a disc of the radius of R1 is cut out from a disc of the radius R2. If disc a disc of the radius of R1 is cut out from a disc of the radius R2. If disc left mass M then find moment of inertia of the disc left about an axis passing through it's centre and perpendicular to the plane
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If the center of the cut out disc is the same as the center of the original disc, then it is simple.
Surface density of the disc = d = M/ [π (R2² - R1²)]
Mass of disc of radius R1 = M1 = d * π R1²
M1 = M * R1²/ [R2² - R1²]
Moment of inertia of disc of radius R1: I1 = M1 R1² /2
I1 = M R1⁴ / [2 (R2² - R1²) ]
Moment of Inertia of the original disc (without cutting anything):
I = (d * π R2²) * R2² /2 = M R2⁴ /[2(R2² - R1²) ]
Moment of inertia of the left out disc:
I2 = I - I1
= M (R2⁴ - R1⁴) /[2 (R2² - R1²)]
= M (R2² + R1²) / 2
Surface density of the disc = d = M/ [π (R2² - R1²)]
Mass of disc of radius R1 = M1 = d * π R1²
M1 = M * R1²/ [R2² - R1²]
Moment of inertia of disc of radius R1: I1 = M1 R1² /2
I1 = M R1⁴ / [2 (R2² - R1²) ]
Moment of Inertia of the original disc (without cutting anything):
I = (d * π R2²) * R2² /2 = M R2⁴ /[2(R2² - R1²) ]
Moment of inertia of the left out disc:
I2 = I - I1
= M (R2⁴ - R1⁴) /[2 (R2² - R1²)]
= M (R2² + R1²) / 2
kvnmurty:
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