# Parallel axis theorem + rotational inertia

• lemonpie
ML^2This doesn't seem to be what you're looking for. :(In that problem, the moment of inertia of the rod about A was givenin the question.

## Homework Statement

Two particles, each with mass m = 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.6 cm and mass M = 1.2 kg. The combination rotates around the rotation axis with angular speed w = 0.30 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?
http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%2010.%20Rotation/Problems/c10x10_12.xform_files/nw0536-n.gif

## Homework Equations

parallel-axis theorem: I = Icom + Mh^2
rotational inertia for thin rod: I = 1/12ML^2
K = 1/2Iomega^2

## The Attempt at a Solution

Inet = Irod1 + Iparticle1 + Irod2 +Iparticle2
Inet = 1/12Md^2 + Md^2 + 1/12M(2d)^2 + M(2d)^2

help!

Think more carefully about applying the parallel axis theorem to the rods.

Inet = 1/12Md^2 + Md^2 + 1/12(2M)(2d)^2 + 2M(2d)^2 ?

In your formula for the parallel axis threorem, h means distance
from O to the center of mass.

This is d/2 for the inner rod and 3d/2 for the outer.

Last edited:
oh. h means distance from O to com. not just d.

Inet = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2 ?

Last edited:
Nope.

You can take each rod separately (as you seem to be trying to do),
or you could treat them as one single rod.
You are muddling these two approaches.

sigh. i guess i would like to treat them as one rod. so this is my answer (i noticed that i wrote a wrong number, so I'm fixing it here):

Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2

i think the particles are okay, but not the rods. is that right?

lemonpie said:
sigh. i guess i would like to treat them as one rod. so this is my answer (i noticed that i wrote a wrong number, so I'm fixing it here):

Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2

i think the particles are okay, but not the rods. is that right?

Yes the particles are fine.
But look at the parallel axis theorem again. you don't seem to be applying
it.
And the mass of the outer rod isn't 2M is it?

Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12(1/2)d^2 + Md^2) + md^2 + (M/12(3d/2)^2 + Md^2) + m(2d)^2

actually, why do you have to use the parallel axis theorem here? why can't you just add up the rotational inertias for rods and particles? someone please help me out soon! thanks!

davieddy said:
In your formula for the parallel axis threorem, h means distance from O to the center of mass.

This is d/2 for the inner rod and 3d/2 for the outer.

David

Last edited:
lemonpie said:
actually, why do you have to use the parallel axis theorem here? why can't you just add up the rotational inertias for rods and particles? someone please help me out soon! thanks!

Because the rotation is about O, not the c of m.

lemonpie said:

## Homework Equations

parallel-axis theorem: I = Icom + Mh^2
rotational inertia for thin rod: I = 1/12ML^2

Icom = 1/12 ML^2
where L is the length of the rod

so... d/2 and 3d/2 are for the Mh^2 not Icom? that seems weird to me. wait let me try.

Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12 d^2 + M(1/2 d)^2) + md^2 + (M/12 d^2 + M(3d/2)^2) + m(2d)^2

lemonpie said:
inet = irod1 + iparticle1 + irod2 + iparticle2 = (m/12 d^2 + m(1/2 d)^2) + md^2 + (m/12 d^2 + m(3d/2)^2) + m(2d)^2

BINGO!

Nearly. But the mass of the rods is M not m.

lemonpie said:
okay, but this problem:
rotation is about point A, and it doesn't require parallel axis theorem.
In that problem the rotational inertia of the rod about point A is given. You don't have to calculate it at all.

lemonpie said:
okay, but this problem:
rotation is about point A, and it doesn't require parallel axis theorem.

sorry! I'm trying to understand!

In that problem, the moment of inertia of the rod about A was given
in the question.

Try this one:
Using the parallel axis theorem, show that the moment of inertia
of a rod about one end is 1/3 M L^2.

Doc Al said:
In that problem the rotational inertia of the rod about point A is given. You don't have to calculate it at all.

You beat me to it!

BTW I think "rotational" inertia is a much better term than
"moment of" inertia. I shall try to adopt it.

David

davieddy said:
In that problem, the moment of inertia of the rod about A was given
in the question.

Try this one:
Using the parallel axis theorem, show that the moment of inertia
of a rod about one end is 1/3 M L^2.
um...

I = 1/12ML^2 + M(1/2L)^2
I = 1/12ML^2 + 1/4ML^2
I = 1/12ML^2 + 3/12ML^2
I = 4/12ML^2
I = 1/3ML^2

fortunately (or unfortunately), there weren't any questions about the parallel axis theorem on the test. i got a B+, not as good as i'd hoped, but considering how little i knew, maybe i should just be thankful. thanks for your help!

lemonpie said:
Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12 d^2 + M(1/2 d)^2) + md^2 + (M/12 d^2 + M(3d/2)^2) + m(2d)^2

Tidy this up by collecting the terms for the rod and the particles.
Then check to see if your answer for the rod agrees with taking one rod
length 2d. mass 2M and applying the formula you have just derived
(immaculately) for a rod rotating about one end namely:

(1/3) (2M) (2d)^2