Question

A thin uniform rod of length \(2\) m cross-sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity \(ω\). If the value of \(ω\) in terms of its rotational kinetic energy \(E\) is \((αE/Ad)^{1/2}\), then the value of \(α\) is

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Moment of inertia of a rod about its centre is \(\frac{ML^2}{12}\), with \(M = Ad L\).

Answer 1

The Correct Option is B

Step 1: Plan:
Think of the rod as two halves, each of length $1$ m, spinning about one end.

Step 2: Steps:
Each half has mass $Ad$ and moment of inertia about its end $\frac{(Ad)(1)^2}{3} = \frac{Ad}{3}$. The two halves together give $I = \frac{2Ad}{3}$.
From $E = \frac12I\omega^2$, $\omega^2 = \frac{2E}{I} = \frac{2E\cdot3}{2Ad} = \frac{3E}{Ad}$. So $\alpha = 3$.

Final Answer:
The value of $\alpha$ is $3$, option (B). \[ \boxed{3} \]