Problem 60 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A particle starting from rest revolves with uniformly increasing speed in a clockwise circle in the \(x y\) plane. The center of the circle is at the origin of an \(x y\) coordinate system. At \(t=0,\) the particle is at \(x=0.0, y=2.0 \mathrm{~m} .\) At \(t=2.0 \mathrm{~s},\) it has made one-quarter of a revolution and is at \(x=2.0 \mathrm{~m}, y=0.0 .\) Determine \((a)\) its speed at \(t=2.0 \mathrm{~s}\) (b) the average velocity vector, and \((c)\) the average acceleration vector during this interval.

Step-by-Step Solution

Answer

(a) Speed = \(\pi \text{ m/s}\); (b) Average Velocity = \((1, -1) \text{ m/s}\); (c) Average Acceleration = \(\left(\frac{\pi}{2\sqrt{2}}, -\frac{\pi}{2\sqrt{2}} \right) \text{ m/s}^2\)."}

Step 1: Determine the radius of the circle

From the initial position at time \( t = 0 \), the particle is at \( (0, 2.0) \). Since this coordinates is on the circumference, the radius \( r \) of the circle is equal to the \( y \)-coordinate: \[ r = 2.0 \text{ m} \]

Step 2: Find the circumference traveled in one-quarter revolution

In one-quarter of a revolution, the arc length traveled by the particle is a quarter of the full circumference of a circle. The circumference \( C \) of the circle is \[ C = 2 \pi r \]. Calculate a quarter of that: \[ \text{Arc length} = \frac{1}{4}\times 2 \pi r = \frac{1}{2} \pi r = \pi \text{ m}\] Given \( r = 2.0 \text{ m} \), \[ \text{Arc length} = \pi \times 2 = 2\pi \text{ m} \]

Step 3: Calculate speed at t = 2.0s

Speed is calculated as the distance traveled over the time taken. The particle travels \( 2\pi \text{ m} \) in 2.0 s. Thus, the speed \( v \) is: \[ v = \frac{2\pi}{2} = \pi \text{ m/s} \]

Step 4: Determine the average velocity vector

The average velocity vector is defined as the total displacement divided by the total time. The displacement is the vector from the initial position \( (0, 2.0) \) to the final position \( (2.0, 0) \): \[ \text{Displacement vector} = (2.0 - 0, 0 - 2.0 ) = (2.0, -2.0) \].The magnitude of this vector is \[ \sqrt{(2.0)^2 + (-2.0)^2} = \sqrt{8} = 2\sqrt{2} \text{ m} \].The average velocity is \[ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} = \left( 1, -1 \right) \text{ m/s} \].

Step 5: Calculate the average acceleration vector

Average acceleration is given by the change in velocity divided by the time. The initial speed is 0 m/s and the final speed is \( \pi \text{ m/s } \) in the direction of displacement \((1, -1)\).Therefore, final velocity vector is: \[ \text{Final velocity vector} = \pi \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) = \left(\frac{\pi}{\sqrt{2}}, -\frac{\pi}{\sqrt{2}} \right) \text{ m/s} \].Change in velocity vector is: \[ \text{Change in velocity vector} = \left(\frac{\pi}{\sqrt{2}}, -\frac{\pi}{\sqrt{2}} \right) - (0,0) = \left(\frac{\pi}{\sqrt{2}}, -\frac{\pi}{\sqrt{2}} \right) \text{ m/s} \].Average acceleration vector is: \[ \text{Average Acceleration} = \frac{\left(\frac{\pi}{\sqrt{2}}, -\frac{\pi}{\sqrt{2}} \right)}{2 \text{ s}} = \left(\frac{\pi}{2\sqrt{2}}, -\frac{\pi}{2\sqrt{2}} \right) \text{ m/s}^2 \].

Key Concepts

Uniform Acceleration

Uniform acceleration occurs when the velocity of an object changes by equal amounts in equal intervals of time. In the context of circular motion, it means that the speed of the particle increases at a constant rate as it travels along the circular path. This is different from the direction change because even if the direction changes, in uniform acceleration, the rate at which speed changes is constant.
In a circular path, a uniformly accelerating particle increases its speed linearly with time. Here, uniform acceleration could be considered as producing an angular speed increase at a constant rate. Given that the initial speed is zero, the object moves with increasing speed meaning every segment of the journey can be described by its increasing velocity.
Imagine a car steadily picking up speed as it goes around a curve, achieving a new speed limit over time. This is how uniform acceleration works. By understanding this, you can predict the speed at any given time and analyze motion along circular paths.

Average Velocity

Average velocity is the displacement divided by the time taken. It's different from average speed because it considers direction. In circular motion, displacement is the shortest distance between the starting and ending points, formed by a straight line.
For the problem at hand, the particle starts at the point (0, 2) and ends at (2, 0), making a quarter of a circle in 2 seconds. The displacement is the straight line connecting these two points, leading to the displacement vector (2, -2) meters.
The magnitude of this line can be calculated using the Pythagorean theorem:

• \[\sqrt{(2)^2 + (-2)^2} = \sqrt{8}\]

The average velocity vector is then the displacement vector divided by the time. So, the average velocity is

• \[\left( \frac{2}{2}, \frac{-2}{2} \right) = (1, -1) \text{ m/s}\]

This vector gives us direction and speed, helping to understand the net motion of the particle over time.

Average Acceleration

Average acceleration is the change in velocity over a period of time. It tells us how quickly velocity is changing. This can be confusing with circular motion, but we will break it down simply.
First, the change in velocity is calculated. In the exercise, the initial speed of the particle is 0 m/s and at 2 seconds, it has a final speed of \(\pi\) m/s in a new direction. The direction is given by the displacement direction: towards (1, -1). Thus, the final velocity can be written as \(\pi\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)\).
The change in velocity vector is the difference

• \[\left(\frac{\pi}{\sqrt{2}}, -\frac{\pi}{\sqrt{2}} \right) - (0,0)\]

Therefore, the average acceleration vector is this change divided by the time interval of 2 seconds:

• \[\left(\frac{\pi}{2\sqrt{2}}, -\frac{\pi}{2\sqrt{2}} \right) \text{ m/s}^2\]

Understanding average acceleration helps quantify how fast the speed and direction of an object are changing, giving a clearer picture of dynamic motion especially in curving paths.