Momemnt of inertia, momentum problem

[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?

W_o = 0.047
I_o = 0.14

R = 0.22
M_s = 0.39

L_o = L_f <----conservation of momentum

I_o =I_f
W_oI_o = W_fI_f

I_f = MR² = 0.39(0.22²)

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

That is what I got but the answer is wrong.

 

[Doc Al]

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

 

[jehan4141]

Would I just add the two moment of inertia together?

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


W_oI_o = W_FI_F

W_oI_o / I_F = W_F

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

 

[Doc Al]

Would I just add the two moment of inertia together?

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

Right!

W_oI_o = W_FI_F

W_oI_o / I_F = W_F

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

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

 

[jehan4141]

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