Question

A thin rod of mass m and length l is suspended from one of its ends

Answer 1

The question is incomplete.

However,

Given:

Mass = m

Length = l

Angular speed = ω

To find:

The rise of the center of mass from it's lowest position.

Solution:

As the center mass is positioned at a point,

There is an exchange of energies between kinetic and potential energy.

1 / 2 m v^2 = m g h

Where,

m - mass

v - velocity

g - gravity

h - height

Solving for h,

Height = v^2 / 2g

Substituting,

Height = ω^2 L^2 / 8 g

Hence, The rise of the center of mass from it's lowest position will be ω^2 L^2 / 8 g.

Answer 2

Explanation:

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