Question

Consider an object moving along the circular trajectory \(\mathbf{r}(t)=\langle A \cos \omega t, A \sin \omega t\rangle,\) where \(A\) and \(\omega\) are constants. a. Over what time interval \([0, T]\) does the object traverse the circle once? b. Find the velocity and speed of the object. Is the velocity constant in either direction or magnitude? Is the speed constant? c. Find the acceleration of the object. d. How are the position and velocity related? How are the position and acceleration related? e. Sketch the position, velocity, and acceleration vectors at four different points on the trajectory with \(A=\omega=1\)

Short answer

1. Based on the given position function, find the time interval for completing one full circle. Answer: \([0, T] = \left[0, \frac{2\pi}{\omega}\right]\) 2. Calculate the velocity and speed of the object, and determine if they are constant. Answer: $$\mathbf{v}(t) = \langle -A\omega \sin \omega t, A\omega \cos \omega t \rangle \quad \text{and} \quad s(t) = A \omega$$ The velocity is not constant, but the speed is constant. 3. Determine the acceleration of the object. Answer: $$\mathbf{a}(t) = \langle -A\omega^2 \cos \omega t, -A\omega^2 \sin \omega t \rangle$$ 4. Describe the relationship between position, velocity, and acceleration. Answer: The velocity function is the derivative of the position function, and the acceleration function is the derivative of the velocity function. Velocity and acceleration are always perpendicular to position, and acceleration always points opposite to the position vector. 5. Explain how to sketch the position, velocity, and acceleration vectors at four different points on the trajectory with \(A=\omega=1\). Answer: Draw the circular path and plot the position vector, \(\mathbf{r}(t)\), from the origin to the point on the circle. Plot the velocity vector, \(\mathbf{v}(t)\), at the tip of the position vector, tangent to the circular path. Plot the acceleration vector, \(\mathbf{a}(t)\), at the tip of the position vector, pointing opposite to the position vector. Repeat this for the four chosen points on the trajectory and keep the lengths of the vectors proportional to their magnitudes.

Step-by-step solution

a. Time interval for completing one full circle

To find the time interval for the object to complete one full circle, we need to find the smallest positive time \(T\) when the object reaches the starting point again, which is \(2\pi\) radians. We know that the angle \(\omega t\) changes with time, and when it changes by \(2\pi\), the object completes a full circle. Therefore: $$\omega t = 2\pi \implies t = \frac{2\pi}{\omega}$$ Thus, the time interval for completing one full circle is: \([0, T] = \left[0, \frac{2\pi}{\omega}\right]\).

b. Velocity and Speed

To find the velocity and speed of the object, we first need to find the derivative of the position function: $$\mathbf{r}(t) = \langle A \cos \omega t, A \sin \omega t \rangle$$ Taking the derivative with respect to \(t\): $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle -A\omega \sin \omega t, A\omega \cos \omega t \rangle$$ The speed of the object is the magnitude of the velocity vector: $$s(t) = \left|\mathbf{v}(t)\right| = \sqrt{(-A\omega \sin \omega t)^2 + (A\omega \cos \omega t)^2} = A \omega$$ Hence, the velocity and speed of the object are given by: $$\mathbf{v}(t) = \langle -A\omega \sin \omega t, A\omega \cos \omega t \rangle \quad \text{and} \quad s(t) = A \omega$$ The velocity is not constant in either direction or magnitude since it depends on the sinusoidal functions of time. However, the speed is constant since it is only proportional to \(A\) and \(\omega\).

c. Acceleration

To find the acceleration of the object, we take the derivative of the velocity function: $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \langle -A\omega^2 \cos \omega t, -A\omega^2 \sin \omega t \rangle$$

d. Relationship between position, velocity, and acceleration

Comparing the position, velocity, and acceleration functions: Position: $$\mathbf{r}(t) = \langle A \cos \omega t, A \sin \omega t \rangle$$ Velocity: $$\mathbf{v}(t) = \langle -A\omega \sin \omega t, A\omega \cos \omega t \rangle$$ Acceleration: $$\mathbf{a}(t) = \langle -A\omega^2 \cos \omega t, -A\omega^2 \sin \omega t \rangle$$ We can see that the velocity function is the derivative of the position function, and the acceleration function is the derivative of the velocity function. Also, velocity and acceleration are always perpendicular to position, with acceleration always pointing opposite to the position vector.

e. Sketching position, velocity, and acceleration vectors

For this task, follow these steps at four different points (such as \(t=0\), \(t=\frac{\pi}{2}\), \(t=\pi\), and \(t=\frac{3\pi}{2}\)) on the trajectory with \(A=\omega=1\): 1. Draw the circular path. 2. Plot the position vector, \(\mathbf{r}(t)\), from the origin to the point on the circle. 3. Plot the velocity vector, \(\mathbf{v}(t)\), at the tip of the position vector, tangent to the circular path. 4. Plot the acceleration vector, \(\mathbf{a}(t)\), at the tip of the position vector, pointing opposite to the position vector. Repeat this for the four chosen points on the trajectory. Make sure to keep the lengths of the position, velocity, and acceleration vectors proportional to their actual magnitudes.

Key concepts

Understanding Velocity in Circular Motion

In circular motion, velocity plays a crucial role. Though students often confuse it with speed, the two are distinct. Velocity, a vector quantity, encapsulates both the direction and magnitude of motion. For an object moving along a circular path, the velocity vector is always tangent to the circle at any given point.

This means that while the object travels in a circular trajectory, its direction is constantly changing, even if the speed remains constant. As derived from the given parametric equations, the velocity of an object in circular motion can be expressed as:

• Velocity Vector:
• Magnitude of the Velocity (Speed): Finding the derivative of the position function provides us with this velocity vector. It reveals how fast the object moves along the path and in which direction at any particular time \(t\).
  • Speed is constant.
  • Velocity is not constant due to its changing direction.
  • Speed is directly linked to the constants \(A\) and \(\omega\).
  Understanding these differences helps students better grasp the concept of circular motion.

Decoding Acceleration in Circular Trajectories

Acceleration in circular motion is often counterintuitive because many students expect it to behave like linear motion. However, in circular paths, acceleration manifests differently. Even if an object's speed is constant, it’s still accelerating due to the continuous change in direction.

In our specific example, acceleration can be described as the derivative of the velocity function. It shows how the velocity vector changes with time. Here's what we can establish:

• Acceleration Vector:
• The magnitude of acceleration is equal to \(A\omega^2\).
• Acceleration is directed inwards, towards the center of the circular path.
• Always perpendicular to the velocity vector and in the opposite direction of the position vector.

Through these characteristics, students can see that acceleration in a circular motion works to maintain the object in its curved path by changing the direction of the velocity.

Parametric Equations in Circular Motion

Parametric equations offer a powerful means to model circular motion mathematically. They allow us to describe positions, velocities, and accelerations at any given time along a circular path in a concise way. In a circular trajectory, the path of motion is traced using the parametric equations for the coordinates:

• Position:
  • \( x(t) = A \cos(\omega t) \)
  • \( y(t) = A \sin(\omega t) \)
  These expressions describe the object's location in terms of a variable \( t \), making the analysis of movement clearer. By varying \( t \), we get information about the object's trajectory over time.
• Determining coordinates at specific time points helps visualize the path.
• Parametric equations are excellent for deriving velocity and acceleration vectors using differentiation.

Time and again, these equations prove vital in understanding motion, allowing us to compute the exact nature of movement for objects following any complex paths, especially circular.