A uniform cylinder 20 cm long, suspended by a steel wire attached to its mid-point so that its long axis is horizontal, is found to oscillate with a period of 2 seconds when the wire is twisted and released. When a small disc, of mass 10 g, is attached to each end the period is found to be 2.3 seconds. Calculate the moment of inertia of the cylinder about the axis of oscillation.
L = Iω
I = ∑(mr^2)
The Attempt at a Solution
angular velocity of cylinder, ω = 2π/2 = 3.14 rad/s
angular velocity of cylinder with discs, ω = 2π/2.3 = 2.73 rad/s
Conservation of angular momentum:
Iω = Iω(1) + 2 (mr^2)ω(1)
3.14I = 2.73I + 2 x (0.01 x 0.1^2) x 2.73
0.41I = 5.46 x 10^-4
I = 1.33 x 10^-3 kg m^2
The correct answer is 6.2 x 10^-4, so I am way off. Any help would be much appreciated. Thanks!