The linear mass density of a thin rod AB of length L varies from A to B as λ(x) = λ_0(1 + x/L), wher

Q: Why can't we just use the standard formula $ML^2/3$?

The standard formula $ML^2/3$ is derived under the strict assumption that the rod is uniform (constant mass density). In this problem, the density increases as you move along the rod, meaning more mass is concentrated further from the axis. Therefore, you must integrate from scratch.

Q: What exactly does $\lambda_0$ represent physically?

If you plug $x=0$ into the density equation $\lambda(x) = \lambda_0(1 + x/L)$, you get $\lambda(0) = \lambda_0$. So, $\lambda_0$ is simply the linear mass density of the rod exactly at the starting end $A$.

Q: Why do we integrate from $0$ to $L$?

We are setting up our coordinate system such that the axis of rotation is at the origin ($x=0$). The rod extends from the axis to a distance $L$. Therefore, to sum up the contributions of all the tiny elements $dx$, we must integrate from the start of the rod ($0$) to the end of the rod ($L$).

Q: Can we use the parallel axis theorem here to save time?

No. The parallel axis theorem requires you to know the moment of inertia about the center of mass. For a non-uniform rod, the center of mass is not at $L/2$, and finding its exact location requires integration anyway. It is much faster to calculate the moment of inertia directly about the end $A$.

Q: How do we know we need to calculate the total mass $M$?

Always look at the options! Your initial integration for $I$ will give you an answer containing $\lambda_0$. However, all the multiple-choice options are expressed in terms of $M$ and $L$. This is a clear signal that you must find a mathematical relationship between $\lambda_0$ and $M$ to substitute it out.

Question

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

Options:

(A) $\frac{5}{12}ML^2$

(B) $\frac{7}{18}ML^2$

(C) $\frac{2}{5}ML^2$

(D) $\frac{3}{7}ML^2$

Sigma Percentile · Accepted Answer

### Visualizing the Non-Uniformity

When dealing with rigid bodies in rotational motion, the moment of inertia is a measure of how difficult it is to change the body's rotational speed. For a uniform rod, we have standard formulas like $\frac{ML^2}{3}$ about its end. However, in this problem, the rod is **non-uniform**.

The linear mass density $\lambda(x)$ is given by $\lambda_0\left(1 + \frac{x}{L}\right)$. This means the rod gets heavier and denser as we move from end $A$ to end $B$. Because the mass is distributed unevenly, we cannot rely on standard formulas. We must return to the fundamental definition of moment of inertia and use calculus.

### Setting Up the Integral

We start by considering a tiny, infinitesimally small element of the rod. Let this element have a length $dx$ and be located at a distance $x$ from the axis of rotation (which passes through end $A$).

The mass of this tiny element is $dm = \lambda(x) dx$.

The moment of inertia of this point-like element about the axis is:
$$dI = dm \cdot x^2$$

Substituting our expression for $dm$, we get:
$$dI = \lambda_0\left(1 + \frac{x}{L}\right) x^2 dx$$

To find the total moment of inertia $I$, we integrate this expression over the entire length of the rod, from $x = 0$ to $x = L$:
$$I = \int_0^L \lambda_0\left(x^2 + \frac{x^3}{L}\right) dx$$

Evaluating this integral is straightforward using the power rule:
$$I = \lambda_0 \left[ \frac{x^3}{3} + \frac{x^4}{4L} \right]_0^L$$

Plugging in the upper limit $L$:
$$I = \lambda_0 \left( \frac{L^3}{3} + \frac{L^3}{4} \right) = \frac{7}{12}\lambda_0 L^3$$

### The Mass Constraint

We have successfully found the moment of inertia, but our answer is in terms of the constant $\lambda_0$. If you look at the options, they are all expressed in terms of the total mass $M$ and length $L$. This means we need to find a relationship between $\lambda_0$ and $M$.

The total mass $M$ of the rod is simply the integral of $dm$ over the entire length:
$$M = \int_0^L dm = \int_0^L \lambda_0\left(1 + \frac{x}{L}\right) dx$$

Let's evaluate this integral:
$$M = \lambda_0 \left[ x + \frac{x^2}{2L} \right]_0^L$$

$$M = \lambda_0 \left( L + \frac{L^2}{2L} \right) = \lambda_0 \left( L + \frac{L}{2} \right) = \frac{3}{2}\lambda_0 L$$

Rearranging this equation to solve for $\lambda_0$, we get:
$$\lambda_0 = \frac{2M}{3L}$$

### The Final Assembly

Now, we have everything we need. We take our expression for $\lambda_0$ and substitute it back into our equation for the moment of inertia $I$:

$$I = \frac{7}{12} \left( \frac{2M}{3L} \right) L^3$$

Simplifying the fraction:
$$I = \frac{14}{36} ML^2$$

Which reduces beautifully to our final answer:
$$I = \frac{7}{18} ML^2$$

This result makes physical sense. A uniform rod would have a moment of inertia of $\frac{1}{3} ML^2$ (which is $\frac{6}{18} ML^2$). Because our rod is denser further away from the axis, more mass is concentrated at a greater distance, leading to a slightly higher moment of inertia of $\frac{7}{18} ML^2$.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Moment of Inertia of an Element}

\text{Substituting Mass Density}

\text{Integrating for Total } I

\text{Finding Total Mass } M

\text{Evaluating Total Mass}

\text{Final Substitution}

\text{The Way Forward}

The Sigma Insight: Moment of Inertia

Visualizing the Non-Uniformity

Setting Up the Integral

The Mass Constraint

The Final Assembly

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