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.)

Question

I know I = (1/12)(M)(L^2) and then I need to use the parallel-axis 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.

Answer 1

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

Answer 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.

Answer 3

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.

Answer 4

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. .

Answer 5

(1/12)(.58)(1^2) = .0483
+
(.58)(.07^2) = .0028

So final answer about .0511 if that helps finding out your error.

Answer 6

You are right. I moved a decimal point and added a wrong (1/2)ML^2 value. Thanks for catching that.

Answer 7

A butterfly at eye level is 25 cm in front of an plane mirror. You are behind the butterfly, 45 cm from the mirror. What is the distance between your eye and the image of the butterfly in the mirror?
1 cm

Answer 8

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