# Rotating rod with little rings that slides out

1. Dec 16, 2011

### amiras

1. The problem statement, all variables and given/known data

A uniform rod of mass M and length L rotates in horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass m are mounted so that they can slide along the rod. Rod rotates with the initial angular speed of ω.

2. Relevant equations

What is the angular speed of the rod after the rings leave it?

3. The attempt at a solution

The answer books says, that it is the same angular speed as the rings would have slided to the both ends of the rod and stuck there. Now knowing this it would be a simple use of conservation of angular momentum.

However I do not understand why the rod rotates with the same angular speed in both of these cases:
1. The little rings slided to the ends of the rod.
2. The little rings slided out completely out of rod.

The moments of inertia at both cases are different, so how come they have the same angular speed?

2. Dec 16, 2011

### Delphi51

It seems to me it is kind of like a car falling off a bridge. If a wheel comes loose and falls separately, they both continue at the original speed. When those rings fly off they take with them the same angular momentum that they had while on the rod, leaving the rod with the same share of angular momentum it originally had.

3. Dec 17, 2011

### amiras

For example why the energy is not conserved, or is not conserved in the way I think it should.

Lets say at the time the rings was at fixed distance r from the center rod rotated at ω₁, then they slided to the ends of the rod L/2 and rod rotates at ω₂.

So the conservation of energy should give:

1/2 I₁ω₁^2 = 1/2 I₂ω₂^2

But it does not gives the right result, the right result is from I₁ω₁ = I₂ω₂

4. Dec 17, 2011

### JHamm

The two rings will move at some angle to their position vector, since there is no torque on them their angular momentum is conserved.

Setting up an energy conservation -
$I_1 =$ moment of inertia of the rod.
$I_2 =$ moment of inertia of each ring at the end of the rod.
$v =$ velocity of each ring after it leaves the rod.
$\omega_1 =$ angular velocity of rod before the rings leave.
$\omega_2 =$ angular velocity of rod after the rings leave.
$$\frac{1}{2}(I_1 + 2I_2)\omega_1^2 = \frac{1}{2}I_1\omega_2^2 + \frac{1}{2}(2m)v^2$$
There is no work done on the rings therefore
$$\frac{1}{2}2I_2\omega_1^2 = \frac{1}{2}2mv^2$$
$$\displaystyle{\dot{..}} I_1\omega_1^2 = I_1\omega_2^2 \Rightarrow \omega_1 = \omega_2$$

Last edited: Dec 17, 2011
5. Dec 17, 2011

### Delphi51

Not all that clear, but in view of the "same angular speed as the rings would have slided to the both ends of the rod", the question must be asking how the speed changes at the instant the rings fly off. NOT in sliding out to the edges of the rod. The speed WOULD change due to them sliding to the ends, but not due to the flying off.