Given:
$$\overrightarrow r(\theta) = (R \cos(θ), R \sin(θ), q θ) \quad ; \quad \theta = \theta (t)$$
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,
$$\overrightarrow {v^2}(\theta) = \left ( -R^2 \sin ^2 (\theta) \left ( \frac{d\theta }{dt} \right )^2, R^2 \cos ^2(\theta) \left ( \frac{d\theta }{dt} \right )^2, q^2 \left ( \frac{d\theta }{dt} \right )^2 \right )$$
Best Answer
Kinetic energy is a scalar. By square we mean dot product: $(x_1,y_1)\cdot (x_2,y_2)=x_1x_2+y_1y_2$.