A torsion pendulum consists of two small point masses, each of mass m = 0.5 kg mounted on the ends of a uniform rod of length l = 0.75 m and of mass M = 2 kg. The system is suspended from a wire of negligible mass attached to the center of the rod and set into torsional oscillation. The torsion constant of the wire is κ = 1.2 N m/rad. (a) [4 points] Question: Derive a symbolic, simplified expression for the moment of inertia I of the torsion pendulum about an axis along the wire in terms of the masses m and M, and the length l of the rod.

I_total = L² (m / 2 + M / 12)

Explanation:

For this exercise we have the system formed by two small masses, point and a rod, they ask us to calculate the moment of inertia, let's use that the moment of inertia is a scalar quantity, therefore additive.

To calculate the moment of inertia of the point mass with respect to the central wire, which is at a distance r = L / 2

           I = m r²

The moment of inertia of a rod with respect to its center is tabulated

          I_rod = 1/12 M L²2

As all the elements rotate around the same point, we can add the moments of inertia

        I_total = 2 I + I_roda

        I_total = 2 m r² + 1/12 M L²

        I_total = 2 m (L / 2)² + 1/12 M L²

        I_total = L² (m / 2 + M / 12)

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

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