Torsion Pendulum Homework: Sphere vs. Cube Period Ratio

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SUMMARY

The discussion focuses on calculating the period ratio of a torsion pendulum using a sphere and a cube as suspended masses. The rotational inertia for the sphere is defined as Is = 110msD², while for the cube, it is Ic = 61mcS². Given that the mass of the sphere (ms) is 0.20 kg and the mass of the cube (mc) is 12 kg, with the assumption that D equals S, the expected ratio of their periods, Tc/Ts, can be derived from their respective moments of inertia. The correct formulas for the moments of inertia are I_s = (m_s * D²) / 10 and I_c = (m_c * S²) / 6.

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Homework Statement


In the experiment, you will study an oscillator called a "torsion pendulum." In this case, the restoring "force" is the torsion constant of the wire that suspends the weight X and the inertial term is the rotational inertia of the suspended mass. You will compare the periods of a suspended sphere and of a suspended cube. The rotational inertia of a sphere is Is = 110msD2 where ms is the mass of the sphere and D is its diameter. The rotational inertia of a cube is Ic = 61mcS2 where mc is the mass of the cube and S is the length of its side. If the cube and the sphere are suspended from the same wire, what is the expected ratio of their periods, Tc/Ts?
Assume that D = S
ms = 020kg
and mc = 12 kg

Homework Equations


The Attempt at a Solution



I'm pretty lost with this one.
 
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I wonder why they gave those moments of inertia. the real values are

I_s = \frac { m_s d^2 } {10}

and

I_c = \frac { m_c s^2 } {6}

Do you know the solution of a simple harmonic oscillator?

F = m\frac {d^2x}{dt^2} = - k x

you don't know the torsion constant, but because you only need the ratio of the periods
that isn't a problem.
 

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