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Angular kinematics at the racetrack

                 

We derived the kinematic relations for motion in linear spaces like motion on the line, in the plane, or in 3D physical reality. However, they work just as well in other spaces, e.g. for a curved racetrack.

Making the identifications below, we can apply the relations unchanged: \[\begin{align} \textrm{Angular position}\ \theta &\longleftrightarrow \textrm{Position}\ s \\ \textrm{Angular velocity}\ \dot{\theta} &\longleftrightarrow \textrm{Velocity}\ \dot{s} \\ \textrm{Angular acceleration}\ \ddot{\theta} &\longleftrightarrow \textrm{Acceleration}\ \ddot{s}. \end{align}\] The angular velocity and acceleration are also commonly referred to as \(\omega\) and \(\alpha\).

We measure the position of the racecar around the circular track by the angle \(\theta\). This is related to the distance \(s\) around the track by \(s=R\theta\). This implies the following relations \[\begin{align} \dot{s} &= R\dot{\theta} = R\omega \\ \ddot{s} &= R\ddot{\theta} = R\alpha \end{align}\] By substitution, we can obtain the angular form of the first kinematic relation \[\theta_T = \theta_0 + \omega T + \frac12 \alpha_0 T^2,\] and similarly \[{\omega_T}^2 - {\omega_0}^2 = 2\alpha_0\left(\theta_T-\theta_0\right).\] Note that \(\theta\) is periodic (\(0\) and \(2\pi\) represent the same location), while \(s\) has no bound.

One distinction between the circle and the line is that the circle closes on itself, so angles are between \(0\) and \(2\pi\).

This means that we need to keep track of how many times we've gone around the circle.

Suppose we leave the starting line with \(v=\SI[per-mode=symbol]{10\pi}{\meter\per\second}\), and the track has radius \(R=\SI{100}{\meter}\). What is our angular position (in \(\si{\radian}\)) around the track after \(\SI{25}{\second}?\)

Suppose you're a half lap behind the leader when they suffer a mechanical failure, dropping their speed to \(v_1 = \SI[per-mode=symbol]{50}{\meter\per\second}\). Meanwhile your speed stays at \(v_2 = \SI[per-mode=symbol]{60}{\meter\per\second}\).

If the lead car is one and a half laps from the finish line when the mechanical failure occurs, will you catch the leader before the end of the race?

Note: surprising though it may seem, it does not matter what the radius of the track is.

What is the minimum acceleration (in \(\si[per-mode=symbol]{\radian\per\second\squared}\)) you'll need to catch the lead car before the end of the race?

Remember: use your results from the previous problems. Again the radius of the track is \(R = \SI{100}{\meter}.\)

In professional races like NASCAR, the driver's starting position is determined by their performance in a time trial. Their trial is reported in terms of the average lap speed \(\langle v_\textrm{avg}\rangle\).

Since the length of the lap is fixed at \(\ell\), we have \(\ell = \langle v\rangle T\) so that the average speed \(\langle v_\textrm{avg}\rangle\) and the total time \(T\) are inversely proportional, \(\langle v_\textrm{avg}\rangle\sim T^{-1}\).

On the first lap of your trial, you underperform slightly, averaging \(\SI{110}{mph}\) though you're aiming for an average speed of \(\langle v_T \rangle = \SI{118}{mph}\).

How fast do you need to go (in mph) in your second lap to hit your goal?

Suppose that on the first lap you skid out on an oil slick so that your average speed is \(\langle v_1\rangle = \SI{61}{mph}\).

How fast (in \(\si{mph}\)) must you go on your second lap to hit your goal of \(\langle v_T\rangle = \SI{120}{mph}\)?

Details

  • \(\langle v_T\rangle\) is your average speed over the two laps of the time trial.

In this quiz we showed how to translate our equations of linear kinematics so as to handle motion in rotational systems. Despite the fundamental difference between the two kinds of spaces—one is infinite and the other is periodic—we only had to make minor changes to map our equations.

As we go on, the ability to remap results from one space to another, or from one dimensionality to another, becomes more important. In many cases, it is the fastest way to make progress on difficult problems.

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