Problem 101 - Physics for Scientists and Engineers with Modern Physics

Question

A car drives at a constant speed around a banked circular track with a diameter of 127 \(\mathrm{m}\) . The motion of the car can be described in a coordinate system with its origin at the center of the circle. At a particular instant the car's accel- eration in the horizontal plane is given by $$\vec{\mathbf{a}}=(-15.7 \hat{\mathbf{i}}-23.2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}$$ (a) What is the car's speed? (b) Where \((x\) and \(y)\) is the car at this instant?

Step-by-Step Solution

Answer

The car's speed is approximately 42.2 m/s, and its position is around (35.2 m, 53.0 m).

Step 1: Understand the Problem

We have a car moving in a circular path with a constant speed. The diameter of the circular track is given as 127 m. The acceleration vector at a certain instant is given as \( \vec{\mathbf{a}}=(-15.7 \hat{\mathbf{i}}-23.2 \hat{\mathbf{j}}) \ \mathrm{m}/\mathrm{s}^{2} \). We need to find the car's speed and its position on this path.

Step 2: Determine the Radius

Since the diameter of the circular track is given, we first find the radius of the circle. The radius \( r \) is half of the diameter: \[ r = \frac{127 \ \mathrm{m}}{2} = 63.5 \ \mathrm{m} \] This will be useful for determining the speed using centripetal acceleration.

Step 3: Calculate the Magnitude of Acceleration

The given acceleration vector can be used to find the total centripetal acceleration using the Pythagorean theorem: \[ a = \sqrt{(-15.7)^2 + (-23.2)^2} = \sqrt{246.49 + 538.24} = \sqrt{784.73} = 28 \ \mathrm{m/s}^2 \] This magnitude represents the centripetal acceleration.

Step 4: Use Centripetal Acceleration to Find Speed

Centripetal acceleration \( a_c \) is related to the speed \( v \) by the formula: \[ a_c = \frac{v^2}{r} \] Solving for speed, we get: \[ v = \sqrt{a_c \cdot r} = \sqrt{28 \cdot 63.5} = \sqrt{1778} \approx 42.2 \ \mathrm{m/s} \] This is the car's speed at that instant.

Step 5: Find the Car's Position

The car is located at the position \( (x, y) \) on the circle such that the acceleration vector points towards the center. The direction of this vector gives us a hint on where the car is on its path. With the horizontal acceleration given by \( \vec{\mathbf{a}}=(-15.7 \hat{\mathbf{i}}-23.2 \hat{\mathbf{j}})\), the direction indicates that the position \((x, y)\) can be determined as directly opposite the vector direction. Applying trigonometry: \( \tan^{-1} \left( \frac{-23.2}{-15.7} \right) \approx 56.3^\circ \) from x-axis. With the radius 63.5 meters, \( \begin{align*} x & = 63.5 \cdot \cos(56.3^\circ) \ y & = 63.5 \cdot \sin(56.3^\circ) \end{align*} \) So the position is approximately \( x \approx 35.2 \ \mathrm{m}, \ y \approx 53.0 \ \mathrm{m} \).

Key Concepts

Centripetal Acceleration

When a car moves along a circular path, it continuously changes direction. This change in direction is due to acceleration directed towards the center of the circle, known as centripetal acceleration.
This acceleration vector keeps the car moving in a circular path. Without it, the car would move off in a straight line. The symbol for centripetal acceleration is usually designated as \( a_c \). It is important to understand its relationship with speed and radius, which are the two main factors that determine its magnitude.
The formula that relates centripetal acceleration to speed and radius is:

• \( a_c = \frac{v^2}{r} \)

where \( v \) is the velocity of the object and \( r \) is the radius of the circular path.
The formula tells us that if either the speed increases or the radius decreases, the centripetal acceleration increases. Conversely, a larger radius or smaller speed will lead to a decrease in centripetal acceleration. Hence, understanding this concept is pivotal when analyzing circular motion dynamics.

Radius Calculation

To solve problems involving circular motion, finding the radius of the path often serves as an essential step. In this case, since the diameter is given, calculating the radius becomes very straightforward.
The radius \( r \) of a circle is half of its diameter; therefore, if the diameter \( D \) is known, the radius can be calculated using the formula:

• \( r = \frac{D}{2} \)

For example, in this exercise, the circular track has a diameter of 127 meters, so the radius would be:

• \( r = \frac{127 \, \text{m}}{2} = 63.5 \, \text{m} \)

This value of the radius is critical when using formulas and equations involving circular motion, especially those calculating speed or centripetal acceleration.
Accurately determining the radius ensures you have precise inputs for other calculations, which affects all subsequent results.

Vector Analysis

Vectors play a significant role in understanding motion, especially in a two-dimensional plane such as circular motion. Vectors possess both magnitude and direction, making them ideal for describing quantities like velocity and acceleration.
In this exercise, the given acceleration vector \( \vec{\mathbf{a}} = (-15.7 \hat{\mathbf{i}} - 23.2 \hat{\mathbf{j}}) \mathrm{m}/\mathrm{s}^2 \) illustrates how a car's acceleration can be divided into its components along the \( x \) and \( y \) axes. The components \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \) represent direction along these axes.
To find the magnitude of a vector, such as an acceleration vector, the Pythagorean theorem can be applied:

• \( a = \sqrt{(-15.7)^2 + (-23.2)^2} \)

Calculating this provides a single value for the vector's magnitude. This computation is essential because it helps in assessing the net effect of acceleration in a particular instance. Further, understanding the direction—indicated by the signs and ratio of components—is equally crucial. It allows us to determine where the vector is pointing, which in circular motion points towards the center of the circle.
Mastery of vector concepts enables the analysis of more complex motion scenarios involving multiple forces and directions.