Relativistic rotational kinetic energy

[johne1618]

Hi,

I was wondering if the formula for rotational energy:

E = 1/2 * L * w

where L is the angular momentum and w is the angular velocity,
is actually correct for relativistic velocities.

Using

L = p * r

and

w = v / r

where

p = the linear momentum = m * v

We get:

E = 1/2 * (m * v * r) * (v/r)

I would have thought that this expression is correct up to relativistic velocity as the expressions for linear momentum, p = m * v, and angular velocity, w = v / r, do not require relativistic modification provided that we acknowledge that m increases as v -> c.

As a corollary, as v -> c, it seems that the maximum rotational energy of a system is:

E = 1/2 * m c^2 (i.e. half the total mass/energy of the system)

John

 

[Bill_K]

Kinetic energy is not a particularly useful concept in special relativity. Neither is the 'relativistic mass.' Always use the rest mass and put in the factors of γ explicitly. The exact formula for the total energy of a particle is E² = m²c⁴ + p²c² where p = γmv. This can be written

E = mc² √(1 + γ²v²/c²)

The kinetic energy will be the total E minus the rest energy:

KE = mc²(√(1 + γ²v²/c²) - 1)

This is still exact. For circular motion, just put in v = ω/r.