/*! 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 16 A merry-go-round rotates from re... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 16

A merry-go-round rotates from rest with an angular acceleration of \(1.50 \mathrm{rad} / \mathrm{s}^{2}\). How long does it take to rotate through (a) the first \(2.00\) rev and (b) the next \(2.00\) rev?

Short Answer

Expert verified
(a) 4.09 seconds, (b) 1.70 seconds.

Step by step solution

01

Understanding the Problem

We're given that the merry-go-round starts from rest and has a constant angular acceleration. We need to find out how long it takes for it to complete the first 2.00 revolutions and then the next 2.00 revolutions.
02

Convert Revolutions to Radians

Since 1 revolution is equivalent to \(2\pi\) radians, we convert the given revolutions to radians.\[2.00 \text{ rev} = 2.00 \times 2\pi \text{ radians} = 4\pi \text{ radians}\] So we need the time for the first \(4\pi\) radians and the next \(4\pi\) radians.
03

Using the Kinematic Equation

To find the time, we use the kinematic equation for angular motion: \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\). Since the initial angular velocity \(\omega_0\) is 0, the equation simplifies to \(\theta = \frac{1}{2}\alpha t^{2}\).
04

Solve for the First 2.00 Revolutions

Substitute \(\theta = 4\pi\) radians and \(\alpha = 1.50 \text{ rad/s}^2\) into the equation: \[4\pi = \frac{1}{2} \times 1.50 \times t^{2}\] \[t^2 = \frac{8\pi}{1.50}\] \[t = \sqrt{\frac{8\pi}{1.50}}\] Calculate \(t\) to find the time for the first 2.00 revolutions.
05

Calculate Time for First 2.00 Revolutions

Calculate \(t\): \[t = \sqrt{\frac{8\times3.1416}{1.50}} \approx \sqrt{16.755} \approx 4.09 \text{ seconds}\] Thus, it takes about 4.09 seconds for the first 2 revolutions.
06

Solve for the Next 2.00 Revolutions

To find the time for the next 2 revolutions, first determine the total time to achieve \(8\pi\) radians: \[8\pi = \frac{1}{2} \times 1.50 \times T^{2}\] \[T^2 = \frac{16\pi}{1.50}\] \[T = \sqrt{\frac{16\pi}{1.50}}\] Calculate \(T\) to find the total time for 4.00 revolutions.
07

Calculate Time for 4.00 Revolutions

Calculate \(T\): \[T = \sqrt{\frac{16\times3.1416}{1.50}} \approx \sqrt{33.51} \approx 5.79 \text{ seconds}\] The total time to complete 4 revolutions is about 5.79 seconds.
08

Determine Time for the Next 2.00 Revolutions

Subtract the time for the first 2 revolutions from the total time for 4 revolutions: \(T_{next} = 5.79 - 4.09 = 1.70\) seconds. Thus, it takes 1.70 seconds for the next 2 revolutions.

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Key Concepts

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

Angular Acceleration
Angular acceleration is a measure of how quickly the angular velocity of an object changes. It is expressed in radians per second squared ( ext{rad/s}^2). This concept is similar to linear acceleration but applies to rotational motion.
For the merry-go-round problem, the angular acceleration is constant at 1.50 ext{rad/s}^2. This means that the rate of change of angular velocity remains the same over time. With angular acceleration, one can determine how fast a rotating object speeds up or slows down over a period.
  • This is crucial when calculating how long it takes for objects to complete certain angular distances.
  • As you deal with problems involving rotation, understanding this value helps in predicting motion and solving equations involving rotational kinematics.
Kinematic Equations
Kinematic equations describe the motion of objects, accounting for their initial conditions, accelerations, and time. When dealing with angular motion, we use these equations just like in linear motion but apply them in terms of ext{radians} and ext{angular velocity}.
The primary kinematic equation used in this problem is:\[\theta = \omega_0 t + \frac{1}{2}\alpha t^2\]where:
  • \(\theta\) is the angular displacement in radians.
  • \(\omega_0\) is the initial angular velocity, which is zero in this case.
  • \(\alpha\) is the angular acceleration (given as 1.50 ext{rad/s}^2).
  • \(t\) is the time.
With the initial angular velocity being zero, the equation simplifies. This eases calculations when finding how long it takes to reach a certain angular position.
Angular Displacement
Angular displacement refers to the change in rotational position and is measured in radians. It indicates how far an object has rotated from its initial position over time.
In this problem, we're interested in how long it takes to reach a specific angular displacement鈥攏amely completing the first 2 revolutions and then the next two.
  • Using the kinematic equation, angular displacement is denoted by \(\theta\).
  • We calculated that \(2.00\) revolutions correspond to \(4\pi\) radians for angular displacement.
This mathematical representation is crucial in solving problems involving angular motion as it standardizes the change in rotational positions.
Revolutions to Radians Conversion
When dealing with rotational motion, it's common to convert between revolutions and radians because equations of motion typically use radians.
  • One revolution is equivalent to \(2\pi\) radians.
  • To convert revolutions to radians, multiply the number of revolutions by \(2\pi\).
For instance, in our problem, \(2.00\) revolutions becomes \(4\pi\) radians. This conversion is critical to properly use kinematic equations since they require angular displacement in radians. This small step ensures that calculations are consistent and align with the mathematical standards of angular motion.

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Most popular questions from this chapter

The flywheel of an engine is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). When the engine is turned off, the flywheel slows at a constant rate and stops in \(20.0 \mathrm{~s}\). Calculate (a) the angular acceleration of the flywheel, (b) the angle through which the flywheel rotates in stopping, and (c) the number of revolutions made by the flywheel in stopping.

A high-wire walker always attempts to keep his center of mass over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of \(70.0 \mathrm{~kg}\) and a rotational inertia of \(15.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the wire. What is the magnitude of his angular acceleration about the wire if his com is \(5.0 \mathrm{~cm}\) to the right of the wire and (a) he carries no pole and (b) the \(14.0 \mathrm{~kg}\) pole he carries has its \(\operatorname{com} 10 \mathrm{~cm}\) to the left of the wire?

A disk, initially rotating at \(120 \mathrm{rad} / \mathrm{s}\), is slowed down with a constant angular acceleration of magnitude \(4.0 \mathrm{rad} / \mathrm{s}^{2} .(\mathrm{a})\) How much time does the disk take to stop? (b) Through what angle does the disk rotate during that time?

A small ball of mass \(0.75 \mathrm{~kg}\) is attached to one end of a 1.25-m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is \(30^{\circ}\) from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of \(200 \pi \mathrm{rad} / \mathrm{s}\). One such flywheel is a solid, uniform cylinder with a mass of \(500 \mathrm{~kg}\) and a radius of \(1.0 \mathrm{~m} .\) (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of \(8.0 \mathrm{~kW}\), for how many minutes can it operate between chargings?
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