College Physics
posted by Ryan on .
Calculate the rotational inertia of a meter stick, with mass 0.58 kg, about an axis perpendicular to the stick and located at the 43 cm mark. (Treat the stick as a thin rod.)
I know I = (1/12)(M)(L^2) and then I need to use the parallelaxis theorem so I need to add Mh^2 but that is not working. I get 0.156 doing this method but it is not correct.

Check your units (are they meters?) and your decimal point.
The center of the stick is at the 0.50 m mark and the moment of inertia about that point is (1/12)ML^2 = 0.0483 kg m^2.
(L = 1.0 m and M = 0.4833 kg)
Using the parallel axis theorem, the moment of inertia about an axis h = 0.07 m away is higher by
M*h^2 = 0.0028 kg/m^2, for a total of 0.4861 kg/m^2 
Ah, I see what I may have done wrong. I was not using .07m for h. Unfortunately, 0.4861 for the overall answer still seems to be wrong.

Oops I found the problem, it seems you were off by a decimal point. I did the math and got .051 for a final answer, thank you for the help.

I made a typo error saying that I used a mass of M = 0.4833 kg. I actually used 0.58 kg
I may have made a computational error somewhere but I don't see where. . 
(1/12)(.58)(1^2) = .0483
+
(.58)(.07^2) = .0028
So final answer about .0511 if that helps finding out your error. 
You are right. I moved a decimal point and added a wrong (1/2)ML^2 value. Thanks for catching that.

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