Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s\(^2\), what is its angular velocity at \(t =\) 2.50 s? (b) Through what angle has the wheel turned between \(t =\) 0 and \(t =\) 2.50 s?

Short Answer

Expert verified
(a) 2.00 rad/s, (b) 4.375 rad

Step by step solution

01

Identify Given Values

From the problem, we have:- Initial angular velocity, \( \omega_0 = 1.50 \, \text{rad/s} \)- Angular acceleration, \( \alpha = 0.200 \, \text{rad/s}^2 \)- Time, \( t = 2.50 \, \text{s} \)
02

Find Final Angular Velocity

We use the equation for angular velocity with constant angular acceleration:\[ \omega = \omega_0 + \alpha t \]Substitute the given values:\[ \omega = 1.50 \, \text{rad/s} + (0.200 \, \text{rad/s}^2)(2.50 \, \text{s}) \]\[ \omega = 1.50 \, \text{rad/s} + 0.50 \, \text{rad/s} \]\[ \omega = 2.00 \, \text{rad/s} \]Thus, the final angular velocity at \( t = 2.50 \, \text{s} \) is 2.00 rad/s.
03

Calculate Total Angular Displacement

Use the angular displacement formula:\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]Substitute the known values:\[ \theta = (1.50 \, \text{rad/s})(2.50 \, \text{s}) + \frac{1}{2}(0.200 \, \text{rad/s}^2)(2.50 \, \text{s})^2 \]\[ \theta = 3.75 \, \text{rad} + \frac{1}{2}(0.200)(6.25) \, \text{rad} \]\[ \theta = 3.75 \, \text{rad} + 0.625 \, \text{rad} \]\[ \theta = 4.375 \, \text{rad} \]Thus, the total angle through which the wheel has turned is 4.375 radians.

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 Vaia!

Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates or spins around a central point or axis. In this context of the bicycle wheel, it describes how fast the wheel is spinning. It is a vector quantity, meaning it has both a magnitude and a direction, but in most simple scenarios such as a bicycle wheel, we focus primarily on the magnitude.
Angular velocity is often represented by the Greek letter omega (\( \omega \)), and it is measured in radians per second (rad/s). Imagine it like the speed of a car, but for a rotating object instead. When a wheel spins faster, its angular velocity increases.

To calculate angular velocity when angular acceleration is constant, you use this equation:
\[ \omega = \omega_0 + \alpha t \]
Where:
  • \( \omega_0 \) is the initial angular velocity (how fast it was spinning to begin with)
  • \( \alpha \) is the angular acceleration
  • \( t \) is the time elapsed

In the original problem, when the bicycle wheel started with an angular velocity of 1.50 rad/s and an angular acceleration of 0.200 rad/s² was applied for 2.5 seconds, its final angular velocity increased to 2.00 rad/s as calculated using the formula. This indicates the wheel spins faster over time due to the constant acceleration.
Angular Acceleration
Angular acceleration is the rate at which angular velocity changes with time. It tells us how quickly the angular speed of a spinning object, like a wheel, is increasing or decreasing. It is represented by the Greek letter alpha (\( \alpha \)) and measured in radians per second squared (rad/s²).

Think of it as how hard you're pushing or pulling on a spinning wheel to change its speed. If a wheel is accelerating, it means the angular velocity of the wheel is either increasing or decreasing.

In our exercise, the bicycle wheel has a constant angular acceleration of 0.200 rad/s². This means every second, the speed at which the wheel spins increases by 0.200 rad/s. This constant acceleration is crucial when using the formula for finding the change in angular velocity over time, as it simplifies the calculation by allowing us to assume a linear increase. Without constant acceleration, the equations would be more complex, involving calculus to account for variable changes over time.
Angular Displacement
Angular displacement is the measure of how much an object has rotated or turned, generally represented by the Greek letter theta (\( \theta \)). It’s akin to distance in linear motion but deals with angles in rotational motion.
Measured in radians, it gives us the total change in the angle of the rotating object from start to finish.

To calculate angular displacement, especially when dealing with constant angular acceleration, you can use the formula:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Where:
  • \( \omega_0 \) is the initial angular velocity
  • \( \alpha \) is the angular acceleration
  • \( t \) is time

Using this formula, you start with the initial rotation and add the effect of acceleration over a specified time.
In this problem, with the given values, the total amount that the wheel rotates in 2.5 seconds comes out to be 4.375 radians. This tells you exactly how far the wheel has turned from its initial position.

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

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is 60.0 cm long and has mass 0.400 kg. (a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod? (b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a 60.0\(^\circ\) angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?

At \(t =\) 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 rad/s\(^2\) until a circuit breaker trips at \(t =\) 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t =\) 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm 1rev/min2. (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A thin, uniform rod is bent into a square of side length \(a\). If the total mass is \(M\), find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (\(Hint\): Use the parallel-axis theorem.)

If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \\( \frac{1}{2} \\)\(MV^2\). However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13% of a person's mass, while the legs and feet together account for 37%. For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm30^\circ\) (a total of 60\(^\circ\)) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

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.

Sign-up for free