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

w1 = 0.047 rad/s  

I1 = 0.060 kg m^2

(A) I2 = I1 + (0.25 x 0.30^2)

= 0.060 + 0.0225

= 0.0825 kg m^2

(B) Li = Lf

0.060 x 0.047 = 0.0825 wf

wf = 0.034 rad/s

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