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QJEEM20Ph0f2a: JEE Main 2020 Physics - Class 11 JEEM20Ph0f2a Solution: A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, …

A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be _________. JEE Main 2020 (Online) 5th September Evening Slot Physics - Rotational Motion Question 120 English

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The given situation is shown in the following figure,

JEE Main 2020 (Online) 5th September Evening Slot Physics - Rotational Motion Question 120 English Explanation

Applying law of conservation of angular momentum about pivotal point,

Li=Lf⇒mvl=Iω

⇒mvl=(13ML2+mL2)ω ..... (i)

Given, m = 0.1 kg, v = 80 ms−1, M = 0.9 kg, l = 1 m

Substituting these values in eq. (i), we get

0.1×80×1=(0.9×123+0.1×12)ω

8=(310+110)ω⇒8=410ω⇒ω=20 rad/s

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