Question

Two identical particles each of mass $m$ are separated by a distance $d$. The axis of rotation passes through the midpoint of $d$ and is perpendicular to the length $d$. If $K$ is the average rotational kinetic energy of the system, then the angular frequency is

• A. $2d\sqrt{\frac{m}{K}}$
• B. $\frac{d}{2}\sqrt{\frac{K}{m}}$
• C. $\frac{2}{d}\sqrt{\frac{K}{m}}$
• D. $\frac{d}{4}\sqrt{\frac{m}{K}}$

Hint:

For two identical masses $m$ separated by a distance $d$, the moment of inertia about the central perpendicular axis is always half the value of a single particle rotating at full distance $d$ (i.e., $I = \frac{1}{2}md^2$). Remembering this standard value helps bypass the distance-halving steps entirely during the exam.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem presents a system of two identical masses $m$ separated by a fixed distance $d$.
The system rotates about an axis that passes through the exact center of the distance separating them and runs perpendicular to the line connecting them.
Given the average rotational kinetic energy $K$, we are required to determine the angular frequency $\omega$ of the system.

Step 2: Key Formula or Approach:
The rotational kinetic energy of a system is given by the formula:
$$K = \frac{1}{2} I \omega^2$$ Where $I$ is the total moment of inertia of the system about the given axis of rotation, and $\omega$ is the angular frequency.
The moment of inertia for a system of point masses is defined as $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of each mass from the axis of rotation.

Step 3: Detailed Explanation:
Since the axis of rotation passes through the midpoint of the distance $d$, each particle of mass $m$ is located at an equal perpendicular distance from the axis:
$$r = \frac{d}{2}$$ Now, calculate the total moment of inertia $I$ of the two-particle system:
$$I = m\left(\frac{d}{2}\right)^2 + m\left(\frac{d}{2}\right)^2$$ $$I = m\left(\frac{d^2}{4}\right) + m\left(\frac{d^2}{4}\right) = \frac{2md^2}{4} = \frac{md^2}{2}$$ Substitute this value of $I$ into the rotational kinetic energy formula:
$$K = \frac{1}{2} \left(\frac{md^2}{2}\right) \omega^2$$ $$K = \frac{md^2}{4} \omega^2$$ Rearrange the equation to isolate $\omega^2$:
$$\omega^2 = \frac{4K}{md^2}$$ Taking the square root on both sides yields the angular frequency $\omega$:
$$\omega = \sqrt{\frac{4K}{md^2}} = \frac{2}{d}\sqrt{\frac{K}{m}}$$

Step 4: Final Answer:
The angular frequency of the system is $\frac{2}{d}\sqrt{\frac{K}{m}}$, which perfectly matches option (C).