The moment of inertia of a uniform-density disk rotating about an axle through i

Question

The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be (1/2)MR2. This result is obtained by using integral calculus to add up the contributions of all the atoms in the disk (see Problem 9.3). The factor of 1/2 reflects the fact that some of the atoms are near the center and some are far from the center; the factor of 1/2 is an average of the square distances. A uniform-density disk whose mass is 15 kg and radius is 0.15 m makes one complete rotation every 0.3 s.

(a) What is the moment of inertia of this disk?
I = kg·m2
(b) What is its rotational kinetic energy?
Krot = J
(c) What is the magnitude of its rotational angular momentum?
Lrot = kg·m2/s

Explanation / Answer

Given Moment of inertia of disk , I = 1/2 MR^2 Mass of the disk , M = 15 kg Radius of the disk , R = 0.15 m Angular velocity = 1 rot / 0.3 s = 2 rad / 0.3 s = 20.93 rad /s a) The moment of inertia of the disk is ,       I = 1/2 (15kg) (0.15m)^2 = 0.16875 kg m^2 b) The rotational Kinetic energy      KErot =   1/2 I 2 = 0.5 *( 0.16875 kg m^2) *(20.93 rad /s)^2                = 37 J c) Magnitude of  rotational angular momentum is        L = I          = (0.16875 kg m^2) * 20.93 rad/s       L = 3.53 kg m^2 /s

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