Inertia: blocks attached to rod of negligible mass

AI Thread Summary
The discussion centers on calculating the moment of inertia (I) for small blocks with mass m attached to a rod of negligible mass. The initial formula proposed is I = m(0.5L)^2 + m(0.5L)^2, leading to I = 0.5L^2 m. Participants clarify that the moment of inertia should not be expressed in meters and emphasize the importance of correctly identifying the mass distribution. A mistake in the placement of variables is noted, highlighting the need for careful notation to avoid errors. Ultimately, the calculation is confirmed to be correct after addressing these concerns.
cantgetno
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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.
Find the formula for I


Homework Equations


I=MR^2


The Attempt at a Solution



I=m(0.5 L)^2 + m(0.5 L)^2
I=2(0.5L)^2 m
I=0.5L^2 m

am i right?
 
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Hi cantgetno! :smile:
cantgetno said:
Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass.
Find the formula for I

I=m(0.5 L)^2 + m(0.5 L)^2
I=2(0.5L)^2 m
I=0.5L^2 m

am i right?

mmm …

i] moment of inertia isn't measured in metres

ii] about which point is I being measured?

iii] what about the mass in the middle? :wink:
 


oh its around the middle
so that makes the mass in the middle mass x 0

metres? m=mass
 
cantgetno said:
oh its around the middle
so that makes the mass in the middle mass x 0

metres? m=mass

D'oh! m is mass … should have spotted that! :redface:

In that case, everything is fine. :biggrin:
 


oh crap
i put 1/2 l m^2 ... I am an idiot

thanks anyway
 
cantgetno said:
oh crap
i put 1/2 l m^2 ... I am an idiot

hee hee :biggrin:

Useful tip: keep your m's in the same place :wink:

for some reason, you shifted them from the left to the right …
I=m(0.5 L)^2 + m(0.5 L)^2
I=2(0.5L)^2 m
… which makes it much easier to make a mistake, and much less easy to spot a mistake. :cry:
 
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