Question of Physics

Question

A thin rod of length l has a non-uniform mass distribution. The linear mass density λ is given by λ = Ax where A is a positive

constant. Find the moment of inertia of the rod about an axis passing through its center of mass.

Answer 1

Notation:

1) Let me denote the length of the rod as L, rather than l, because the latter looks like 1.

2) All definite integrals below are between 0 and the length L.

3) Let g be gravitation acceleration.

First calculate the position c of the center of gravity along the rod of length L.

It is the point around which the total torque is zero.

The torque of a point of mass dm at position r from the center of gravity is the product

r g dm

In terms of position x from the beginning of the rod:

r = x - c,

dm = A x dx

The total torque is the sum of all the mass points along the rod and that has to be 0:

∫(x - c) g A x dx = 0

A g ∫(x² - cx) dx = 0

L³/3 - cL²/2 = 0

c = 2L/3

Now we calculate the moment of inertia I.

The moment of inertia dI of a point mass dm of length dx at distance r from the center of gravity is

dI = r²dm

In terms of the x-coordinate

r² = (x - c)² (regardless whether x < c or x > c)

dm = A x dx

The whole moment of inertial is the sum over all the points of mass:

I = ∫(x - c)² A x dx = A ∫(x³ - 2cx² + c²x)dx = A (L⁴/4 - 2cL³/3 + c²L²/2)

Now substitute the above calculated c = 2L/3:

I = A(L⁴/4 - 4L⁴/9 + 4L⁴/18) = AL⁴/36

Just for interest we can calculate the mass M of the whole rod:

M = ∫Axdx = AL²/2

Then we can write the moment of inertia as

I = ML²/18