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

a) about an axis passing through its left end by using the integration.

b) solve part a) by using the result of the previous question (Question 2) and the parallel axis theorem.

Question 2 is (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.

a)

The moment of inertia dI of a point mass dm of length dx at distance x is

dI = x²dm

dm = Axdx

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

I = ∫x²Axdx = A∫x³dx = AL⁴/4

Now we calculate the mass M of the whole rod:

M = ∫Axdx = AL²/2

Then we can write

I = ML²/2

b)

The solution to Question 2) was

- The center of gravity c = 2L/3.

- The moment of inertial I_c around the center of gravity is ML²/18.

Using the parallel axis theorem:

I = I_c + M(2L/3)² = ML²/18 + 4ML²/9 = ML²/2