JEE Main · PhysicsHard

$A uniform rod of length $L$ and mass $M$ is pivoted at one end and oscillates as a physical pendulum. A small bead of mass $m$ is attached to the rod at a distance $x$ from the pivot. What is the value of $x$ that minimizes the period of oscillation?$

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Correct answer: A — $L/\sqrt{3}$

Solution

The period of a physical pendulum is

T = 2 π I / (M + m) g d_{c m}

. By substituting the moment of inertia

I = M L^{2} /3 + m x^{2}

and the center of mass distance

d_{c m} = (M L /2 + m x) / (M + m)

, we minimize

T

with respect to

x

to find

x = L / 3