A solid disk rotates in the horizontal plane at an angular velocity of 0.047 rad

Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.047 rad/s with respect to an axis at its center. The moment of inertia of the disk is 0.060 kg · m2. From above, a small object of mass m = 0.25 kg is dropped straight down onto this rotating disk at a distance of 0.30 m from the axis. The small object is considered as a particle rotating around the center of the disk.Use the conservation of angular momentum (Li = Lf) to

a)      Determine the total moment of inertia of the system disk-object after you drop the object.

b)      What is the new angular velocity of the disk with the object on it?

Explanation / Answer

Initial angular velocity w1 = 0.047 m/s
Initial moment of inertia I1 = 0.060 kg-m^2

Mass of ring m = 0.25 kg
Radius of ring r = 0.30 m
Moment of inertia of ring = mr^2 = 0.25 * 0.30^2 = 0.0225 kg-m^2
Total moment of inertia of ring + disk I2 = I1 + 0.0225 kg-m^2
Or I2 = 0.063 + 0.0225 = 0.0855
Let w2 = final angular velocity

By conservation of angular momentum,
I1 * w1 = I2 * w2
Or 0.060 * 0.047 = 0.0855 * w2
Or w2 = 0.060 * 0.047/0.0855
= 0.033 rad/s

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