Discussion
A particle of unit mass is moving on a plane. Its trajectory, in polar coordinates, is given by \(r(t) = t^{2}, \theta(t) = t\), where \(t\) is time. The kinetic energy of the particle at time \(t = 2\) is
Answer: C
The kinetic energy is 16.
In polar coordinates the velocity has a radial part \(\dot{r}\) and a transverse part \(r\dot{\theta}\). At t = 2:
\[ \begin{aligned} \dot{r} &= 2t = 4, \quad r\dot{\theta} = t^{2} \times 1 = 4 \\ KE &= \tfrac{1}{2}(4^{2} + 4^{2}) = 16 \end{aligned} \]
Using only the radial velocity gives 8, and forgetting the factor 1/2 gives 32; neither is offered. 4 and 12 do not follow from these velocity components, and 24 would need one of them to be larger than it is.
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