Question

If the length of a thin uniform rod is 'L' and the radius of gyration of the rod about an axis perpendicular to its length and passing through one end is K, then K:L=

• A. 1:$\sqrt{3}$
• B. 1:$\sqrt{2}$
• C. 1:3
• D. 1:2

Hint:

Remember the moments of inertia for common shapes. For a thin rod: - About the center: $I_c = ML^2/12$. - About one end: $I_{end} = ML^2/3$. You can derive the second from the first using the parallel axis theorem: $I_{end} = I_c + M(L/2)^2 = ML^2/12 + ML^2/4 = ML^2/3$.

Answer 1

The Correct Option is A

Step 1: State the formula for the moment of inertia of a rod about its end.
The moment of inertia ($I$) of a thin uniform rod of mass $M$ and length $L$ about an axis perpendicular to the rod and passing through one of its ends is given by: \[ I = \frac{1}{3}ML^2. \]

Step 2: State the definition of the radius of gyration.
The radius of gyration ($K$) is defined as the distance from the axis of rotation to a point where the entire mass of the body could be concentrated without changing its moment of inertia. The relationship is: \[ I = MK^2. \]

Step 3: Equate the two expressions for the moment of inertia to find K.
\[ MK^2 = \frac{1}{3}ML^2. \] The mass $M$ cancels out. \[ K^2 = \frac{L^2}{3}. \] Taking the square root, we get: \[ K = \frac{L}{\sqrt{3}}. \]

Step 4: Find the required ratio K:L.
\[ \frac{K}{L} = \frac{L/\sqrt{3}}{L} = \frac{1}{\sqrt{3}}. \] Therefore, the ratio $K:L$ is $1:\sqrt{3}$.