A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity ω. If A is the cross-section and D is the density of material of rod. Find its rotational K.E. :
Answer & explanation
Correct answer: option 1
Moment of inertia for rod = I
I = (ML2 / 12)
K.E. = (1/2)Iω2
= (1/2) × (ML2 / 12)ω2
= [(ML2ω2) / (24)]
Density : D = [(mass) / (volume)]
D = (M/V) = (M / A.L)
M = D.A.L
KE = [(DAL3ω2) / (24)]