Conservation of Momentum

A small mass is thrown with a velocity of 2 m / s \SI[per-mode=symbol]{2}{\meter\per\second} toward a rod, a distance of / 3 \ell/3 from its center. The rod is hinged at the centre.

You are given that the mass is 6 kg \SI{6}{\kilo\gram} , and its length = 3 m \ell = \SI{3}{\meter} and the mass of the small object is 3 kg . \SI{3}{\kilo\gram}. Upon collision, the small mass sticks to the rod.

Find the angular velocity of the rod after this impact (in rad / s \si[per-mode=symbol]{\radian\per\second} ).


The answer is 0.8.

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1 solution

Md Zuhair
Aug 6, 2016

Using Conservation of Rotational Momentum :-

m v L / 3 = ( M L 2 / 12 + m L 2 / 9 ) ω mvL/3 = (ML^2/12 + mL^2/9) * \omega ,,, Find w after putting the values we will get ... ω = 0.2222 \omega = 0.2222

The question did not state whether the small object would stick to the rod or not. I assumed that it did not and obtained a different answer.

A Former Brilliant Member - 4 years, 10 months ago

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Sorry but this sum is wrong i have made a new one which is completely right. THis is the link https://brilliant.org/problems/rotating-with-it/?ref_id=1248045

Md Zuhair - 4 years, 10 months ago

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