JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Rotational Mechanics

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 ____.

Answer: 18

Solution

Linear impulse:

P = mω r_{c m} = mω (\frac{3 L}{2})

. Angular impulse about O:

(\frac{3 L}{2} + x) P = (\frac{m L ^{2}}{12} + m (\frac{3 L}{2})^{2}) ω = \frac{7 m L ^{2}}{3} ω

. Dividing:

\frac{3 L}{2} + x = \frac{7 L}{3} \times \frac{2}{3} = \frac{14 L}{9}

. Solving:

x = \frac{14 L}{9} - \frac{3 L}{2} = \frac{L}{18}

. Therefore,

n = 18

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Integer answer

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JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Rotational Mechanics

Solution

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