JEE Advanced 2024 Paper 1 · Q13 · Angular Momentum

Question

A thin uniform rod of length $L$ and certain mass is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse $P$ is imparted to the rod at a distance $x = L/n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of $n$ is _____.

SolveKar AI · Accepted Answer

Let the rod, of mass $m$ and length $L$, lie along the line through O immediately after the impulse, with its near end attached by a string of length $L$ to O. So the rod's centre of mass is at distance $L + L/2 = 3L/2$ from O. The impulse $P$ is applied perpendicular to the rod at distance $x = L/n$ from the mid-point (centre). Linear impulse: $P = m\,v_{cm}$, with $v_{cm}$ being the CM speed.

For the system to subsequently rotate rigidly around O (rod-and-string colinear), all points of the rod must share the same angular velocity $\omega$ about O. Then $v_{cm} = (3L/2)\omega$.

Angular impulse about O: $P\cdot(3L/2 + x) = I_O\,\omega$, where $I_O = I_{cm} + m(3L/2)^2 = mL^2/12 + 9mL^2/4 = (mL^2/12)(1 + 27) = 7mL^2/3$.

From $P = mv_{cm} = m(3L/2)\omega$: $\omega = 2P/(3mL)$. Substituting: $P(3L/2 + L/n) = (7mL^2/3)\cdot 2P/(3mL) = 14LP/9$. Cancel $P$: $3L/2 + L/n = 14L/9$, so $L/n = 14L/9 - 3L/2 = (28 - 27)L/18 = L/18$. Hence $n = 18$.

JEE Advanced 2024 Paper 1 · Q13 · Angular Momentum

Solution