$$\overrightarrow r(\theta) = (R \cos(θ), R \sin(θ), q θ) \quad ; \quad \theta = \theta (t)$$

I find the velocity by taking the derivative of position $$\overrightarrow v(\theta) = \left ( -R \sin (\theta) \frac{d\theta }{dt}, R \cos (\theta) \frac{d\theta }{dt}, q \frac{d\theta }{dt} \right )$$

I now need to find the kinetic energy (mass = $m$). To square the velocity, do I just square each component? As in,

$$\overrightarrow {v^2}(\theta) = \left ( -R^2 \sin ^2 (\theta) \left ( \frac{d\theta }{dt} \right )^2, R^2 \cos ^2(\theta) \left ( \frac{d\theta }{dt} \right )^2, q^2 \left ( \frac{d\theta }{dt} \right )^2 \right )$$

  • 2
    $\begingroup$ Kinetic energy is $\frac{1}{2}mv^2$ where $v$ is the speed. The speed may be calculated from the velocity vector by taking the norm. Squaring the coordinates of a vector to get a new vector is a very unnatural operation that is not likely to show up anywhere. (Except direct products of rings in abstract algebra.) $\endgroup$ – arctic tern Oct 16 '16 at 4:57

Kinetic energy is a scalar. By square we mean dot product: $(x_1,y_1)\cdot (x_2,y_2)=x_1x_2+y_1y_2$.


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