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A cylindrical rod of uniform density is located wi

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Answer #1

Rotational kinetic energy = 0.5*I*w^2

w = 2*pi/T = 2*pi/0.02

w = 100*pi

I = moment of inertia of rod along x-axis

I = m*L^2/12 + m*R^2/4

I = 6*0.4^/12 + 6*0.06^2/4 = 0.0854

So,

KEr = 0.5*0.0854*(100*pi)^2 = 4214.32 J

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