# Angular speed of rod with 2 masses seems EZ but NOT

1. Oct 20, 2004

### Elbhi

Why cant I get this obviously KE is conserved so I use
I(initial) x Angular speed(initial)=I(final) x Angular speed(final)

A uniform rod of mass 3.05×10-2kg and length 0.440m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.230kg are mounted so that they can slide along the rod.
They are initially held by catches at positions a distance 4.90×10-2m on each side from the center of the rod, and the system is rotating at an angular velocity 25.0 rev/min. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

What is the angular speed of the system at the instant when the rings reach the ends of the rod?

I have tried so many different combinations of inertia and still im not getting the right answer

I know angular velocity (final)=( I(initial)xAngular velocity(initial) ) / I(final)

For I initial I think I=1/12 (M)(L^2)+ MR^2 of the two masses giving me

1) FOR ROD 1/12(3.05×10-2kg)( 0.440m)^2

2) FOR 2 Masses 2(0.230kg)( 4.90×10-2m^2)

3) I add 1 & 2 together then multiply by 25rev/min.

4) For I(final) since the masses are now at the end of the rod, the inertia of the rod changes to 1/3MR^2

5) I add 4 to 2 and I think that should get me I(final)

6) divide 3/5

Am I on the right track at least!!!! Appreciate any feedback thanks

2. Oct 20, 2004

### sal

I just skimmed this, but I couldn't help noticing you said:
In (4) you need to compute a new I for the rings, not the rod, and then add the value for (4) to (1) to get I(final). Right?

Maybe, on the other hand, you just said it wrong and did it right ... I didn't actually grind numbers, sorry.

3. Oct 20, 2004

### Tide

BTW, Elbhi, you're talking about conservation of angular momentum and NOT conservation of kinetic energy.