Calculating Moment of Inertia for a System of Small Blocks on a Clamped Rod

AI Thread Summary
To calculate the moment of inertia for a system of small blocks on a clamped rod, the formula I = Σmr² is applied, where r is the distance from the axis of rotation. The blocks are positioned at 1/4L, 1/4L, and 3/4L from the axis. The individual contributions to the moment of inertia are calculated as m(1/4L)² for the first two blocks and m(3/4L)² for the third block. Upon simplifying the equation, the total moment of inertia is found to be 11/16mL². The confusion arises from the addition of squared distances rather than the linear distances.
elsternj
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Homework Statement


Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass

Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end.
Express your answer in terms of the given quantities.



Homework Equations



I = \summr2


The Attempt at a Solution



Now I know the answer is 11/16mL2 but I am having trouble figuring out how to get to that answer.

the first mass is 1/4L from the axis, the second mass is also 1/4L from the axis and the third mass is 3/4L from the axis.

m(1/4L)2+m(1/4L)2+m(3/4L)2

how does this become 11/16mL2?
 
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uhh... just open up the brackets and add!
 
can i see that step by step? my math isn't the greatest but 1/4+1/4+3/4 does not equal 11/16
 
Its (1/4)^2 + (1/4)^2 + (3/4)^2
 

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