Homework Help: Oscillation and Moment of inertia

1. Feb 28, 2012

tempneff

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

$$\omega^2=\frac{I}{mgd}\hspace{15pt}\omega=\frac{2\∏}{T}\\I=\frac{1}{3}M_{rod}L^2+\frac{2}{5}M_{sphere}R^2(L+R)^2$$Center of Mass$$=\frac{m_1d_1+m_2d_2....m_nd_n}{m_{total}}$$
For d I used the distance from the center of mass to the pivot point$$\frac{L}{2}+R$$
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?

Last edited: Feb 28, 2012
2. Feb 29, 2012

ehild

Check your equations. Neither the one for the moment of inertia nor that for d (distance of CM from the pivot) are correct.

ehild

3. Feb 29, 2012

tempneff

Okay so this is my final equation, but I'm still not getting it right. What am I doing wrong.
$\omega^2=\frac{\frac{1}{3}L^2+\frac{2}{5}M_{sph}R^2+M_{sph}(L+R)^2}{M_{total}g(\frac{\frac{1}{2}LM_{rod}+RM_{sph}}{M_{total}})}$

4. Mar 1, 2012

ehild

The CM is still wrong. How far is the centre of the sphere from the pivot?
The mass of the rod is missing in the numerator.

ehild