The radius of gyration of a uniform rod of length $\ell$, about an axis passing through a point $\frac{\ell}{4}$ away from the centre of the rod, and perpendicular to it is:

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Rotational Mechanics

The radius of gyration of a uniform rod of length

ℓ

, about an axis passing through a point

\frac{ℓ}{4}

away from the centre of the rod, and perpendicular to it is:

(A) $\frac{1}{8} ℓ$
(B) $\frac{7}{48} ℓ$ ✓ Correct
(C) $\frac{1}{4} ℓ$
(D) $\frac{3}{8} ℓ$

Answer: (B)

Solution

The rod's moment of inertia about its centre is

\frac{M ℓ ^{2}}{12}

, and the axis lies a distance

d = \frac{ℓ}{4}

away, so the parallel-axis theorem gives

I = \frac{M ℓ ^{2}}{12} + M (\frac{ℓ}{4})^{2} = \frac{M ℓ ^{2}}{12} + \frac{M ℓ ^{2}}{16} = \frac{7 M ℓ ^{2}}{48}

. Writing

I = M K^{2}

identifies the radius of gyration as

K = \frac{7}{48} ℓ

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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