Question

The position functions of objects \(A\) and \(B\) describe different motion along the same path for \(t \geq 0\). a. Sketch the path followed by both \(A\) and \(B\). b. Find the velocity and acceleration of \(A\) and \(B\) and discuss the differences. c. Express the acceleration of A and \(B\) in terms of the tangential and normal components and discuss the differences. $$A: \mathbf{r}(t)=\langle\cos t, \sin t\rangle, B: \mathbf{r}(t)=\langle\cos 3 t, \sin 3 t\rangle$$

Short answer

Question: Compare the motion of objects A and B, focusing on their paths, velocity, and acceleration components. Answer: Both object A and B follow the same circular path, but object B completes the path 3 times faster than object A, due to its higher velocity. The acceleration of object B is 9 times stronger than object A, with both objects having constant acceleration directed towards the center of motion.

Step-by-step solution

1. Visualize the paths followed by both A and B

First, we need to visualize the path followed by A and B in terms of their position functions. Object A: $$\mathbf{r}(t)=\langle\cos t, \sin t\rangle$$ Object B: $$\mathbf{r}(t)=\langle\cos 3 t, \sin 3 t\rangle$$ Both of these functions trace a circle in the plane. For object A, it moves around the unit circle in a clockwise direction with one full revolution completed after 2π. For object B, it moves around the unit circle 3 times faster than A, completing one full revolution after (2π)/3.

2. Compute the velocity and acceleration of A and B

Next, we need to compute the velocity and acceleration functions of both A and B. For object A: Velocity: $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt}\langle\cos t, \sin t\rangle = \langle -\sin t, \cos t\rangle$$ Acceleration: $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \frac{d}{dt}\langle -\sin t, \cos t\rangle = \langle -\cos t, -\sin t\rangle$$ For object B: Velocity: $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt}\langle\cos 3t, \sin 3t\rangle = \langle -3\sin 3t, 3\cos 3t\rangle$$ Acceleration: $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \frac{d}{dt}\langle -3\sin 3t, 3\cos 3t\rangle = \langle -9\cos 3t, -9\sin 3t\rangle$$

3. Calculate tangential and normal components of acceleration for A and B

To compute the tangential and normal components of acceleration for A and B, we can use the following formulas: Tangential Component: $$a_T = \frac{d}{dt}\left(\frac{1}{||a||}\right)$$ Normal Component: $$a_N = \sqrt{||a||^2 - a_T^2}$$ For object A, we have: Tangential Component: $$a_T^{(A)} = 0$$ (since the magnitude of vector A is always constant) Normal Component: $$a_N^{(A)} = ||\mathbf{a}(t)|| = ||\langle -\cos t, -\sin t\rangle|| = \sqrt{(-\cos t)^2 + (-\sin t)^2} = 1$$ For object B, we have: Tangential Component: $$a_T^{(B)} = 0$$ (since the magnitude of vector B is always constant) Normal Component: $$a_N^{(B)} = ||\mathbf{a}(t)|| = ||\langle -9\cos 3t,-9\sin 3t\rangle|| = \sqrt{(-9\cos 3t)^2 + (-9\sin 3t)^2} = 9$$

4. Discuss differences

Comparing the results for A and B, we can infer the following differences: - Both objects move along the same unit circle, but object B completes the circular path 3 times faster than object A. - The velocity of object A is slower compared to object B, as indicated by their respective velocity functions. - The acceleration components for object A indicate that the acceleration is constant and always points towards the center of motion. Meanwhile, object B has an acceleration that is 9 times stronger than object A and has identical behavior in terms of being always directed towards the center of motion. In conclusion, both object A and B follow the same circular path, but A circles this path slower than B with 1/3 of velocity as B. Additionally, the acceleration of object B is much stronger (9 times) than object A, making it complete the path 3 times faster.

Key concepts

Velocity Function

When discussing motion along a circular path, such as the movements of objects A and B, understanding the velocity function is crucial. The velocity function represents the rate of change of position over time. For a circular motion, velocity is not only about speed but also direction.
For object A, given by the position function \[ \mathbf{r}(t) = \langle \cos t, \sin t \rangle, \]the velocity function is derived by differentiating the position with respect to time. It yields \[ \mathbf{v}(t) = \langle -\sin t, \cos t \rangle. \]
Notice that the velocity vector rotates with the same angle as displacement but is perpendicular to the radius. It remains tangent to the circular path.

• This results in a constant speed on the circle's edge.
• The direction changes as the object maintains circular motion.

For object B, with position \[ \mathbf{r}(t) = \langle \cos 3t, \sin 3t \rangle, \]the velocity function becomes:\[ \mathbf{v}(t) = \langle -3\sin 3t, 3\cos 3t \rangle. \]
This velocity is three times the magnitude of object A's velocity. The higher coefficients reflect that object B moves faster around the circle, completing the path three times more quickly.

• Object B's motion is faster, evident from its velocity's factor of 3.
• Both velocity vectors are perpendicular to their respective radius vectors, ensuring tangential motion.

Understanding these velocity functions help explain how objects move swiftly and consistently along a circular orbit.

Acceleration Function

The acceleration function for circular motion signifies how the velocity changes over time: in magnitude, direction, or both. It is attained by differentiating the velocity function.
For object A, the acceleration function derived from its velocity is:\[ \mathbf{a}(t) = \langle -\cos t, -\sin t \rangle. \]
This demonstrates that acceleration is also perpendicular to the path and directed towards the circle's center, thereby indicating a centripetal nature.

• The acceleration does not change in magnitude, remaining constant and facilitating circular movement.
• This points to the role of centripetal acceleration in maintaining the object's circular trajectory.

For object B, we differentiate its velocity:\[ \mathbf{a}(t) = \langle -9\cos 3t, -9\sin 3t \rangle. \]
Object B's acceleration is nine times that of object A's. This is due to moving 3 times faster, implying stronger centripetal force requirements.

• The acceleration here indicates a tighter fast-moving orbit due to a larger centripetal force.
• Both accelerations highlight the necessary inward force to sustain their respective circular paths.

The study of these accelerations underscores how different speed settings in circular motion demand adjustments in force to ensure a stable path, underscoring the fundamental relationship between velocity and acceleration.

Tangential and Normal Components

In circular motion, acceleration can be broken down into tangential and normal components. These components help us understand different parts of the acceleration.
The tangential component, \( a_T \), measures changes in speed. Both objects A and B have \( a_T = 0 \).This reveals that their speed does not change along the path, only the direction changes, conforming to uniform circular motion.
The normal component, \( a_N \), provides insight into how force is directed inward to maintain circular motion.
For object A:- The normal component is \( a_N^{(A)} = 1 \), which is constant.- This confirms a steady centripetal force acting to keep object A moving in a circle.
For object B:- The normal component \( a_N^{(B)} = 9 \) explains why object B requires much more force, a result of its higher speed.

• This component is always directed towards the center of the circle.
• It plays a critical role in sustaining each object's circular progression along its unique path speeds.

By separating acceleration into these components, we deepen our understanding of how different forces act on objects moving in circular paths, establishing the groundwork for more complex kinetic analysis.