A very light rigid rod with a length of 1.88 m extends straight out from one end

Question

A very light rigid rod with a length of 1.88 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. I_P = I_CM + MD^2 (a) Determine the period of oscillation. [Suggestion: Use the parallel-axis theorem equation given above. Where D is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object.] (b) By what percentage does the period differ from the period of a simple pendulum 1 m long?

Explanation / Answer

(a) Moment of inertia of the rod is zero since its a very light rod.

Moment of inertia of the meter stick about its center of mass,

ICM = M * 12 / 12 = M/12

Using parallel axis theorem, the moment of inertia of the stick about the pivot,

Ip = ICM + MD2

Here, D2 = (0.52 + 1.882) = 3.7844 m2

So,

Ip = (M/12) + 3.7844M = 3.8677M

Now, the period of this compund pendulum is,

Tc = 2 (I / MgD)1/2

=> Tc = 2 * [3.8677 / (M * 9.81 * 3.78441/2)]1/2 = 2.83 s

(b) Period of a simple pendulum of 1 m long, Ts = 2 (L / g)1/2 = 2 (1 / 9.81)1/2 = 2.01 s

So, percentage difference = (Tc - Ts) * 100 / Ts = (2.83 - 2.01) * 100 / 2.01 = 40.8 %

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