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Chapter 9: Problem 68

A computer disk drive is turned on starting from rest and has constant angular acceleration. If it took \(0.0865 \mathrm{~s}\) for the drive to make its second complete revolution, (a) how long did it take to make the first complete revolution, and (b) what is its angular acceleration, in rad/s \(^{2}\) ?

Short Answer

Expert verified
The time it takes to make the first complete revolution is \(0.04325 s\). And the angular acceleration is given by \( \alpha = 2 * 2\pi / (0.04325 s)^2 rad/s^2\).

Step by step solution

01

Convert revolution to radians

Given that one complete revolution is equal to \(2\pi\) radians, the number of radians for the second complete revolution is \(4\pi\) radians.
02

Find the time for the first revolution

Since the angular acceleration is constant, the time taken for each revolution should be increasing linearly. This means the time for the first revolution is half of the total time given. Hence, the time to complete the first revolution \(t_1\) is \(0.0865 s / 2 = 0.04325 s\).
03

Find the angular acceleration

To find the angular acceleration, we'll use the following equation from rotational kinematics: \( \theta = \omega_0 t + 0.5 \alpha t^2\), where \( \theta\) is the total angular displacement, \( \omega_0\) is the initial angular velocity, \( \alpha\) is the angular acceleration, and \( t\) is the time. The drive starts from rest, so \( \omega_0 = 0\). After the first revolution, \( \theta = 2\pi\) and \( t = t_1 = 0.04325 s\). Solving for \( \alpha\) we find: \( \alpha = 2\theta / t^2 = 2 * 2\pi / (0.04325 s)^2\).
04

Calculate the angular acceleration

Substitute the values into the equation to get the numerical value of \( \alpha\).

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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 the rate of change of angular velocity over time. It's similar to how linear acceleration describes the rate of change of velocity. In rotational motion, angular acceleration helps us understand how quickly or slowly an object spins up or down. In this exercise, the disk drive starts from rest and spins up, which means its angular velocity increases consistently over time. Since there's constant angular acceleration, we use a straightforward approach to find its value by applying kinematic equations. This clarifies how quickly the drive reaches a new speed given the initial state of zero velocity.
Revolutions to Radians Conversion
When dealing with rotational motion, it's essential to convert between different units of angle measurement. One full revolution equates to a full circle, which in radians, is expressed as \(2\pi\) radians. Understanding this conversion is crucial, especially when solving problems like the one in the exercise. Here, for two complete revolutions, the angular displacement is \(4\pi\) radians.
  • 1 revolution = \(2\pi\) radians
  • 2 revolutions = \(4\pi\) radians
This conversion helps simplify angular calculations, enabling you to apply them directly into kinematic equations.
Rotational Motion
Rotational motion involves objects rotating or spinning around an axis. It can describe everything from bicycle wheels to orbiting planets. Just like linear motion is characterized by displacement, velocity, and acceleration, rotational motion also has its counterparts: angular displacement, angular velocity, and angular acceleration.
  • Angular Displacement: Measured in radians, indicates the angle covered.
  • Angular Velocity: The rate of change of angular position, measured in radians per second.
  • Angular Acceleration: Describes how angular velocity changes over time, with the first equation from the exercise emphasizing this relation.
Understanding rotational motion helps dissect the behavior of rotating systems, and makes solving problems involving these principles manageable.
Kinematics Equations
Kinematics equations provide a framework to calculate the various parameters of motion鈥攚hether linear or rotational. In this problem, the equation \( \theta = \omega_0 t + 0.5 \alpha t^2\) is pivotal. It connects angular displacement (\(\theta\)), initial angular velocity (\(\omega_0\)), time (\(t\)), and angular acceleration (\(\alpha\)). For a disk starting from rest, the initial angular velocity is zero, simplifying the equation to \( \theta = 0.5 \alpha t^2\). This is instrumental when there's a need to solve for the unknown variable, such as angular acceleration in this specific problem.
  • Solve for \(\alpha\): \( \alpha = \frac{2\theta}{t^2} \)
These equations enable you to determine any missing value in a motion scenario, making them an essential tool in physics.

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Most popular questions from this chapter

A uniform disk with radius \(R=0.400 \mathrm{~m}\) and mass \(30.0 \mathrm{~kg}\) rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=(1.10 \mathrm{rad} / \mathrm{s}) t+\left(6.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

Three small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point onefourth of the length from one end.

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

A flywheel with radius \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2} .\) For a point on the rim of the flywheel, what are the magnitudes of the tangential, radial, and resultant accelerations after \(2.00 \mathrm{~s}\) of acceleration?

At \(t=3.00 \mathrm{~s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of \(50.0 \mathrm{~m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{~s}\) and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{~s} ?\) (d) At what time will the radial acceleration equal \(g ?\)
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