Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligib

Q: Why is the work done equal to the rotational kinetic energy?

By the Work-Energy Theorem, the net work done on a system equals its change in kinetic energy. Since the rod starts from rest and reaches an angular speed $\omega$, the work done is exactly $\frac{1}{2}I\omega^2$.

Q: Why does minimum work imply minimum moment of inertia?

The final angular speed $\omega$ is given as uniform (constant for our comparison). In the formula $W = \frac{1}{2}I\omega^2$, $W$ is directly proportional to $I$. Thus, minimizing $I$ minimizes $W$.

Q: How do we know $I$ is minimum at the Center of Mass?

The Parallel Axis Theorem states $I = I_{com} + Md^2$. Since $Md^2$ is always positive or zero, $I$ is strictly minimized when $d = 0$, which means the axis passes exactly through the Center of Mass.

Q: Can I solve this without knowing the Center of Mass property?

Yes! You can write the expression for $I$ as a function of $x$: $I(x) = 0.3x^2 + 0.7(1.4 - x)^2$. To find the minimum, take the derivative with respect to $x$ and set it to zero: $dI/dx = 0$. Solving this will give you the exact same $x = 0.98$ m.

Q: Does the negligible mass of the rod affect the answer?

Yes, significantly. Because the rod has negligible mass, we only consider the two point masses when calculating the moment of inertia and the center of mass. If the rod had mass, we would have to include its contribution, which would shift the COM.

Question

Two point masses of $0.3$ kg and $0.7$ kg are fixed at the ends of a rod of length $1.4$ m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

Options:

(A) $0.42$ m from mass of $0.3$ kg

(B) $0.70$ m from mass of $0.7$ kg

(C) $0.98$ m from mass of $0.3$ kg

(D) $0.98$ m from mass of $0.7$ kg

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Masses

Imagine you are holding a rod with two weights attached to its ends. One is a lighter $0.3$ kg mass, and the other is a heavier $0.7$ kg mass. The rod itself is practically weightless, serving only as a rigid bridge of $1.4$ m between the two masses.

Your task is to spin this rod like a baton. But here is the catch: you want to be lazy. You want to find the exact pivot point that requires the absolute minimum amount of work to reach a specific spinning speed.

### The Physics Connection: Work and Inertia

To understand how to minimize our effort, we need to translate 'work' into the language of rotational dynamics. According to the Work-Energy Theorem, the work done on a system equals its change in kinetic energy.

Since we are starting from rest and reaching a final angular velocity $\omega$, the work done is exactly equal to the final rotational kinetic energy. Mathematically, this is expressed as $W = \frac{1}{2} I \omega^2$.

Here is the brilliant part: the target angular velocity $\omega$ is fixed. Therefore, the only way to minimize the work $W$ is to minimize the moment of inertia $I$. We are no longer looking for minimum work; we are hunting for the **minimum moment of inertia**!

### The Master Key: The Parallel Axis Theorem

How do we find the axis that gives the smallest possible moment of inertia? This is where the Parallel Axis Theorem comes to our rescue. The theorem states that $I = I_{com} + Md^2$.

In this equation, $I_{com}$ is the moment of inertia about the Center of Mass, $M$ is the total mass, and $d$ is the distance from the Center of Mass to our chosen axis. Because $Md^2$ can never be negative, the absolute minimum value for $I$ occurs when $d = 0$.

This is a profound realization! The universe is telling us that a body naturally 'wants' to rotate about its Center of Mass. To minimize our work, we simply need to place our axis exactly at the Center of Mass of the system.

### The Final Calculation: Pinpointing the Sweet Spot

Now, the physics problem has transformed into a simple geometry problem. We just need to locate the Center of Mass of our two-particle system. Let's set our origin at the lighter mass, $m_1 = 0.3$ kg.

The formula for the Center of Mass is $x_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Since $m_1$ is at the origin ($x_1 = 0$), the formula simplifies beautifully to $x_{com} = \frac{m_2 L}{m_1 + m_2}$.

Let's plug in our numbers. We have $x_{com} = \frac{0.7 	imes 1.4}{0.3 + 0.7}$. Notice how the denominator perfectly sums up to $1.0$, making our calculation a breeze.

Multiplying $0.7$ by $1.4$ gives us exactly **$0.98$ m**. Therefore, the optimal axis of rotation is located $0.98$ m away from the $0.3$ kg mass. This perfectly matches option (c).

### The Calculus Alternative: A Rigorous Proof

What if you forgot the Center of Mass trick during the exam? Don't panic; calculus always has your back. You can write the moment of inertia as a function of the distance $x$ from the first mass.

The equation becomes $I(x) = 0.3x^2 + 0.7(1.4 - x)^2$. To find the minimum of this function, we take the derivative with respect to $x$ and set it to zero: $\frac{dI}{dx} = 0$.

Differentiating gives $2(0.3)x - 2(0.7)(1.4 - x) = 0$. Solving this linear equation will lead you straight back to $x = 0.98$ m. While calculus is a powerful safety net, recognizing the physical significance of the Center of Mass saves precious time in competitive exams like JEE!

Select Answer:

Visualized Solution

Visualizing the Setup

Work-Energy Theorem

Condition for Minimum Moment of Inertia

Center of Mass Formula

Substituting Values

Final Calculation

Conclusion

Alternative Method: Calculus

The Sigma Insight: Moment of Inertia

The Setup

The Physics Connection

The Master Key

The Final Calculation

The Calculus Alternative

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