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Physics · Rotational Dynamics

JEE Advanced 2024 — Paper 1 — Question 30

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

Question figure

Answer: 18

Numerical answer — enter this value.

Step-by-step solution

Conservation of angular momentum about O.

P(L+L2+x)=[mL212+m(3L2)2]ωP\left(L+\frac{L}{2}+x\right)=\left[\frac{m L^{2}}{12}+m\left(\frac{3 L}{2}\right)^{2}\right] \omega

P(3L2+x)=73mL2ω\begin{gathered} P\left(\frac{3 L}{2}+x\right)=\frac{7}{3} m L^{2} \omega \end{gathered} vcm=32ω L\begin{gathered} \mathrm{v}_{\mathrm{cm}}=\frac{3}{2} \omega \mathrm{~L} \end{gathered}

Conservation of linear momentum P=mvvcm\mathrm{P}=\mathrm{mv} \mathrm{v}_{\mathrm{cm}} \quad...

(iii) Pm=32[37{P(3L2+x)}1mL]\frac{P}{m}=\frac{3}{2}\left[\frac{3}{7}\left\{P\left(\frac{3 L}{2}+x\right)\right\} \frac{1}{m L}\right]

14L=92(3L+2x)14 L=\frac{9}{2}(3 L+2 x)

⇒L=18x\Rightarrow L=18 x x=L/18x=L / 18

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Angular Momentum and its Conservation

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