Conservation of energy

Classical Mechanics Level 2

A rod of mass M M and length L L is hinged at its end and is in horizontal position initially. It is then released to fall under gravity. Find the angular speed of rotation of rod when the rod becomes vertical.

ω = 2 g L \omega = \sqrt {\frac{{2g}}{L}} ω = 3 g 2 L \omega = \sqrt {\frac{{3g}}{2L}} ω = 3 g L \omega = \sqrt {\frac{{3g}}{L}} ω = g 3 L \omega = \sqrt {\frac{{g}}{3L}}

Rohit Gupta by Rohit Gupta , 34, India

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1 solution

Rohit Ner
Feb 10, 2016

Applying conservation of energy at the highest and the bottom most point we get
M g L 2 = 1 2 . M L 2 3 ω 2 ω = 3 g L Mg\frac{L}{2}=\frac{1}{2}.\frac{M{L}^2}{3}{\omega}^2\\\huge\color{#3D99F6}{\boxed{\omega=\sqrt{\frac{3g}{L}}}}

5 Helpful 0 Interesting 0 Brilliant 1 Confused

What about the Kinetic Energy, Or the Translation K.E

Md Zuhair - 4 years, 2 months ago

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If you want to take into account the translation kinetic energy, you have to take the moment of inertia around the centre of gravity and not the centre of rotation. This will give the same result.

Younes Bouhafid - 1 year, 6 months ago

Yes the rod also posseses a kinetic energy at bottom

Sarthak Shiv - 3 years, 7 months ago

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