(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A pendulum is constructed of a solid metal sphere of

mass M = 4.00 kg, attached to a thin metal rod of mass

m = 1.00 kg and length L = 40.0 cm. The pivot point for

the pendulum is at the upper end of the thin rod. The

pendulum oscillates through a small angle with a period

of T = 1.40 s. Find the radius of the sphere.

2. Relevant equations

[tex]\omega^2=\frac{I}{mgd}\hspace{15pt}\omega=\frac{2\∏}{T}\\I=\frac{1}{3}M_{rod}L^2+\frac{2}{5}M_{sphere}R^2(L+R)^2[/tex]Center of Mass[tex]=\frac{m_1d_1+m_2d_2....m_nd_n}{m_{total}}[/tex]

For d I used the distance from the center of mass to the pivot point[tex]\frac{L}{2}+R[/tex]

3. The attempt at a solution

I have filled in all known values leaving only R, then solved for R using the quadratic formula.

My professor gave use 10.8 cm as an answer. I am getting ~30cm. This should be straight forward right? Chug and Plug?

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# Oscillation and Moment of inertia

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