Moment of inertia of a meter stick

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a meter stick with a mass of 0.2 kg, bored at the 10 cm mark for hanging. Participants express confusion about using the parallel axis theorem and the appropriate equations for a meter stick, as it is not a cylinder. The correct approach involves using the definition of moment of inertia through integration, specifically from -0.01 m to 0.09 m, with the linear mass density calculated as λ = 0.2 kg/m. After correcting the bounds and arithmetic, the moment of inertia is determined to be 0.48 kg·m². The conversation emphasizes the importance of verifying calculations and understanding the geometry of the object.
Pablo
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Homework Statement



A meter stick has a mass of 0.2kg. A small hole is bored in it at the 10cm mark so the meter stick can be hung from a horizontal nail. The moment of inertia of the meter stick around an axis at the 10cm mark is

Homework Equations



m = 0.2kg
I = mr^2

The Attempt at a Solution


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So I know I am trying to find the moment of inertia of a meter stick 0.1m from one of its edges. My first thought was to use the parallel axis theorem and think of the meter stick as a cylinder. However, I don't think it is a cylinder. I don't know any equation for a meter stick shape, so I am not sure how to get started.
 
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You might do best to start with the definition of mass moment of inertia in terms of an integral. The integration will run from -0.1 m to 0.9 m.
 
How could I use the integral of r^2 dm if I don't know the length of the rod? I can't find λ
 
Pablo said:
My first thought was to use the parallel axis theorem and think of the meter stick as a cylinder. However, I don't think it is a cylinder. I don't know any equation for a meter stick shape, so I am not sure how to get started.
you have no information about the thickness of the stick, so assume that is negligible.
 
haruspex said:
you have no information about the thickness of the stick, so assume that is negligible.
Would I use the definition of moment of inertia using calculus or is there another way. I am very confused, and I am not sure how to get started.
 
Pablo said:
How could I use the integral of r^2 dm if I don't know the length of the rod? I can't find λ
Well, it is a meter stick... :wink:
 
gneill said:
Well, it is a meter stick... :wink:

Ok so given that the length is 1 meter, I know λ = 0.2 / 1 = 0.2. I also know the moment of inertia is the integral of x^2 dm from -0.01 to 0.09, equivalent to λ * [ (0.09^3 / 3) - (-0.01^3 / 3)] = 0.000049. This answer is still not correct.

EDIT:
My bounds were incorrect. I got 0.48, thanks!
 
Check your limit values. Is 0.01 m really 10 cm?
 
Check your arithmetic, and then consider the possibility that the answers offered to you may not include the correct answer.
 
  • #10
You might start with the moment of inertia of a rod about its center of mass = m L^2 /12
 
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