A classical mechanics problem by Shivam Yaduvanshi

A uniform thin rod AB is equipped at both ends with the hooks and is supported by a frictionless horizontal table . initially the rod is hooked at a to fixed pin C about which it rotates with a constant angular velocity w1. suddenly end B of the rod hits pin D and gets hooked to pin D causing end A to be released determine the magnitude of angular velocity w2 of the rod in its subsequent rotation about D.

2w1 3w1 none w1/2 w1/3

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3 solutions

Sanjay Shukla
Dec 2, 2019

applying linear and angular impulse momentum theorem. linear -: -mVncm - (-mVcm) =I Vncm= new velocity of COM sign convention-: up=+ down=- hence-: -mLw2/2 + mLw1/2 =I angular impulse theorem about the first end of rod-: -mVncmL/2 + mL^2w2/12 -
(-mL^2w1/3)=IL hence-: -mL^2w2/4 + mL^2w2/12 + mL^2w1/3=LI I= -mLw2/6 + mLw1/3 equating with previous equation-: -mLw2/2 + mLw1/2 = -mLw2/6 + mLw1/3 hence mLw1/6 = mLw2/3 hence -: w2=w1/2

-(ml^2)w1/12+ml^2w1/4=ml^2w2/3

Could you kindly explain this further?

Kunal Mehta - 2 years, 8 months ago

W1/2 is correct answer

freaky1 town - 2 years, 3 months ago

How?? Please explain more

Ikshit Gupta - 2 years, 3 months ago
Devansh Kapri
Sep 25, 2018

Iw1 = mw2l^2 - Iw2

How u got this ? I'm getting iw2-mw2l^2

Madhura K - 1 year, 7 months ago

I used I(w1) = m(w1)l^2 - I(w2)

Lionel Messi - 1 year, 5 months ago

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