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An automobile, starting from rest, has a linear acceleration to the right whose magnitude is 0.800 m/s2 (see the figure). During the next 20.0 s, the tires roll and do not slip. The radius of each wheel is 0.330 m. At the end of this time, what is the angle through which each wheel has rotated?

Short Answer

Expert verified
The wheels rotate through approximately 484.85 radians.

Step by step solution

01

Understand the problem

We are given an automobile starting from rest with a constant acceleration of 0.800 m/s² for 20 seconds. The wheels do not slip, meaning linear motion can be related to rotational motion. We need to find the angle each wheel has rotated through at the end of this period.
02

Calculate linear distance traveled

Using the kinematic equation for linear motion, calculate the distance traveled by the automobile. Since it starts from rest, the equation is \( d = \frac{1}{2} a t^2 \). Here, \( a = 0.800 \text{ m/s}^2 \) and \( t = 20.0 \text{ s} \). Substitute these values to get \( d = \frac{1}{2} \times 0.800 \times (20.0)^2 \).
03

Solve for linear distance

Perform the calculations: \( d = \frac{1}{2} \times 0.800 \times 400 = 160 \text{ m} \). Thus, the automobile travels a linear distance of 160 meters.
04

Relate linear distance to angular distance

The relationship between linear distance (\( d \)) and angular distance (\( \theta \)) is given by the formula \( \theta = \frac{d}{r} \) where \( r \) is the radius of the wheel. Here, \( r = 0.330 \text{ m} \). Substitute the known values, \( \theta = \frac{160}{0.330} \).
05

Calculate angle in radians

Compute the angular displacement: \( \theta = \frac{160}{0.330} \approx 484.85 \text{ radians} \). Therefore, the wheels rotate through approximately 484.85 radians.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Acceleration
Linear acceleration is the rate at which the velocity of an object changes with respect to time. It is a vector quantity, meaning it has both magnitude and direction. In our exercise, the car moves to the right with an initial linear acceleration of 0.800 m/s². Since the vehicle starts from rest, this acceleration will increase the velocity over a period.
To find the linear distance covered during acceleration, we use the formula derived from Newton's second law:
  • The formula: \( d = \frac{1}{2} a t^2 \)
  • Where \( d \) is distance, \( a \) is acceleration, and \( t \) is time.
In the problem, the duration \( t \) is given as 20 seconds. By plugging in the values, we calculated the car travels 160 meters. This distance is crucial as it provides a direct path to understanding its rotational impact through the wheels.
Understanding linear acceleration helps in not only solving physics problems but also in real-life mechanics, such as determining how quickly a vehicle can speed up.
Angular Displacement
Angular displacement refers to the angle through which a point, line, or body is rotated in a specified direction about a specified axis. Unlike linear displacement, which deals with straight-line motion, angular displacement involves movement along a circular path.
In relation to the wheel of an automobile, as the car covers a linear distance, each wheel rotates about its axis, covering an angular distance. This angular displacement can be calculated using:
  • The formula: \( \theta = \frac{d}{r} \)
  • Where \( \theta \) is angular displacement in radians, \( d \) is the linear distance traveled, and \( r \) is the radius of the wheel.
Given the wheel radius of 0.330 meters and the linear distance of 160 meters, the wheels rotate through approximately 484.85 radians. Linking linear distance to angular displacement is key in understanding how far a wheel has turned.
This concept is critical in disciplines like engineering where rotating bodies are analyzed or in everyday applications like estimating tire wear.
Kinematic Equations
Kinematic equations are the tools physicists use to predict and describe the motion of objects. They offer a mathematical framework for understanding the relationships between velocity, acceleration, time, and displacement.
In the context of our exercise, these equations help us link linear motion with rotational motion. The primary equation we used for linear displacement was:
  • \( d = \frac{1}{2} a t^2 \)
Here, we seamlessly moved from this linear perspective to examining the rotational motion of the car's wheels. By using the linear distance covered and the radius, we employed:
  • \( \theta = \frac{d}{r} \)
Kinematic equations are powerful because they allow for predictions about how an object behaves over time, given particular constraints or forces. These equations are foundational in both physics education and practical engineering designs, helping to solve problems from simple classroom exercises to complex machinery dynamics.

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Most popular questions from this chapter

The earth has a radius of \(6.38 \times 10^{6} \mathrm{m}\) and turns on its axis once every \(23.9 \mathrm{h}\). (a) What is the tangential speed (in \(\mathrm{m} / \mathrm{s}\) ) of a person living in Ecuador, a country that lies on the equator? (b) At what latitude (i.e., the angle \(\theta\) in the drawing) is the tangential speed one-third that of a person living in Ecuador?

Suppose you are riding a stationary exercise bicycle, and the electronic meter indicates that the wheel is rotating at \(9.1 \mathrm{rad} / \mathrm{s}\). The wheel has a radius of \(0.45 \mathrm{m} .\) If you ride the bike for \(35 \mathrm{min},\) how far would you have gone if the bike could move?

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The angular open spaces between the three propeller blades are each \(\pi / 3\) rad \(\left(60.0^{\circ}\right) .\) The vertical drop of the arrow may be ignored. There is a maximum value \(\omega\) for the angular speed of the propeller, beyond which the arrow cannot pass through an open space without being struck by one of the blades. Find this maximum value when the arrow has the lengths \(L\) and speeds \(v\) shown in the following table. $$ \begin{array}{lll} & L & v \\ (\mathrm{a}) & 0.71 \mathrm{m} & 75.0 \mathrm{m} / \mathrm{s} \\ \hline(\mathbf{b}) & 0.71 \mathrm{m} & 91.0 \mathrm{m} / \mathrm{s} \\ \hline(\mathbf{c}) & 0.81 \mathrm{m} & 91.0 \mathrm{m} / \mathrm{s} \\ \hline \end{array} $$

An electric fan is running on HIGH. After the LOW button is pressed, the angular speed of the fan decreases to \(83.8 \mathrm{rad} / \mathrm{s}\) in \(1.75 \mathrm{s}\). The deceleration is \(42.0 \mathrm{rad} / \mathrm{s}^{2} .\) Determine the initial angular speed of the fan.

A racing car, starting from rest, travels around a circular turn of radius \(23.5 \mathrm{m}\). At a certain instant, the car is still accelerating, and its angular speed is \(0.571 \mathrm{rad} / \mathrm{s}\). At this time, the total acceleration (centripetal plus tangential) makes an angle of \(35.0^{\circ}\) with respect to the radius. (The situation is similar to that in Interactive Figure \(8.12 b .\) ) What is the magnitude of the total acceleration?

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