Moment of inertia of a rod: axis not through the centre!?


by wizzle
Tags: axis, centre, inertia, moment
wizzle
wizzle is offline
#1
Mar28-09, 05:13 PM
P: 26
1. The problem statement, all variables and given/known data

A meter stick of mass 0.44 kg rotates, in the horizontal plane, about a vertical
axis passing through the 30 cm mark. What is the moment of inertia of the stick?
(Treat it as a long uniform rod)


2. Relevant equations

I know that for long uniform rods with length L, if the axis is through the centre, the moment of inertia is (1/12)ML^2. If the axis is through the end, it's (1/3)ML^2.

3. The attempt at a solution

I thought it might work to act as though there were two different weights and splitting the mass according to how far each was from the axis since it's a uniform rod (left side = .3 * .44 kg) (right side = .7*.44kg)

Calling the left side, 30 cm to the left of the axis, A, and the right side of the rod, located 70 to the right of the axis, B, here's what I came up with:

Ia: (1/3)(0.132)*(0.3)^2 = 3.96 x 10^-3 kg*m^2

Ib: (1/3)(.308)(.70)^2 = 5.031 x 10^-2 kg*m^2

I = Ia+Ib = 5.43 x 10^-2 kg*m^2

Does that seem logical? Any input would be greatly appreciated.
Thanks!
-Lauren
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rock.freak667
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#2
Mar28-09, 05:29 PM
HW Helper
P: 6,214
Why not just find the moment of inertia through the centre using (1/12)ML2 and then move the axis using the parallel axis theorem?
wizzle
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#3
Mar28-09, 05:38 PM
P: 26
Hmm...the parallel axis theorem hey? Unfortunately I don't know what that is, but I'm looking it up. Are you referring to Iz=Icom+Md^2?

rock.freak667
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#4
Mar28-09, 05:58 PM
HW Helper
P: 6,214

Moment of inertia of a rod: axis not through the centre!?


Quote Quote by wizzle View Post
Hmm...the parallel axis theorem hey? Unfortunately I don't know what that is, but I'm looking it up. Are you referring to Iz=Icom+Md^2?
Yes that would be it, where Icom is the moment of inertia through the centre and d is the distance from the centre to the new axis.
wizzle
wizzle is offline
#5
Mar28-09, 06:30 PM
P: 26
Great! I calculated I through the centre to be (1/12)ML^2 = (1/12)(.44 kg)(1 m)^2 = 3.67 x 10^-2 kg*m^2.

I then used Iz = Icom + Md^2 = (3.67 x 10^-2 kg*m^2)+(.44 kg)(.2 m)^2 = 5.43 x 10^-2 kg*m^2.

I don't know how to gauge whether that is a reasonable answer. Any chance you can spot a mistake?

Thanks again!


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