Question

Calculate the moment of inertia of a rod whose linear density changes from $\rho$ to $\eta\rho$ from the thinner end to the thicker end.

The mass of the rod is equal to $M$ and length $L$. Consider the axis of rotation perpendicular to the rod and passing through the thinner end. Express your answer in terms of $M$, $L$ and $\eta$.

• A. $\frac{1}{6}ML^2\frac{3\eta+1}{\eta+1}$
• B. $\frac{1}{3}ML^2\frac{3\eta+1}{\eta+1}$
• C. $\frac{1}{6}ML^2\frac{\eta+1}{3\eta+1}$
• D. $\frac{1}{3}ML^2\frac{\eta+1}{3\eta+1}$

Answer 1

Answer: (A)

$\frac{1}{6}ML^2\frac{3\eta+1}{\eta+1}$

Answer 2

Let the linear density of the rod at a distance $x$ from the thinner end be $\lambda(x)$. Given that the density at the thinner end ($x=0$) is $\rho$ and at the thicker end ($x=L$) is $\eta\rho$. Assuming a linear variation of density with distance, we can write $\lambda(x)$ as: $\lambda(x) = \rho + \frac{(\eta\rho - \rho)}{L}x = \rho + \frac{(\eta-1)\rho}{L}x$

The total mass $M$ of the rod is given by integrating the linear density over its length: $M = \int_0^L \lambda(x) dx = \int_0^L \left(\rho + \frac{(\eta-1)\rho}{L}x\right) dx$ $M = \rho \left[x + \frac{(\eta-1)}{L}\frac{x^2}{2}\right]_0^L = \rho \left(L + \frac{(\eta-1)L^2}{2L}\right)$ $M = \rho \left(L + \frac{(\eta-1)L}{2}\right) = \rho L \left(1 + \frac{\eta-1}{2}\right) = \rho L \left(\frac{2+\eta-1}{2}\right)$ $M = \rho L \frac{\eta+1}{2}$

From this, we can express $\rho$ in terms of $M$, $L$, and $\eta$: $\rho = \frac{2M}{L(\eta+1)}$

The moment of inertia $I$ of the rod about an axis perpendicular to the rod and passing through the thinner end ($x=0$) is given by: $I = \int_0^L x^2 dm = \int_0^L x^2 \lambda(x) dx$ $I = \int_0^L x^2 \left(\rho + \frac{(\eta-1)\rho}{L}x\right) dx = \rho \int_0^L \left(x^2 + \frac{\eta-1}{L}x^3\right) dx$ $I = \rho \left[\frac{x^3}{3} + \frac{\eta-1}{L}\frac{x^4}{4}\right]_0^L = \rho \left(\frac{L^3}{3} + \frac{(\eta-1)L^4}{4L}\right)$ $I = \rho \left(\frac{L^3}{3} + \frac{(\eta-1)L^3}{4}\right) = \rho L^3 \left(\frac{1}{3} + \frac{\eta-1}{4}\right)$ $I = \rho L^3 \left(\frac{4 + 3(\eta-1)}{12}\right) = \rho L^3 \left(\frac{4 + 3\eta - 3}{12}\right) = \rho L^3 \frac{3\eta+1}{12}$

Now, substitute the expression for $\rho$: $I = \left(\frac{2M}{L(\eta+1)}\right) L^3 \left(\frac{3\eta+1}{12}\right)$ $I = \frac{2M L^3}{L(\eta+1)} \cdot \frac{3\eta+1}{12} = \frac{2M L^2}{\eta+1} \cdot \frac{3\eta+1}{12}$ $I = \frac{1}{6} ML^2 \frac{3\eta+1}{\eta+1}$