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Chapter 10: Problem 70

A wheel, starting from rest, rotates with a constant angular acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). During a certain \(3.00 \mathrm{~s}\) interval, it turns through \(90.0\) rad. (a) What is the angular velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval? (b) How long has the wheel been turning before the start of the \(3.00\) s interval?

Short Answer

Expert verified
(a) The angular velocity at the start is 27.0 rad/s.\n(b) The wheel was turning for 10.5 seconds before the 3-second interval.

Step by step solution

01

Understand the Problem

We are given that a wheel starts from rest and has a constant angular acceleration of \(2.00 \, \mathrm{rad/s}^2\). During a \(3.00\, \mathrm{s}\) interval, it rotates \(90.0\, \mathrm{rad}\). We need to find (a) the initial angular velocity at the start of the \(3.00\, \mathrm{s}\) interval and (b) the time the wheel has been turning before this interval began.
02

Use the Angular Displacement Formula

The formula for angular displacement \(\theta\) with constant angular acceleration \(\alpha\) is \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), where \(\omega_0\) is the initial angular velocity and \(t\) is the time period. Substitute \(\theta = 90.0\, \mathrm{rad}\), \(\alpha = 2.00\, \mathrm{rad/s}^2\), and \(t = 3.00\, \mathrm{s}\) into the equation. We have: \[ 90 = \omega_0 \times 3 + \frac{1}{2} \times 2 \times 3^2 \].
03

Solve for Initial Angular Velocity

Simplify the equation from Step 2: \[ 90 = 3\omega_0 + 9 \]. Reordering gives \( 3\omega_0 = 81 \), thus \( \omega_0 = \frac{81}{3} = 27.0 \, \mathrm{rad/s} \). This is the angular velocity at the start of the \(3.00\, \mathrm{s}\) interval.
04

Use Angular Velocity Formula to Find Total Time

The formula for angular velocity with constant acceleration is \( \omega = \omega_0 + \alpha t \). Since the wheel starts from rest at the beginning of its motion, we have \( \omega = \alpha \times T \) where \( T \) is the total time of rotation. Use \( \omega = 27.0\, \mathrm{rad/s} \) and \( \alpha = 2.00\, \mathrm{rad/s}^2 \): \( 27 = 2T \).
05

Solve for Total Time of Rotation

From \( 27 = 2T \), we get \( T = \frac{27}{2} = 13.5 \mathrm{s} \). This is the total time the wheel has been in motion since it started from rest.
06

Calculate Time Before 3 Seconds Interval

Since the \(3.00\, \mathrm{s}\) interval occurs at the end of this total time, subtract this time from the total time: \( T_{\text{before}} = 13.5\, \mathrm{s} - 3.0\, \mathrm{s} = 10.5\, \mathrm{s} \).

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Key Concepts

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

Angular Displacement Formula
In rotational motion, angular displacement is the measure of the angle through which an object rotates around a fixed axis. To calculate angular displacement when there is constant angular acceleration, we use the formula: \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]This equation helps us determine how far an object has rotated within a specific time period. Here,
  • \(\theta\) is the angular displacement,
  • \(\omega_0\) is the initial angular velocity,
  • \(t\) is the time interval, and
  • \(\alpha\) is the constant angular acceleration.
In the exercise, the wheel rotates through \(90.0\) radians in \(3.00\) seconds under an acceleration of \(2.00 \, \mathrm{rad/s}^2\). By substituting these values into the formula, students can find the initial angular velocity needed to start this rotation.
Angular Velocity
Angular velocity is a measure of how fast an object is rotating. It's comparable to linear velocity, but in the context of rotational motion. When an object starts from rest and speeds up with constant acceleration, the angular velocity at any point in time can be determined using:\[ \omega = \omega_0 + \alpha t \]Here,
  • \(\omega\) is the final angular velocity,
  • \(\omega_0\) is the initial angular velocity, and
  • \(\alpha\) is the angular acceleration,
  • \(t\) is the time.
In the given problem, we use this formula to determine how fast the wheel is rotating at the start of a \(3.00 \, \mathrm{s}\) interval, after accelerating steadily for some time. This provides essential understanding of how velocity changes in rotary systems.
Constant Acceleration
Constant acceleration means that the rate of change of velocity remains the same over time. In the case of angular acceleration, it implies that the speed of rotation increases steadily. The benefit of constant acceleration is that it allows us to use specific equations of motion to predict future behavior of the rotating object. Key aspects of constant acceleration include:
  • Linear and angular equations of motion share similarities, providing a way to understand rotations using familiar linear concepts.
  • Steady changes make calculations predictable and reliable.
By knowing this constant acceleration, one can determine key details such as how fast an object will be moving after some time or how long it took to reach a certain velocity, as illustrated in the exercise.
Rotational Motion
Rotational motion involves objects moving around a central pivot or axis and is a pivotal concept in physics. It's directly analogous to translational motion, which deals with linear movement. In rotational motion:
  • The position is expressed as angular displacement.
  • Speed is measured as angular velocity.
  • Acceleration is referred to as angular acceleration.
Understanding rotational motion is vital due to its application in everyday mechanical systems, like wheels and gears. In the exercise, the wheel's motion provides a clear example of such rotation, illustrating how these concepts are used to decipher the movement of rotating objects.

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Most popular questions from this chapter

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period \(T\) of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of \(T=0.033 \mathrm{~s}\) that is increasing at the rate of \(1.26 \times 10^{-5} \mathrm{~s} / \mathrm{y} .\) (a) What is the pulsar's angular acceleration \(\alpha ?\) (b) If \(\alpha\) is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year 1054 . Assuming constant \(\alpha\), find the initial \(T\).

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length \(55.0 \mathrm{~m}\). At the instant it makes an angle of \(35.0^{\circ}\) with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle \(\theta\) is the tangential acceleration equal to \(g\) ?

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.0 \mathrm{~s}\), the angular velocity of the wheel is \(5.0 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=20 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=40 \mathrm{~s}\) ?

A record turntable rotating at \(33 \frac{1}{3}\) rev/min slows down and stops in \(30 \mathrm{~s}\) after the motor is turned off. (a) Find its (constant) angular acceleration in revolutions per minute-squared. (b) How many revolutions does it make in this time?

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just before it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.)
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