Problem 107 - Fundamentals of Physics

Question

A particle \(P\) travels with constant speed on a circle of radius \(r=\) \(3.00 \mathrm{~m}\) (Fig. \(4-56)\) and completes one revolution in \(20.0 \mathrm{~s}\). The particle passes through \(O\) at time \(t=0 .\) State the following vectors in magnitudeangle notation (angle relative to the positive direction of \(x\) ). With respect to \(O\), find the particle's position vector at the times \(t\) of (a) \(5.00 \mathrm{~s}\), (b) \(7.50 \mathrm{~s}\), and \((\mathrm{c}) 10.0 \mathrm{~s}\). (d) For the \(5.00 \mathrm{~s}\) interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

Step-by-Step Solution

Answer

Position vectors: (a) 3m @ 90°, (b) ≈3m @ 135°, (c) 3m @ 180°. Displacement: ≈4.24m @ 225°. Average velocity: ≈0.85m/s @ 225°. Velocities: (f) -0.94m/s, (g) -0.942m/s. Accelerations: (h) -0.296m/s², (i) 0.296m/s².

Step 1: Determine the Angular Velocity

Given that the particle completes one revolution in 20 seconds, we can find the angular velocity \( \omega \) using the formula \( \omega = \frac{2\pi}{T} \), where \( T = 20 \text{ s} \). Therefore, \( \omega = \frac{2\pi}{20} = \frac{\pi}{10} \text{ rad/s} \).

Step 2: Calculate Position for 5.00 s

The angular position \( \theta \) at \( t = 5 \text{ s} \) is determined by \( \theta = \omega t = \frac{\pi}{10} \times 5 = \frac{\pi}{2} \). At this angle, the position vector is \( r(\cos(\theta)\hat{i} + \sin(\theta)\hat{j}) = 3(0\hat{i} + 1\hat{j}) = 3\hat{j} \). Thus, its angle is 90° (relative to the positive x-axis).

Step 3: Calculate Position for 7.50 s

For \( t = 7.5 \text{ s} \), \( \theta = \omega t = \frac{\pi}{10} \times 7.5 = \frac{3\pi}{4} \). At this position, the vector is \( 3(\cos\frac{3\pi}{4}\hat{i} + \sin\frac{3\pi}{4}\hat{j}) = 3\left(-\frac{\sqrt{2}}{2}\right)\hat{i} + 3\left(\frac{\sqrt{2}}{2}\right)\hat{j} \). The angle from the positive x-axis is 135°.

Step 4: Calculate Position for 10.0 s

For \( t = 10 \text{ s} \), \( \theta = \omega t = \frac{\pi}{10} \times 10 = \pi \). The position is then \( 3(-1\hat{i} + 0\hat{j}) = -3\hat{i} \), corresponding to an angle of 180° relative to the positive x-axis.

Step 5: Calculate Displacement from 5.00 s to 10.00 s

Displacement is the final position minus the initial position. From 10 s, the position is \(-3\hat{i}\), and from 5 s it's \(3\hat{j}\). The displacement is \(-3\hat{i} - 3\hat{j} \). The magnitude is \( \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2} \), and the angle is \(225°\) relative to the x-axis (since it is directly pointing from positive j to negative i and j coordinates).

Step 6: Compute Average Velocity Over the Interval

Average velocity \( \mathbf{v}_{avg} \) is the displacement divided by time; hence \( \mathbf{v}_{avg} = \frac{-3\hat{i} - 3\hat{j}}{5} \text{ m/s} = -0.6\hat{i} - 0.6\hat{j} \text{ m/s} \). The magnitude is \( 0.6\sqrt{2} \approx 0.848 \text{ m/s} \), and the direction is 225°.

Step 7: Find Velocity at the Beginning of the Interval

Velocity at any time is given by \( \mathbf{v} = r\omega(-\sin(\theta)\hat{i} + \cos(\theta)\hat{j}) \). At \( t = 5\text{ s} \), \( \theta = \frac{\pi}{2} \), so \( \mathbf{v} = 3 \times \frac{\pi}{10} (-1\hat{i}) = -\frac{3\pi}{10}\hat{i} \approx -0.942\hat{i} \text{ m/s} \), directed left.

Step 8: Find Velocity at the End of the Interval

At \( t = 10\text{ s} \), \( \theta = \pi \). Thus, the velocity is \( 3 \times \frac{\pi}{10}(0\hat{i} + (-1)\hat{j}) = -\frac{3\pi}{10}\hat{j} \approx -0.942\hat{j} \text{ m/s} \), directed downward.

Step 9: Determine Acceleration at the Beginning of Interval

The acceleration \( \mathbf{a} = -r\omega^2(\cos(\theta)\hat{i} + \sin(\theta)\hat{j}) \). At \( t = 5\text{ s} \), \( \theta = \frac{\pi}{2} \), leading to \( \mathbf{a} = -3 \times \left(\frac{\pi}{10}\right)^2 (0\hat{i} + 1\hat{j}) = -\frac{3\pi^2}{100}\hat{j} \approx -0.296\hat{j} \text{ m/s}^2 \), directed inward towards the center.

Step 10: Determine Acceleration at the End of Interval

At \( t = 10\text{ s} \), with \( \theta = \pi \), \( \mathbf{a} = -3 \times \left(\frac{\pi}{10}\right)^2 (-1\hat{i} + 0\hat{j}) = \frac{3\pi^2}{100}\hat{i} \approx 0.296\hat{i} \text{ m/s}^2 \), also directed inward towards the center.

Key Concepts

Angular Velocity

Angular velocity is crucial when dealing with circular motion. It is the measure of how fast an object rotates or revolves around a central point. Imagine a car on a circular track. Angular velocity determines how quickly the car completes a lap. The formula to calculate angular velocity is \( \omega = \frac{2\pi}{T} \), where

• \( \omega \) is the angular velocity in radians per second
• \( \pi \) is approximately 3.14159
• and \( T \) is the time taken for one complete revolution, in seconds

In our example, the particle completes one revolution in 20 seconds, so its angular velocity is \[ \omega = \frac{2 \pi}{20} = \frac{\pi}{10} \text{ rad/s}. \]Thus, every second, the particle changes its angular position by \( \frac{\pi}{10} \) radians.

Position Vector

The position vector is like drawing a line from the center of the circle to where our object (particle) is on the circle. It points directly at the particle's location at any given time. To find the position vector at a specific time, use the formula:\[ r(\cos(\theta)\hat{i} + \sin(\theta)\hat{j}) \]where

• \( r \) is the radius of the circle
• and \( \theta \) is the angular position

For example, at 5 seconds, the angular position is \( \frac{\pi}{2} \). Therefore, the position vector is \( 3(0\hat{i} + 1\hat{j}) = 3\hat{j} \), making the particle directly above the center of the circle, in the direction of the y-axis.Remember to always express angles in terms of their relationship to the positive x-axis.

Average Velocity

Average velocity is the overall change in position divided by the total time interval. Unlike instantaneous velocity, it considers the start and end rather than every moment in between.Calculate it using the formula:\[ \mathbf{v}_{avg} = \frac{\Delta \mathbf{s}}{\Delta t}, \]where

• \( \Delta \mathbf{s} \) is the displacement, or the straight-line distance between starting and ending positions
• and \( \Delta t \) is the change in time

For the 5-second interval from 5 to 10 seconds, we found that the particle moved from \( 3\hat{j} \) to \(-3\hat{i} \).Thus, the displacement is \(-3\hat{i} - 3\hat{j} \),resulting in the average velocity of\[ \mathbf{v}_{avg} = \frac{-3\hat{i} - 3\hat{j}}{5} \approx -0.6\hat{i} - 0.6\hat{j} \, \text{m/s}. \]

Centripetal Acceleration

Centripetal acceleration is all about keeping the particle moving in a circle. It always points toward the center of the circle, ensuring that the particle doesn't fly off in a straight line.By formula, it is:\[ \mathbf{a} = -r\omega^2(\cos(\theta)\hat{i} + \sin(\theta)\hat{j}) \]For circular motion, \( \omega \) and the angle \( \theta \) will change depending on the time under consideration. For example, at 5 seconds, \( \theta \) is \( \frac{\pi}{2} \), so the centripetal acceleration is:\[ \mathbf{a} = -3 \left( \frac{\pi}{10} \right)^2 \hat{j} \approx -0.296\hat{j} \, \text{m/s}^2. \]This acceleration serves to change the direction of the velocity, keeping the particle on its circular path without changing its speed.