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A solid disk rotates at an angular velocity of 0.049 rad/s with respect to an axis perpendicular to the disk at its center. The moment of intertia of the disk is 0.20 kgm2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

2007-03-20 13:26:21 · 1 answers · asked by mootle 1 in Science & Mathematics Physics

1 answers

The final spin rate is 0.035 rad/s.

Here's how that result is obtained:

There are two key assumptions,

(A) that the dropped sand effectively adheres to the rotating disk, staying in place and merely increasing the full moment of inertia. It brings no angular momentum in with it; and

(B) that angular momentum is conserved. That means that the final product (M.I.)*(ang. velocity) is equal to the initial value before the sand was dropped onto the disk.

The moment of inertia M_s of the sand is 0.50 kg * (0.4)^2 m^2 = 0.08 kgm^2. So the final moment of inertia M_f is 0.28kgm^2 That is 1.4 times larger than it was initially.

Then, by conservation of angular momentum, the final spin rate will be REDUCED by that same 1.4 factor. It will then be:

(1/1.4) (0.049) rad/s = 0.035 rad/s.

QED

LIve long and prosper.

2007-03-20 13:36:52 · answer #1 · answered by Dr Spock 6 · 0 0

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