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in Rigid Body Dynamics by (49.6k points)
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Calculate the kinetic energy of a body, rotating about an axis with uniform angular velocity.

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Rotational Kinetic Energy

Suppose a body is rotating with uniform angular velocity co relative to the axis YY’. Every particle of the body will have same angular velocity but different linear velocity. Suppose, a particle of mass m1 is at a distance r1 from the rotational axis. Since, the linear velocity of each particle is the multiplication of the angular velocity with the distance from the rotational axis. Hence, linear velocity of the particle.
v1 = r1ω
and its kinetic energy

Fig: Rotational kinetic energy

Similarly, if the masses of other particles are m2, m3,… and distances from the rotational axis are r2, r3,… then their kinetic energy will be \(\frac{1}{2}\)m2r22ω2\(\frac{1}{2}\) m3r32ω2, … respectively since, kinetic energy is a scalar quantity, hence the total kinetic energy will be the sum of kinetic energies of all the particles.
ER = \(\frac{1}{2}\) m1r12ω2 + \(\frac{1}{2}\) m2r22ω2 + \(\frac{1}{2}\) m3r32ω2 + ………
\(\frac{1}{2}\) (m1r12 + m2r22 + m3r32 + …………) ω2
\(\frac{1}{2}\) (Σ mr2) ω2
But, Σ mr2 , is the moment of inertia I of the figid body about the axis
Therefore; ER = \(\frac{1}{2}\) Iω2 …………… (1)
This is the formula for kinetic energy of uniform rotation. It is clear from this that the linear kinetic energy \(\left[\frac{1}{2} m v^{2}\right]\) of a body is half the multiplication of mass m and linear velocity v2. Similarly, the rotational kinetic energy \(\left[\frac{1}{2} I \omega^{2}\right]\) is half the multiplication of the moment of inertia I and angular velocity ω2.
Keeping ω = 1 in equation (1) ;
I = 2E
Hence, if a body is moving with unit angular velocity along an axis then the moment of inertia of the body along the axis would be double its rotational kinetic energy. In other words rotational kinetic energy would be half of moment of inertia.
If any body is moving along its axis and also performing linear motion (e.g., moving motor or cycle’s wheel); then its total kinetic energy will be equal to the sum of rotational kinetic energy and linear kinetic energy.
Etotal = \(\frac{1}{2}\) Iω2 + \(\frac{1}{2}\) Mv2 ………….. (2)

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