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

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a system of small blocks positioned on a clamped rod. The problem specifies the configuration of the blocks and the axis about which the moment of inertia is to be computed, focusing on the relationship between mass, distance from the axis, and the resulting moment of inertia expression.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of moment of inertia using the formula I = Σmr², with specific attention to the distances of the blocks from the axis. There is an exploration of how to simplify the expression involving the squares of the distances.

Discussion Status

Some participants are attempting to clarify the steps involved in the calculation, while others are questioning the arithmetic involved in reaching the final expression. There is an ongoing dialogue about the interpretation of the mathematical operations needed to arrive at the answer.

Contextual Notes

Participants express varying levels of confidence in their mathematical skills, which may affect their ability to follow the calculations presented. There is an emphasis on ensuring clarity in the steps taken to compute the moment of inertia.

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 = [tex]\sum[/tex]mr2


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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