Rigid Body Relativistic Motion

Classical Mechanics Level pending

Consider a rigid rod of length L L and rest mass M M uniformly distributed uniformly along the length of the rod. One end of the rod is hinged at the origin and the rod rotates about the Z-axis at a constant angular speed ω \omega . Compute the relativistic kinetic energy of the rod. The result comes out to be:

T = M c 2 ( π A B − C ) \mathcal{T} = Mc^2\left(\frac{\pi^A}{B}-C\right)

Where T \mathcal{T} denotes the kinetic energy of the rod. Enter your answer as A + B + C \boxed{A+B+C} , where A A and B B and C C are positive integers. Take:

ω = c 2 L \omega = \frac{c}{2L}

Where c c is the speed of light.

Hint: Start with the expression for the relativistic kinetic energy for a point mass.

Bonus: Try to derive the classical result using the final expression that you obtain.


The answer is 5.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Steven Chase
Apr 26, 2020

Some preliminaries:

d m = M d r L v = r ω dm = \frac{M \, dr}{L} \\ v = r \omega

Relativistic KE for a segment of length d r dr :

d T = d m c 2 ( 1 1 − r 2 ω 2 c 2 − 1 ) = M d r L c 2 ( 1 1 − r 2 ω 2 c 2 − 1 ) d \mathcal{T} = dm \, c^2 \, \Big( \frac{1}{\sqrt{1 - \frac{r^2 \omega^2}{c^2}}} - 1 \Big) = \frac{M \, dr}{L} \, c^2 \, \Big( \frac{1}{\sqrt{1 - \frac{r^2 \omega^2}{c^2}}} - 1 \Big)

Integrating from r = 0 r = 0 to r = L r = L results in:

T = M c 2 L [ c w s i n − 1 ( ω L c ) − L ] \mathcal{T} = \frac{M c^2}{L} \Big[\frac{c}{w} sin^{-1} \Big(\frac{\omega L }{c} \Big) - L \Big]

Plugging in ω = c 2 L \omega = \frac{c}{2L} in the relativistic case:

T = M c 2 L [ 2 L s i n − 1 ( 1 2 ) − L ] = M c 2 [ 2 π 6 − 1 ] = M c 2 [ π 3 − 1 ] \mathcal{T} = \frac{M c^2}{L} \Big[2 L sin^{-1} \Big(\frac{1 }{2} \Big) - L \Big] = M c^2 \Big[2 \frac{\pi}{6} - 1\Big] = M c^2 \Big[ \frac{\pi}{3} - 1\Big]

Now, for the low-energy case, we can use an approximation for arcsine, assuming that the argument is small.

s i n − 1 x ≈ x + x 3 6 sin^{-1} x \approx x + \frac{x^3}{6}

Plugging in:

T = M c 2 L [ c w ( ω L c + ω 3 L 3 6 c 3 ) − L ] = M c 2 L ( ω 2 L 3 6 c 2 ) = M ( ω 2 L 2 6 ) = 1 2 M L 2 3 ω 2 = 1 2 I r o d ω 2 \mathcal{T} = \frac{M c^2}{L} \Big[\frac{c}{w} \Big(\frac{\omega L }{c} + \frac{\omega^3 L^3 }{6 c^3} \Big) - L \Big] \\ = \frac{M c^2}{L} \Big( \frac{\omega^2 L^3 }{6 c^2} \Big) \\ = M \Big( \frac{\omega^2 L^2 }{6} \Big) = \frac{1}{2} \frac{M L^2}{3} \omega^2 = \frac{1}{2} I_{rod} \, \omega^2

Thus, in the low-energy limit, the relativistic expression boils down to the ordinary kinetic energy expression for a rod rotating about one end.

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