Rotating Rod Kinetic Energy

Classical Mechanics Level 3

A uniform thin rod of mass M = 1 M = 1 has one end fixed at the origin, but it can otherwise rotate about the origin. The rod is parametrized as follows:

x = r cos ⁡ ( θ ) sin ⁡ ( ϕ ) y = r sin ⁡ ( θ ) sin ⁡ ( ϕ ) z = r cos ⁡ ( ϕ ) 0 ≤ r ≤ 1 x = r \, \cos (\theta) \, \sin (\phi) \\ y = r \, \sin (\theta) \, \sin (\phi) \\ z = r \, \cos (\phi) \\ 0 \leq r \leq 1

At a particular instant, the angles and their time rates of change are:

θ = π 3 ϕ = π 4 θ ˙ = 5 ϕ ˙ = 7 \theta = \frac{\pi}{3} \\ \phi = \frac{\pi}{4} \\ \dot{\theta} = 5 \\ \dot{\phi} = 7

At this instant, what is the rod's kinetic energy?


The answer is 10.25.

Steven Chase by Steven Chase , 37, USA

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1 solution

Square of velocity of the particles of the rod as a function of distance r from the origin is 61.5 r 2 r^2 . Hence kinetic energy of an element of length dr at a distance r from the origin is 1 2 \dfrac{1}{2} (61.5) r 2 r^2 dr. Integrating this within the limits from 0 to 1 we get the K.E. of the rod equal to 10.25

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