# Momemnt of inertia, momentum problem

1. Oct 23, 2011

### jehan4141

A solid disk rotates in the horizontal plane at an angular velocity of 0.047 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.14 kg·m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance 0.22 m from the axis. The sand in the ring has a mass of 0.39 kg. After all the sand is in place, what is the angular velocity of the disk?

Wo = 0.047
Io = 0.14

R = 0.22
Ms = 0.39

Lo = Lf <----conservation of momentum

Io =If
WoIo = WfIf

If = MR2 = 0.39(0.222)

0.047(0.14) = 0.39(0.222)(Wf)
Wf = 0.35 ????

That is what I got but the answer is wrong.

2. Oct 23, 2011

### Staff: Mentor

The final moment of inertia is not just that of the sand--don't forget the disk itself.

3. Oct 23, 2011

### jehan4141

Would I just add the two moment of inertia together?

IF = 0.14 +[0.39(0.222)]
IF ~ 0.158876

WoIo = WFIF

WoIo / IF = WF

WF = 0.047(0.14) / 0.158876
WF ~ 0.414?

4. Oct 23, 2011

### Staff: Mentor

Right!

Everything's good except the very last line. (You misplaced a decimal point.)

5. Oct 23, 2011

### jehan4141

Thank you!!! yes I did miss a decimal point. thank you!