A machine part is made from a uniform solid disk of radius R = 0.169 m and mass

Question

A machine part is made from a uniform solid disk of radius R = 0.169 m and mass M = 5.83 kg. A hole of radius R/2 is drilled into the disk, with the center of the hole at a distance R/2 from the center of the disk (the diameter of the hole spans from the center of the disk to its outer edge). What is the moment of inertia of this machine part about the center of the disk?

A machine part is made from a uniform solid disk of radius R = 0.169 m and mass M = 5.83 kg. A hole of radius R/2 is drilled into the disk, with the center of the hole at a distance R/2 from the center of the disk (the diameter of the hole spans from the center of the disk to its outer edge). What is the moment of inertia of this machine part about the center of the disk?

Explanation / Answer

Moment of inertia of solid disk around it's center is 1/2*m*r^2

1/2*5.83*.169^2=.0833

For rotation on it's edge:
I = Icm + MD^2
Icm = (1/2)MR^2
D = R
I = (1/2)MR^2 + MR^2
I = (3/2)MR^2

to find the mass of this, area of it over area of total will be in the ratio of their radius's squared.
r1^2/r2^2=m1/m2
(0.169/2)^2/(0.169^2)=x/5.83
(0.169/2)^2/(0.169^2)*5.83=1.4575

so (3/2)*1.4575*(.169/2)^2=.0156

.0833-.0156=.0677

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