Question

A circular trajectory An object moves clockwise around a circle centered at the origin with radius 5 m beginning at the point (0,5). a. Find a position function \(\mathbf{r}\) that describes the motion if the object moves with a constant speed, completing 1 lap every \(12 \mathrm{s}\). b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{-t}\).

Short answer

Question: In an exercise, an object moves in a circular trajectory with the origin as the center and a radius of 5 meters. The initial position of the object is (0, 5). In the first part of the exercise, the object moves with a constant speed and completes one lap in 12 seconds. In the second part, the object moves with a speed given by \(e^{-t}\). Please write down the position functions for both parts. Answer: In the first part of the exercise, when the object moves with a constant speed, the position function is: $$ \begin{cases} x(t) = 5 \cos\left(\dfrac{\pi}{6} t + \dfrac{\pi}{2}\right) \\ y(t) = 5 \sin\left(\dfrac{\pi}{6} t + \dfrac{\pi}{2}\right) \end{cases} $$ In the second part of the exercise, when the object moves with a speed given by \(e^{-t}\), the position function is: $$ \begin{cases} x(t) = 5 \cos\left(-\dfrac{1}{5}e^{-t} + \dfrac{\pi}{2} + \dfrac{1}{5}\right) \\ y(t) = 5 \sin\left(-\dfrac{1}{5}e^{-t} + \dfrac{\pi}{2} + \dfrac{1}{5}\right) \end{cases} $$

Step-by-step solution

Part a: Constant speed

To find the position function for an object moving in a circle with a constant speed, we can use the parametric equations for a circle, which are given by $$ \begin{cases} x = R \cos(\omega t + \theta) \\ y = R \sin(\omega t + \theta) \end{cases} $$ where \(R\) is the radius, \(\omega\) is the angular speed, \(t\) is the time and \(\theta\) is the initial angle. In our case, initial angle \(\theta = \dfrac{\pi}{2}\) as the object is initially at \((0, 5)\). The angular speed can be computed as follows: $$ \omega = \dfrac{2\pi}{T} $$ where \(T\) is the time taken to complete one lap. Here, \(T=12\)s, so $$ \omega = \dfrac{2\pi}{12} = \dfrac{\pi}{6} $$ Now, plugging the values of \(R\), \(\omega\), and \(\theta\) in the parametric equations, we get the position function: $$ \begin{cases} x(t) = 5 \cos\left(\dfrac{\pi}{6} t + \dfrac{\pi}{2}\right) \\ y(t) = 5 \sin\left(\dfrac{\pi}{6} t + \dfrac{\pi}{2}\right) \end{cases} $$

Part b: Speed given by \(e^{-t}\)

To find the position function for this case, first, we will find the relationship between the speed and the angular speed using the formula \(v = R\omega\). The given speed function is \(v(t) = e^{-t}\). Therefore, $$ R\omega(t) = e^{-t} $$ Dividing by \(R=5\)m, we get $$ \omega(t) = \dfrac{1}{5}e^{-t} $$ Now, we have to integrate \(\omega(t)\) with respect to time to find the angle \(\theta(t)\): $$ \theta(t) = \int \dfrac{1}{5} e^{-t} dt = -\dfrac{1}{5}e^{-t} + C $$ At \(t=0\), \(\theta(0) = \dfrac{\pi}{2}\), so $$ C = \dfrac{\pi}{2} + \dfrac{1}{5} $$ Thus, the angle \(\theta(t)\) is given by $$ \theta(t) = -\dfrac{1}{5}e^{-t} + \dfrac{\pi}{2} + \dfrac{1}{5} $$ Finally, we can write the parametric equations for x and y as $$ \begin{cases} x(t) = 5 \cos\left(-\dfrac{1}{5}e^{-t} + \dfrac{\pi}{2} + \dfrac{1}{5}\right) \\ y(t) = 5 \sin\left(-\dfrac{1}{5}e^{-t} + \dfrac{\pi}{2} + \dfrac{1}{5}\right) \end{cases} $$

Key concepts

Angular Speed

Angular speed is a measure of how fast an object moves around a circular path. It is often denoted by the symbol \( \omega \) and is expressed in radians per second. This concept is crucial when dealing with objects moving in circular motion because it tells us how quickly the direction of the object's position vector changes.
For an object moving in a circle with a radius \( R \), the angular speed \( \omega \) can be calculated using the formula:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the period, or the time taken to complete one full circle or lap. If the object completes one circle in 12 seconds, as in the given problem, the angular speed becomes:\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \]

• It helps determine how quickly an object is rotating.
• It is a constant in situations where speed does not change over time.

Understanding angular speed is essential in predicting the movement path of objects in circular trajectories.

Position Function

The position function is used to describe the location of a point moving in a specific path over time. For an object in circular motion, this function is typically expressed using parametric equations, with time \( t \) being the parameter. This means that the positions \( x(t) \) and \( y(t) \) are functions of time and are based on a circle's geometry.
In circular motion with constant speed, these parametric equations look like:\[\begin{cases}x(t) = R \cos(\omega t + \theta) \y(t) = R \sin(\omega t + \theta)\end{cases}\]

• \( R \) is the circle's radius.
• \( \omega \) is the angular speed.
• \( \theta \) is the initial angle at \( t = 0 \).

For example, if an object starts at point (0,5) with radius 5 m, initial angle \( \theta = \frac{\pi}{2} \), and \( \omega = \frac{\pi}{6} \), the position is given by:

• \( x(t) = 5 \cos \left(\frac{\pi}{6} t + \frac{\pi}{2}\right) \)
• \( y(t) = 5 \sin \left(\frac{\pi}{6} t + \frac{\pi}{2}\right) \)

The position function allows us to visualize and predict where the object will be at any given time during its motion.

Circular Motion

Circular motion refers to the movement of an object along the perimeter of a circle. This motion can occur with constant speed, known as uniform circular motion, or with changing speed.
In uniform circular motion, although the object's speed remains constant, its velocity changes due to the continuous change in direction. The position functions will then depend heavily on the constant angular speed and radius of the circle.

• It involves both a radial distance (radius) and a changing angle (angular movement).
• Even if the speed is constant, the velocity vector changes direction continuously.

Non-uniform circular motion, as in part b of the problem, involves changing speed, modeled by a different position function where speed evolves as \( v(t) = e^{-t} \). For such cases, angular speed \( \omega(t) = \frac{1}{5}e^{-t} \) changes over time, and integrating \( \omega \) helps find the angle \( \theta(t) \). This integration gives:\[ \theta(t) = -\frac{1}{5}e^{-t} + \frac{\pi}{2} + \frac{1}{5} \]By incorporating varying angular speed into the parametric equations, one can describe the dynamic trajectory of the object over time. Circular motion is fundamental in physics and engineering, especially in systems involving wheels, gears, or celestial orbits.