A solid disk, which has a radius of 0.070 m and a mass of 0.88 kg, is rotating w

Question

A solid disk, which has a radius of 0.070 m and a mass of 0.88 kg, is rotating with an angular frequency of 12 rad/s about an axis perpendicular to its center. A thin ring, which has a radius of 0.040 m and a mass of 1.8 kg, is dropped straight down onto the disk so that they are concentric and rotating together. (Note: The ring is not rotating before the collision.)
a. What is the moment of inertia of the disk before the collision?




b. What is the moment of inertia of the disk and ring system after the collision?


c. What is the angular frequency of the system after the collision? Assume that the system is isolated so that angular momentum is conserved.

Explanation / Answer

moi of the disc is 1/2mr2

moi disc = .5 *.88 * .072 = .002156 kg m2/s2

moi of the system is moi1 + moi2

moi disc = .002156

moi ring mr2

= 1.8 * .042 = .00288

moi of system = .00288 + .002156 = .005036 kg m2/s2

use conservation of momentum. find the initial momentum first

L initial = 1/2 mr2

Li = .5 *.88 * .072 * 12

Li = .025872

now L of the system equals Li + L of the dropped ring

L of the dropped ring equals 0 since its not rotating

so inital = final.

.025872 = (moi1 + moi2)

.025872 = (.005036)

= 5.137 rad/s

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