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

[PLAIN]http://www.smartphysics.com/images/content/mechanics/ch15/momentofinertia2new.png [Broken]

What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

m_{rod}(hence denoted as m_{r}) = 7.11 kg

m_{sphere}(hence denoted as m_{s}) = 35.55 kg

r_{rod}(this is the length of the rod, and it will hence be denoted as r_{r}) = 5.8 m

r_{sphere}(this is the radius of the sphere, hence denoted as r_{s}) = 1.45 m

2. Relevant equations

[tex]I = I_{center of mass} + md^2[/tex] (parallel axis theorem)

[tex]I_{sphere} = \frac{2}{5}mr^2[/tex]

[tex]I_{rod at end} = \frac{1}{3}ml^2[/tex]

3. The attempt at a solution

I figured that the moment of inertia must include both the moment of a rod spinning at its edge and the moment resulting from the parallel axis theorem. This leads to the first part.

[tex] \frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2 [/tex]

Doing something similar to the sphere gives a similar equation

[tex] \frac{2}{5}m_s(r_s)^2 + m_s(r_s/2)^2 [/tex]

Then I just combine these together, if the hyperphysics page is anything to go by (http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html). This yields the following.

[tex] \frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2 + \frac{2}{5}m_s(r_s)^2 + m_s(r_s/2)^2[/tex]

However, this isn't correct. Why is this, and how can I fix this?

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# Homework Help: Moment of Inertia for Combination of Rod and Ball

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