A uniform solid disk of mass m = 2.93 kg and radius r = 0.200 m rotates about a

Question

A uniform solid disk of mass m = 2.93 kg and radius r = 0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 5.91 rad/s (a) Calculate the magnitude of the angular momentum of the disk when the axis of rotation passes through its center of mass. kg m2/s (b) What is the magnitude of the angular momentum when the axis of rotation passes through a point midway between the center and the rim? How do you calculate the moment of inertia about an axis that does not pass through the center of mass? kg·m2/s

Explanation / Answer

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part A :

Angular momentum L = I W

I is Moment of inertia = 0.5 mr^2

W is angular velocity = 2pif = 5.91 rad/s

so

L = (0.5 * 2.93 * 0.2^2) *(5.91)

L = 0.346 Kg m^2/s
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part B:

again L = I W

here I is the moment of inertia whose axis passes thorugh mid way

L = (0.5 m r^2 + m r1^2) * W

r1 = r/2 = 0.2/2 = 0.1 m

L = (0.5* 2.93* 0.2^2 + 2.93 * 0.1^2) * 5.91

L = 0.5194 kg m^2/s
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