Question

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↬A thin uniform rod of mass m and length l is hinted at the lower and to a level floor and stand vertically. It is now allowed to fall, then its upper end will strike the floor with the velocity?​

Answer 1

Explanation:

Given that a thin uniform rod of mass m and length l is hinted at the lower and to a level floor and stand vertically. When it is standing vertically it's mass is 1/2 of the distance above the ground.

Now the uniform rod is allowed to fall. We need to find its upper end will strike the floor with the velocity.

When the uniform rod is allowed to falls or when it falls on the ground; it's potential energy convert into kinetic energy.

0 + 1/2 Iω²

1/2 × ml²/3 × (v/l)²

Potential energy at G is 1/2 mgl

So,

1/2 mv²/3 = 1/2 mgl

v = √(3gl)

Answer 2

Answer:

When the rod is standing vertically, its center of mass is 1/2

Distance above the ground. So it has a potential energy. When it falls to the ground, its potential energy is converted into kinetic energy.

So using law of conservation of energy. We have,

$mg \frac{1}{2} = \frac{1}{2} w {}^{2} . \\ = mg \frac{1}{2} = \frac{1}{2} ( \frac{1}{3} ml {}^{2} )( \frac{v}{l} ) {}^{2} . \\ v = \sqrt{3gl} .$