Oscillation and Moment of inertia

[tempneff]

Homework Statement

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.

Homework 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]

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?

 

[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

 

[tempneff]

Okay so this is my final equation, but I'm still not getting it right. What am I doing wrong.
[itex]\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}})}[/itex]

 

[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

 

[asteriatic]

I would like to first commend you for attempting to solve this problem using the relevant equations. However, it seems that you may have made a mistake in your calculations. It is important to double check your work and make sure that all units are consistent.

Additionally, it is important to understand the physical meaning behind the equations you are using. In this case, the moment of inertia is a measure of an object's resistance to rotational motion, and it depends on the mass distribution and the distance from the axis of rotation. The equation for the moment of inertia that you have used is for a solid sphere, but in this case, the sphere is attached to a rod and the mass is not uniformly distributed.

I would recommend using the parallel axis theorem, which states that the moment of inertia of an object can be calculated by adding the moment of inertia of the object about its center of mass to the product of its mass and the square of the distance between the center of mass and the axis of rotation. This will give you a more accurate value for the moment of inertia in this scenario.

Additionally, it may be helpful to draw a diagram and label all the known values to better visualize the problem. And as always, make sure to check your units and use the correct formula for the given situation.

In conclusion, while chugging and plugging may be a useful approach in some cases, it is important to have a thorough understanding of the concepts and equations being used to solve a problem. I would recommend reviewing the relevant concepts and equations and double checking your calculations to get a more accurate answer.