Homework Answers
given
m = 0.65 kg
L = 0.70 m
w = 3500 rpm
= 3500*2*pi/60 rad/s
= 366.5 rad/s
moment of inertia of rod about it center, I = m*L^2/12
= 0.65*0.7^2/12
= 0.0265 kg.m^2
Rotational kinetic energy = (1/2)*I*w^2
= (1/2)*0.0265*366.5^2
= 1780 J <<<<<<<---------------Answer
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