Question

The linear mass density of a thin rod AB of length L varies from A to B as $\lambda(x) = \lambda_0 (1 + \frac{x}{L})$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

Answer 1

I = \frac{7}{18}ML^2

Answer 2

• Compute total mass $M = \frac{3}{2}\lambda_0 L$ and express $\lambda_0 = \frac{2M}{3L}$.
• Write the moment of inertia $I = \lambda_0\int_0^L x^2\left(1+\frac{x}{L}\right) dx$.
• Evaluate integrals: $\int_0^L x^2dx=\frac{L^3}{3}$ and $\int_0^L x^3dx=\frac{L^4}{4}$.
• Simplify to obtain $I = \frac{7\lambda_0L^3}{12}$ and substitute $\lambda_0$ to get $I = \frac{7ML^2}{18}$.