Question

A rod of mass 'm' and length 'l' pivoted at P when its angular speed of rotation is ω=√(g/l). find its 1)total kinetic energy 2)rotational kinetic energy 3)translational kinetic energy

Answer 1

Answer:

kinetic energy of rotation = $\frac{1}{2}$CM$w^{2}$

$KE_{tran} = \frac{1}{8} mgt$

$KE_{total} = \frac{7}{24} mgt$

Explanation:

(a) kinetic energy of rotation :

             Rotational energy also known as angular kinetic energy is defined as: The kinetic energy due to the rotation of an object and is part of its total kinetic energy. Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.    

kinetic energy of rotation = $\frac{1}{2}$CM$w^{2}$

(b) kinetic energy of translation : = $\frac{1}{2} mV^{2} CM$

             Translational kinetic energy of a body is equal to one-half the product of its mass, m, and the square of its velocity, v, or 1/2mv2. For a rotating body the moment of inertia, I, corresponds to mass, and the angular velocity (omega), ω, corresponds to linear, or translational, velocity.

$KE_{tran} = \frac{1}{2} (\frac{1}{3} ml^{2} )((\sqrt{\frac{xg{g^{2} } } ))^{2}$

$KE_{tran} = \frac{1}{8} mgt$

(c) Total kinetic energy :

$KE_{rot} + KE_{tran}$

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

$KE_{total} = \frac{7}{24} mgt$

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