Moment of Inertia - Non uniform density

[JerS]

Homework Statement

Hey, need a quick bit of help, Maths Physics end of year exam and I can't find how to calculate the moment of inertia of a non uniform bod thin rod.

I'll post the question here for reference -

1. A rod AB of length 3m has non-uniform density x  15 − 2xkg/m where x measures
distance (in m) from the end A of the rod.
a. Find the moment of inertia of the rod about an axis perpendicular to the rod through the end
A.
b. Use your result to find the kinetic energy of the rod when it is rotating about the given axis
with angular velocity 2rad / s.
c. Deduce the moment of inertia of the rod about an axis parallel to the given axis but passing
through the centre of mass of the rod.
[Use the result in question 1 above for the location of the centre of mass.]

but I really just need the formula,

Thanks tons!

Homework Equations

I = Sigma m.r^2 ?

The Attempt at a Solution

Around 3 A4 pages of pointlessness

 

[berkeman]

Homework Statement

Hey, need a quick bit of help, Maths Physics end of year exam and I can't find how to calculate the moment of inertia of a non uniform bod thin rod.

I'll post the question here for reference -

1. A rod AB of length 3m has non-uniform density x  15 − 2xkg/m where x measures
distance (in m) from the end A of the rod.
a. Find the moment of inertia of the rod about an axis perpendicular to the rod through the end
A.
b. Use your result to find the kinetic energy of the rod when it is rotating about the given axis
with angular velocity 2rad / s.
c. Deduce the moment of inertia of the rod about an axis parallel to the given axis but passing
through the centre of mass of the rod.
[Use the result in question 1 above for the location of the centre of mass.]

but I really just need the formula,

Thanks tons!

Homework Equations

I = Sigma m.r^2 ?

The Attempt at a Solution

Around 3 A4 pages of pointlessness

Welcome to the PF. You will use an integration instead of the Sigma sum. Just break the rod up into little mass pieces dm, with a linear mass density, and a small length dx. Use the equation that you are given for the linear mass density (units are kg/m), and express the each little mass piece dm in terms of that density and the small length piece dx.

Then use the formula that you allude to with your "I = Sigma m.r^2", but use an integration over x, of dm*x^2. Does that help?

 

[nlightner]

.

Hi there,

Calculating the moment of inertia for a non-uniform density object can be a bit tricky, but it is definitely doable. The formula you mentioned, I = Σm.r^2, is correct for calculating the moment of inertia for a point mass. However, for a non-uniform density object like a thin rod, we need to use a different formula.

The moment of inertia for a thin rod with non-uniform density can be calculated using the following formula:

I = ∫r^2dm

Where r is the distance from the axis of rotation to a small element of mass dm. This formula essentially takes into account the varying density of the rod along its length. In your case, the density function is given as 15-2x, so for each small element of mass, we can write dm = (15-2x)dx.

Using this, we can rewrite the formula as:

I = ∫r^2(15-2x)dx

To solve this integral, we first need to find an expression for r in terms of x. Since the rod has a length of 3m, we can say that r = 3-x. Substituting this in the integral, we get:

I = ∫(3-x)^2(15-2x)dx

Expanding this and solving the integral, we get:

I = 25m^4

This is the moment of inertia for the rod about an axis perpendicular to the rod through the end A.

For part (b) of the question, we can use the formula for rotational kinetic energy, which is given by:

K = 1/2Iω^2

Where ω is the angular velocity of the rod. Substituting the value of I that we calculated above and the given value of ω, we get:

K = 1/2(25)(2)^2 = 50J

For part (c), we can use the parallel axis theorem, which states that the moment of inertia about an axis parallel to the given axis but passing through the center of mass is given by:

Icm = I + Md^2

Where M is the total mass of the object and d is the distance between the two axes. In this case, we know the value of I from part (a) and we can calculate the center of mass of the rod using the formula:

xcm =