Circular Motion/Conservation of energy problem

[ahsanxr]

Homework Statement

[PLAIN]http://img180.imageshack.us/img180/8643/mechanicsquestion.jpg

Homework Equations

Conservation of Energy

The Attempt at a Solution

I cannot seem to understand the motion involved, especially because the string is inextensible. That means that while the string is taut, if P moves upwards, for example, Q will have to go downwards and vice versa. I tried to do it by considering conservation of energy on P only. I got the thing the question asked to show but with an extra 2gsin(theeta)sin(Beeta) at the end.

So help me out please?

 

[pgardn]

Homework Statement

[PLAIN]http://img180.imageshack.us/img180/8643/mechanicsquestion.jpg

Homework Equations

Conservation of Energy

The Attempt at a Solution

I cannot seem to understand the motion involved, especially because the string is inextensible. That means that while the string is taut, if P moves upwards, for example, Q will have to go downwards and vice versa. I tried to do it by considering conservation of energy on P only. I got the thing the question asked to show but with an extra 2gsin(theeta)sin(Beeta) at the end.

So help me out please?

I have a way of looking at it. But the geometry involved will be your own.

I interpret this problem as follows:

Both masses or rings will have the same rotational Kinetic energy if the massless "rod or string" stays taut once they both start to slide on the big frictionless ring. And both rings will have the same change in height above a specified point of your choosing in the y-direction. So basically one little ring mass is gaining gravitational U energy, while the other loses the same amount of gravitational U energy, while both gain the same amount of rotational kinetic energy.

Now finding a good reference point in the y-direction, and then the change in y, looks like some geometry needs to be done. You can find omega (angular speed) for both the ring masses if you establish what change the change in y is. And then its just a matter of moving variables around and maybe converting between linear and rotational stuff as the conservation of energy allows you to set this all up.

This is how I see the problem without doing it.

 

[tiny-tim]

Hi ahsanxr!

(have a beta: β and a theta: θ )

That means that while the string is taut, if P moves upwards, for example, Q will have to go downwards and vice versa. I tried to do it by considering conservation of energy on P only. I got the thing the question asked to show but with an extra 2gsin(theeta)sin(Beeta) at the end.

This is just geometry … you simply have to find the difference in height from the equilibrium position.

Your basic method is correct, but I don't understand where your sines came from.

Show us the part of your calculation that involves the heights. ​

 

[ahsanxr]

Thanks for replying, guys. Pgardn, that's how I interpreted it too, but I probably messed up in the geometry. Tiny-tim, here's how I did it:

I took the initial position as the point where θ = 0. Taking the centre of the circle as the reference point for PE, the height of P is acosβ. Then I moved P up as it is in the diagram so that the remaining angle inside the triangle is (β-θ) and the height above the reference point of P is now acos(β-θ).

Using conservation of energy:

Initial PE = Final PE + Final KE

mgacosβ = mgacos(β-θ) + 0.5mv^2

expanding cos(β-θ), dividing both sides by m, and multiplying both sides by 2 we get

2gacosβ - 2gacosβcosθ - 2gasinβsinθ = v^2

Putting v = rθ(dot), where θ(dot) is the angular speed,

(a^2)θ(dot)^2 = 2gacosβ(1-cosθ) - 2gasinβsinθ

Dividing both sides by a,

aθ(dot)^2 = 2gacosβ(1-cosθ) - 2gasinβsinθ

Is what I'm getting. PLEASE point out my error. It will probably be a really stupid one :P

 

[tiny-tim]

Hi ahsanxr!

I took the initial position as the point where θ = 0. Taking the centre of the circle as the reference point for PE, the height of P is acosβ. Then I moved P up as it is in the diagram so that the remaining angle inside the triangle is (β-θ) and the height above the reference point of P is now acos(β-θ).

Using conservation of energy:

Initial PE = Final PE + Final KE

mgacosβ = mgacos(β-θ) + 0.5mv^2

expanding cos(β-θ), dividing both sides by m, and multiplying both sides by 2 we get

2gacosβ - 2gacosβcosθ - 2gasinβsinθ = v^2 …

Yes, that's fine, except your conservation of energy equation must deal with P and Q together …

(conservation of energy only works if there are no external forces, other than the PE force … if you consider P and Q together, that's true, but if you consider P on its own, the tension in PQ is an external force)

… so you'll end up with a cos(β-θ) and a cos(β+θ).

 

[ahsanxr]

Yup, I got it. I asked my teacher this question and he told me the same thing. The gain in KE is equal to the net change in the PE's of P and Q. We can't simply consider one of the rings.

Thanks for your help.

And pgardn, I think what you're saying might be wrong. If P gains the same amount of PE that Q loses, where will the gain in KE come from? There must be a net change in PE.