/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A car on a circular track accele... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A car on a circular track accelerates from rest. (a) The car experiences (1) only angular acceleration, (2) only centripetal acceleration, (3) both angular and centripetal accelerations. Why? (b) If the radius of the track is \(0.30 \mathrm{~km}\) and the magnitude of the constant angular acceleration is \(4.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2},\) how long does the car take to make one lap around the track? (c) What is the total (vector) acceleration of the car when it has completed half of a lap?

Short Answer

Expert verified
(a) Both angular and centripetal accelerations; (b) 117.5 seconds; (c) 21.1 m/s².

Step by step solution

01

Analyze Accelerations

When a car starts accelerating on a circular track, it initially has only angular acceleration because it starts from rest and needs to build up its angular velocity. As it begins to move in a circle, it experiences centripetal acceleration to change the direction of the velocity. Thus, the car experiences both angular and centripetal accelerations.
02

Determine the Time to Complete a Lap

First, we need to find the final angular velocity when the car completes a lap. Given the radius of the track is 0.3 km or 300 meters. The angular distance for one lap is \( 2\pi \) radians. Using the kinematic equation for angular motion \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), where \( \omega_0 \) is initial angular velocity, \( \alpha \) is angular acceleration, and \( t \) is time. Since \( \omega_0 = 0 \), the equation simplifies to \( \theta = \frac{1}{2} \alpha t^2 \). Solving for \( t \) when \( \theta = 2\pi \), we have \( 2\pi = \frac{1}{2} \times 4.5 \times 10^{-3} \times t^2 \). Solving for \( t \), \( t = \sqrt{\frac{2 \times 2\pi}{4.5 \times 10^{-3}}} \).
03

Calculate the Time

Calculate \( t \) using the equation: \( t = \sqrt{\frac{4\pi}{4.5 \times 10^{-3}}} \). This results in \( t \approx 117.5 \) seconds.
04

Finding Total Acceleration at Half Lap

At half a lap, \( \theta = \pi \). The angular velocity \( \omega \) can be found using \( \omega = \alpha t \). For half a lap, \( t' = t/2 \approx 58.75 \) seconds. Hence, \( \omega = 4.5 \times 10^{-3} \times 58.75 \approx 0.264 rad/s \). The centripetal acceleration \( a_c = \omega^2 \times r \). Substitute the values: \( a_c \approx 0.264^2 \times 300 \approx 21.05 \ m/s^2 \). The angular acceleration \( a_t = \alpha \times r = 4.5 \times 10^{-3} \times 300 = 1.35 \ m/s^2 \).
05

Calculate Vector Acceleration

The total acceleration is a vector sum of tangential and centripetal accelerations. Use \( a = \sqrt{a_t^2 + a_c^2} \). Compute: \( a = \sqrt{1.35^2 + 21.05^2} \approx 21.1 \ m/s^2 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It is a fundamental concept when describing rotational motion. If you've ever watched a spinning top start its spin, you've witnessed angular acceleration in action. In the context of the circular track problem, the car accelerates from rest, meaning it initially has no angular velocity. But as the engine works, the car's wheels begin turning, increasing its speed along the path. This change in angular velocity is what we call angular acceleration.

Key points about angular acceleration:
  • Characterized by the symbol \( \alpha \).
  • Measured in \( \text{rad/s}^2 \) (radians per second squared).
  • Relates to linear acceleration through the radius \( r \) by the equation \( a_t = r \times \alpha \), where \( a_t \) is the tangential acceleration.
Angular acceleration allows us to understand how fast an object is picking up speed as it rotates around a point. In the problem, the car's angular acceleration is given as \( 4.5 \times 10^{-3} \text{rad/s}^2 \). This tells us how quickly the car ramps up its rotational speed as it traverses the circular path.
Centripetal Acceleration
Centripetal acceleration is crucial for any object traveling in a circular path. It is what keeps the car on the track and directs it towards the center of the circle. Unlike angular acceleration, which changes the speed of an object, centripetal acceleration changes the direction of the velocity vector, helping the object maintain a circular path.

Some important aspects of centripetal acceleration include:
  • Symbolized as \( a_c \).
  • Calculated using the formula \( a_c = \omega^2 \times r \), where \( \omega \) is the angular velocity.
  • Measured in \( \text{m/s}^2 \) (meters per second squared).
  • Provides the inward force necessary for circular motion.
In our car example, as the vehicle speeds up, its centripetal acceleration increases. This ensures the car can follow the curve of the track smoothly. For a car traveling at an angular velocity of \( 0.264 \text{rad/s} \) on a radius of 300 meters, the centripetal acceleration would be about \( 21.05 \text{m/s}^2 \).
Vector Acceleration
Vector acceleration is a combination of linear (tangential) and centripetal accelerations, taking vector direction into account. Since both of these accelerations are perpendicular, they can be combined using the Pythagorean theorem to find the total acceleration. This gives us a comprehensive understanding of the acceleration the car experiences.

When combining accelerations, keep in mind:
  • Resultant acceleration is found using \( a = \sqrt{a_t^2 + a_c^2} \).
  • \( a_t \) is the tangential (linear) acceleration.
  • Vector diagrams can help visualize how these two components combine.
In the exercise, the vector sum of tangential and centripetal acceleration gives a total acceleration of roughly \( 21.1 \text{m/s}^2 \). This value represents the entire force the car feels as it moves halfway around the track.
Angular Kinematics
Angular kinematics is the study of motion with respect to rotation, similar to how linear kinematics applies to straight-line motion. It's essential for predicting how the position, velocity, and acceleration change over time for rotating objects. Just as linear kinematics uses position, velocity, and acceleration, angular kinematics uses angular displacement, velocity, and acceleration.

Key formulas in angular kinematics include:
  • \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) (angular displacement).
  • \( \omega = \omega_0 + \alpha t \) (angular velocity).
  • Similar to linear kinematics, but with rotational terms.
For our car, starting from rest means initial angular velocity \( \omega_0 = 0 \). The car's angular position after time \( t \) can be found using the displacement formula. This tells us how far the car has rotated on the track. Understanding angular kinematics allows us to predict the car's position and speed at any time during its motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 100 -kg object is taken to a height of \(300 \mathrm{~km}\) above the Earth's surface. (a) What is the object's mass at this height? (b) What is the object's weight at this height?

Just an instant before reaching the very bottom of a semicircular section of a roller coaster ride, the automatic emergency brake inadvertently goes on. Assume the car has a total mass of \(750 \mathrm{~kg}\), the radius of that section of the track is \(55.0 \mathrm{~m},\) and the car entered the bottom after descending vertically (from rest) \(25.0 \mathrm{~m}\) on a frictionless straight incline. If the braking force is a steady \(1700 \mathrm{~N}\), determine (a) the car's centripetal acceleration (including direction), \((\mathrm{b})\) the normal force of the track on the car, (c) the tangential acceleration of the car (including direction), and (d) the total acceleration of the car.

A man has a mass of \(75 \mathrm{~kg}\) on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" \(10 \%\) of his body weight?

The acceleration due to gravity near a planet's surface is known to be \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). If the escape speed from the planet is \(8.42 \mathrm{~km} / \mathrm{s},\) (a) determine its radius. (b) Find the mass of the planet. (c) If a probe is launched from its surface with a speed twice the escape speed and then coasts outward, neglecting other nearby astronomical bodies, what will be its speed when it is very far from the planet? (Neglect any atmospheric effects also.) (d) Under these launch conditions, at what distance will its speed be equal to the escape speed?

(a) Could a circular pie be cut such that all of the wedge-shaped pieces have an arc length along the outer crust equal to the pie's radius? (b) If not, how many such pieces could you cut, and what would be the angular dimension of the final piece?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.