a rotational problem

a. By parallel axis theorem,

moment of inertia of the combination about O, I

= [1/3 Ml^2 + M(l/2)^2] + [1/3 Ml^2 + M(3l/2)^2] + ml^2 + m(2l)^2

= (19M/6 + 5m)l^2

b. Kinetic energy of rotation about O

= 1/2 Iw^2

= 1/2 (19M/6 + 5m)l^2w^2

(a) moment of inertia of rod with length 2l about O = M(2l)^2/3 = (8/3)Ml^2
moment of inertia of the two masses about O
= ml^2 + m(2l)^2 = 5ml^2

Hence, total moment of inertia
= [(8/3)Ml^2 + 5ml^2] = [8M/3 + 5m].l^2

(b) kinetic energy = (1/2)[8M/3 + 5m].l^2.w^2