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QJEEM18Phbca1: JEE Main 2018 Physics - Class 11 JEEM18Phbca1 Solution: A thin rod MN, free to rotate in the vertical plane aboutthe…

A thin rod MN, free to rotate in the vertical plane aboutthe fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to : (see figure)

JEE Main 2018 (Online) 15th April Evening Slot Physics - Rotational Motion Question 187 English

Multiple Choice Options

  • A sin⁡α
  • B sin⁡α
  • C cos⁡α
  • D cos⁡α

Step-by-Step Verified Solution

When the end M is released and rod makes angle α with horizontal, the displacement of centre of mass is L2sin⁡α.

Now, we know mgL2sin⁡α=12Iω2

Here, I=mL23; therefore,

mgL2sin⁡α=12mL23ω2

⇒ω2=3gsin⁡αL⇒ω=3gsin⁡αL⇒ω∝sin⁡α

JEE Main 2018 (Online) 15th April Evening Slot Physics - Rotational Motion Question 187 English Explanation

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