# Moment of interia about axis d

## Homework Statement

Calculate by direct integration the moment of inertia for a thin rod of mass M and length L about an axis located distance d from one end.

## Homework Equations

I = Ʃmir2i

Which can be rewritten as int (r2)dm as Δm -> 0

Since we cannot integrate over a mass, we need to switch to dx, which is dm = M/L dx

## The Attempt at a Solution

The question is asking us to find the moment of inertia as it rotates on an axis d from one end.

Normally, if we are trying to find the moment of inertia of an "infinitely thin" rod rotating on one end, the moment of inertia is (1/3)ML2

However, we have to manually integrate this problem because we have mass on both sides of the axis of rotation.

I took two integrals- one on one side of d, and on the other.

First side:
(M/d) int(x2)dx (with bounds 0 to d)

(M/d)(x3/3) (with bounds 0 to d)

= Md2/3

The second side:
(M/L-d) int(x2) dx (with bounds from d to L)

(M/L-d) (x3/3) (with bounds from d to L)

(M/L-d) ( (L3/3) - (d3/3))

Then I added the two moments of inertia together, to get:

(Md2/3) + (M/L-d)((L3/3) - (d3/3))

This answer is not correct, and I was wondering where i went wrong conceptually/mathematically. Homework is due soon, so any immediate help is greatly appreciated.

ehild
Homework Helper

## Homework Statement

Calculate by direct integration the moment of inertia for a thin rod of mass M and length L about an axis located distance d from one end.

## Homework Equations

I = Ʃmir2i

Which can be rewritten as int (r2)dm as Δm -> 0

Since we cannot integrate over a mass, we need to switch to dx, which is dm = M/L dx

## The Attempt at a Solution

The question is asking us to find the moment of inertia as it rotates on an axis d from one end.

Normally, if we are trying to find the moment of inertia of an "infinitely thin" rod rotating on one end, the moment of inertia is (1/3)ML2

However, we have to manually integrate this problem because we have mass on both sides of the axis of rotation.

I took two integrals- one on one side of d, and on the other.

First side:
(M/d) int(x2)dx (with bounds 0 to d)

(M/d)(x3/3) (with bounds 0 to d)

= Md2/3

The second side:
(M/L-d) int(x2) dx (with bounds from d to L)

(M/L-d) (x3/3) (with bounds from d to L)

(M/L-d) ( (L3/3) - (d3/3))

Then I added the two moments of inertia together, to get:

(Md2/3) + (M/L-d)((L3/3) - (d3/3))

This answer is not correct, and I was wondering where i went wrong conceptually/mathematically. Homework is due soon, so any immediate help is greatly appreciated.

The density is M/L. Why did you use M/d and M/L -d as density? The second one is even dimensionally wrong.

You should integrate simply from -d to L-d.

ehild

I thought you used d and L-d because that was the length over which you were integrating.

It's always just over the total length?

ehild
Homework Helper
I thought you used d and L-d because that was the length over which you were integrating.

The density of both pieces are the same M/L.

It's always just over the total length?
You are supposed to get the moment of inertia by direct integration of the square of the distance from the pivot from one end to the other one. You can replace one integral with the sum of other two, but why?

ehild

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Thank you! I understand a bit better now.

this did not help at all lol. so no one knows the answer?

ehild
Homework Helper
Why do you think that the thread was no help? What is your question?
Read the Forum rules. We must not not give out full solutions.

ehild

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