Question

Let M and L be the mass and length of thin uniform rod respectively. In first case, axis of rotation is passing through centre and perpendicular to length of rod. In second case axis of rotation is passing through one end and perpendicular to length of rod. The ratio of radius of gyration in first case to second case is

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{8}\)
• D. \(\frac{1}{6}\)

Hint:

Use I = ML^2/12 about the centre and ML^2/3 about the end, with I = M k^2.

Answer 1

The Correct Option is B

Step 1: Use the parallel axis theorem:
$I_2=I_1+M\left(\dfrac L2\right)^2=\dfrac{ML^2}{12}+\dfrac{ML^2}{4}=\dfrac{ML^2}{3}$. So $I_1/I_2=\dfrac14$.

Step 2: Convert to radii:
Since $k\propto\sqrt I$ for the same mass, $\dfrac{k_1}{k_2}=\sqrt{\dfrac{I_1}{I_2}}=\sqrt{\dfrac14}$.

Step 3: Pick:
$\dfrac12$, option B.

Final Answer:
The ratio of moments is 1/4, so the ratio of radii is 1/2. \[ \boxed{\text{(B) }\dfrac12} \]