A light rod of length ℓ = 1.00 m rotates about an axis perpendicular to its length and passing through its center as in the figure below. Two particles of masses m_{1} = 4.45 kg and m_{2} = 3.00 kg are connected to the ends of the rod.

(a) Neglecting the mass of the rod, what is the system's kinetic energy when its angular speed is 2.20 rad/s? The correct answer is not zero. J

(b) Repeat the problem, assuming the mass of the rod is taken to be 2.30 kg.

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(a) the kinetic energy is given by the sum of the rotational kinetic energies for each of the masses. KE = (1/2)Ia*w^2 + (1/2)Ib*w^2 KE = (1/2)w^2[Ia + Ib]

For point masses the moments of inertia are: Ia = ma*R^2 Ib = mb*R^2

KE = (1/2)w^2R^2[ma + mb]

w = 2.5 R = 1/2 ma = 4 and mb = 3

KE = (1/2)(2.2)^2(1/2)^2[4.45 + 3] KE = 4.51 J

(b) The KE of the masses will be the same so we only need calculate the kinetic energy contribution from the rod.