Moment of Inertia of a rigid rotating object.

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SUMMARY

The discussion focuses on calculating the moment of inertia of a system consisting of two point masses (M1 = 2.53 kg and M2 = 5.16 kg) attached to a massless rod of length 2.36 m. The distance from mass M1 to the center of mass was determined to be 1.58 m. The moment of inertia was calculated using the formula I = 1/3 ML², confirming the correctness of the approach without the need for integrals or calculus.

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asz304
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Homework Statement



Two small masses are attached to a massless rod of length 2.36 m as shown. Mass M1 is 2.53 kg and mass M2 is 5.16 kg. A) What is the distance, x, from mass M1 to the centre of mass of this system? B) What is the moment of inertia of this system about an axis that passes through the centre of mass and is perpendicular to the rod?


M1------------------(.)----------M2
{--------- x--------}. {---(L-x)-}

(.) is the axis of rotation.

The Attempt at a Solution



I found the answers for question A) using the formula of center of mass and my point of origin was M1.

X = 1.58 m
L-X = 0.78 m

How do I find the moment of inertia if the axis isn't at the middle? If I need to use integrals, can some please show me how? I didn't do integrals yet. thanks
 
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asz304 said:
How do I find the moment of inertia if the axis isn't at the middle? If I need to use integrals, can some please show me how? I didn't do integrals yet. thanks
There's no need for any calculus. Treat the small masses as point masses. What's the moment of inertia of a point mass at some distance from an axis?
 
Oh. thanks that made lots of sense. I used I = 1/3 ML^2 , and I got the right answer when I checked the computer.
 

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