Problem 23 - Thomas Calculus

Question

Motion along a circle Each of the following equations in parts (a) \((\) e) describes the motion of a particle having the same path, namely the unit circle \(x^{2}+y^{2}=1 .\) Although the path of each particle in parts \((a)-(e)\) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions. \(\begin{array}{l}{\text { 1) Does the particle have constant speed? If so, what is its contant }} \\ {\text {speed? }} \\ {\text { ii) Is the particle's acceleration vector always orthogonal to its }} \\ {\text { velocity vector? }}\end{array}\) \(\begin{array}{l}{\text { iii) Does the particle move clockwise or counterclockwise }} \\ {\text { around the circle? }} \\ {\text { iv) Does the particle begin at the point }(1,0) ?}\end{array}\) \(\begin{array}{l}{\text { a. } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geqq 0} \\ {\text { b. } \mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0} \\ {\text { c. } \mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0} \\ {\text { d. } \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { e. } \mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0}\end{array}\)

Step-by-Step Solution

Answer

In parts (a)-(b), the speed is constant; both move counterclockwise starting at (1,0) with orthogonal acceleration to velocity.

Step 1: Check Constant Speed - Part (a)

For the particle described by \( \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j} \), the velocity is obtained as the derivative: \( \mathbf{v}(t) = \frac{d}{dt}((\cos t)\mathbf{i} + (\sin t)\mathbf{j}) = -\sin t \mathbf{i} + \cos t \mathbf{j} \). The speed is the magnitude of velocity: \( |\mathbf{v}(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2} = 1 \). Thus, the speed is constant and equals 1.

Step 2: Check Orthogonality - Part (a)

The acceleration is the derivative of velocity, \( \mathbf{a}(t) = \frac{d}{dt}(-\sin t \mathbf{i} + \cos t \mathbf{j}) = -\cos t \mathbf{i} - \sin t \mathbf{j} \). The dot product of velocity and acceleration, \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (-\sin t)(-\cos t) + (\cos t)(-\sin t) = 0 \). Hence, the acceleration vector is always orthogonal to the velocity vector.

Step 3: Determine Rotation Direction - Part (a)

The velocity \( \mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \) indicates the direction of motion. Since \( \cos t \) corresponds to the \( y \)-component and \( -\sin t \) to the \( x \)-component, the particle moves counterclockwise around the circle.

Step 4: Check Initial Point - Part (a)

At \( t = 0 \), the position is \( \mathbf{r}(0) = (\cos 0) \mathbf{i} + (\sin 0) \mathbf{j} = \mathbf{i} \). So the particle begins at the point (1,0).

Step 5: Repeat Step 1 for Constant Speed - Part (b)

For \( \mathbf{r}(t)=\cos(2t)\mathbf{i} + \sin(2t)\mathbf{j} \), the velocity is \( \mathbf{v}(t) = -2\sin(2t)\mathbf{i} + 2\cos(2t)\mathbf{j} \). Speed is \( |\mathbf{v}(t)| = \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2} = 2 \). Thus, the speed is constant at 2.

Step 6: Repeat Step 2 for Orthogonality - Part (b)

Acceleration, \( \mathbf{a}(t) = -4\cos(2t)\mathbf{i} - 4\sin(2t)\mathbf{j} \). Dot product, \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (-2\sin(2t))(-4\cos(2t)) + (2\cos(2t))(-4\sin(2t)) = 0 \). Therefore, velocity and acceleration vectors are orthogonal.

Step 7: Determine Rotation Direction - Part (b)

The velocity components \( -2\sin(2t) \) (\( x \)) and \( 2\cos(2t) \) (\( y \)) indicate the particle moves counterclockwise around the circle.

Step 8: Check Initial Point - Part (b)

At \( t = 0 \), the position is \( \mathbf{r}(0) = (\cos 0)\mathbf{i} + (\sin 0)\mathbf{j} = \mathbf{i} \), so the particle begins at (1,0).

Step 9: Assess More Steps for Other Parts (c)-(e)

Repeat similar analysis for parts, (c), (d), and (e) to determine if the speed is constant, direction, orthogonality, and the starting point. Follow similar computations by taking the derivatives and evaluating the necessary initial conditions.

Key Concepts

Constant Speed

In circular motion, when a particle travels at a constant speed, it means its speed (which measures how fast it is going along the circle) does not change over time.
This is typical for a particle that moves smoothly around a circle, without accelerating or decelerating.

To determine the constant speed, we look at the velocity vector, which is the derivative of the position function, and find its magnitude.

• In Part (a), the velocity is \(\mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j}\)
• The speed is the magnitude of the velocity, \(|\mathbf{v}(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2} = 1\)

Hence, the particle moves at a constant speed of 1.
This means the particle keeps moving around the circle at the same pace without speeding up or slowing down.

In Part (b), a similar calculation reveals the speed to be constant at 2.
Checking the functional forms for each, by consistently taking derivatives to find velocity, and checking magnitudes, allows us to ensure this constancy.

Orthogonal Vectors

Two vectors are orthogonal if they are at right angles to each other.
This orthogonality is a crucial aspect of ensuring a particle moves in a perfect circle with consistent dynamics.
The velocity and acceleration vectors being orthogonal means that the force exerted by the acceleration is always perpendicular to the direction of motion.

• For the particle in Part (a), the acceleration vector is derived from the velocity as \(\mathbf{a}(t) = -\cos t \mathbf{i} - \sin t \mathbf{j}\)
• If the dot product between velocity \(\mathbf{v}(t)\) and acceleration \(\mathbf{a}(t)\) is zero, \(\mathbf{v}(t) \cdot \mathbf{a}(t) = 0\), then they are orthogonal.

This zero dot product confirms that the acceleration does not change the speed of the particle but only the direction, a characteristic of uniform circular motion.
Whenever you check this condition, it assures that the particle's energy concerning circular movement remains purely kinetic without gaining or losing any.

Counterclockwise Motion

When analyzing a particle's motion direction, knowing if it moves clockwise or counterclockwise helps understand its trajectory and effects.
The direction of velocity components often reveals the rotation direction.

• For example, in Part (a), the velocity vector \(\mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j}\) shows that the particle moves around the circle counterclockwise.
• The positive direction of \( \cos t \mathbf{j} \) indicates an upward movement through the y-component, characteristic of counterclockwise motion.

In general, in the xy-plane, positive \( y \) changes during circular paths usually signal a counterclockwise rotation around a circle.
This type of motion is vital to consider in scenarios like circular tracks or planetary rotations, where the relative direction determines the dynamic forces exerted.

Initial Position

The initial position of the particle is where it starts on the circle at time \( t = 0 \).
This position influences how we interpret the particle's path and dynamics.

• For instance, in Part (a), at \( t = 0 \), the position \( \mathbf{r}(0) = \cos(0) \mathbf{i} + \sin(0) \mathbf{j} = \mathbf{i} \) locates the particle at point (1, 0), which is the rightmost point on the unit circle.
• This means the particle starts its journey from there and follows a predictable cycle as defined by the function.

Understanding initial conditions is crucial when modeling cycles or predicting behaviors over time in physics and engineering.
Analyzing the starting point helps determine if the path or returned observations match expected results based on these initial conditions.