_m in terms of m, M and v_M (the speed of M an instant before the collision) - (hint: see, for example: http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html#c1). That will enable you to determine a) whether it is possible and b) if it is, what m has to be.

AM

 

[Eitan Levy]

I think you mean that the second bead will just barely reach the top of the loop, with zero velocity at that point. Why do you think that is not possible? (I presume that second bead is resting at the bottom, prior to the collision.)

Because then the normal that will be applied on it will have to be negative (becuase mv^2/R=ΣF).

 

[Doc Al]

Because then the normal that will be applied on it will have to be negative (becuase mv^2/R=ΣF).

The normal force at what point? And why can't the normal be whatever it needs to be? (It's a loop, which can exert a sideways force in either direction, as required.)

 

[Eitan Levy]

The normal force at what point? And why can't the normal be whatever it needs to be? (It's a loop, which can exert a sideways force in either direction, as required.)

We were taught that that in order for a body to complete a full loop it has to have a minimum velocity of √(gR) when it reaches the top, because if it doesn't then the normal will have to be negative in order for it to fit the mv^2/R=ΣF formula. (That's what the textbook says and that's what I have used so far).

 

[Doc Al]

We were taught that that in order for a body to complete a full loop it has to have a minimum velocity of √(gR) when it reaches the top, because if it doesn't then the normal will have to be negative in order for it to fit the mv^2/R=ΣF formula. (That's what the textbook says and that's what I have used so far).

Ah, now I see the issue. That minimum velocity is for something on a track to maintain contact as it goes through the loop. A different problem! Here, the bead is threaded on a wire, so you don't have to worry about it falling off the "track". You don't even have to worry about the normal force at all.

 

[Eitan Levy]

Ah, now I see the issue. That minimum velocity is for something on a track to maintain contact as it goes through the loop. A different problem! Here, the bead is threaded on a wire, so you don't have to worry about it falling off the "track". You don't even have to worry about the normal force at all.

So why would they ask the second question? I get a correct answer there if I consider the normal force.

 

[Doc Al]

So why would they ask the second question? I get a correct answer there if I consider the normal force.

For the second question, you do have to have a minimum speed so that the normal force is zero. (When I said that you didn't have to worry about the normal force at all, I meant for the first question.)

 

[Eitan Levy]

For the second question, you do have to have a minimum speed so that the normal force is zero. (When I said that you didn't have to worry about the normal force at all, I meant for the first question.)

Why do I have to consider it in the second question but in the first I can just ignore it? This is what I can't understand.

 

[Doc Al]

Why do I have to consider it in the second question but in the first I can just ignore it? This is what I can't understand.

The first question only requires that the second bead make it to the top. No constraint on the normal force. The bead just needs to make it to the top.

But the second question explicitly requires that the force exerted by the loop (the normal force) be zero at the top. Now you must consider it.

 

[Eitan Levy]

The first question only requires that the second bead make it to the top. No constraint on the normal force. The bead just needs to make it to the top.

But the second question explicitly requires that the force exerted by the loop (the normal force) be zero at the top. Now you must consider it.

I am aware of that, but how is the bead able to make it to there physically in the first question?

 

[Doc Al]

I am aware of that, but how is the bead able to make it to there physically in the first question?

As long as it has enough speed after the collision, it will make it to the top. What's to stop it? (As already mentioned, the bead is threaded on the loop, so it can't fall off.)

Maybe this will help: Assuming there is no friction, mechanical energy is conserved. The bead will keep sliding up the loop until all its kinetic energy is converted to gravitational potential energy.

 

[Kavya Chopra]

If you've ever played loop the loop, in the game, we have to give a ball some velocity (kinetic energy) initially so that it can complete the loop. It's the same principle here.
Say, your friend challenges you to make it to the top of the loop. You don't need to do the full loop, just reach the topmost point. It reaches the topmost point and then just...stops. It has no more velocity to propel itself further. Also, a normal force is balancing the gravitational force(mg) exerted downwards as it doesn't move. And by the third law, the bead too, is exerting a normal force on the loop.

Now, in the second case, the bead exerts no force. So the normal force must be 0, and mg must be acting downwards. But hey- the loop is circular, and mg is acting towards the centre. So, why don't we equate it to the centripetal force to find out how much velocity it has so that our condition of no normal force is fulfilled?
And as a bonus, since the bead has a certain velocity at the top in this case, it comes down again. And your loop is completed.
Basically, in the first case, at the top the bead only has gravitational potential energy
But in the second, it has both gravitational potential energy AND kinetic energy.