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Chapter 9: Problem 66

A computer disk drive is turned on starting from rest and has constant angular acceleration. If it took 0.0865 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
(a) 0.0611 s; (b) 3386.52 rad/s虏.

Step by step solution

01

Understanding the Problem

We are given that a disk drive starts from rest with constant angular acceleration and takes 0.0865 s to complete its second revolution. We need to calculate the time taken for the first revolution and the angular acceleration.
02

Using the Angular Kinematics Formula

The equation for angular displacement under constant angular acceleration is \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \). Since it starts from rest \( \omega_i = 0 \) and the angle for one revolution is \( 2\pi \) radians.
03

Writing Equations for First and Second Revolutions

For the first revolution, \( 2\pi = \frac{1}{2} \alpha t_1^2 \). For the first and second revolutions combined, \( 4\pi = \frac{1}{2} \alpha t_2^2 \). We know \( t_2 = t_1 + 0.0865 \) s.
04

Solving for Angular Acceleration

We can solve the equations simultaneously to find \( \alpha \), the angular acceleration. Substituting \( t_2 = t_1 + 0.0865 \) into the second equation: \( 4\pi = \frac{1}{2} \alpha (t_1 + 0.0865)^2 \).
05

Simplifying and Solving for \( t_1 \)

From the first equation: \( \alpha = \frac{4\pi}{t_1^2} \). Substitute into the combined equation gives: \( 4\pi = 2\pi (\frac{t_2^2}{t_1^2}) \), solve for \( t_1 \) gives \( t_1 \approx 0.0611 \) s.
06

Calculating Angular Acceleration

Using \( t_1 \approx 0.0611 \) s in the first formula for \( \alpha \), \( \alpha = \frac{4\pi}{(0.0611)^2} = 3386.52 \text{ rad/s}^2 \).

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Key Concepts

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

Constant Angular Acceleration
Understanding constant angular acceleration is crucial to solving this problem. This term describes how an object's rotational speed changes at a constant rate over time. In layman's terms, it means that with each passing second, the disk drive spins faster and faster by the same amount.
For example, if the angular acceleration of the disk is 3386.52 rad/s two, as calculated in the solutions, it means that the rotational speed increases by this amount every second.
  • Constant angular acceleration implies predictable rotational motion.
  • It allows us to use simple equations to solve complex problems.
This concept is akin to linear acceleration in everyday experiences, like a car speeding up at a constant rate.
Angular Kinematics
Angular kinematics involves the equations used to describe rotational motion, similar to linear motion equations used for objects moving in a straight line. Here, however, we're dealing with angles, angular speed, and angular acceleration instead of distances, speeds, and linear accelerations.
The main formula used in this problem is:\[\theta = \omega_i t + \frac{1}{2} \alpha t^2\]where:
  • \(\theta\): Angular displacement (in radians)
  • \(\omega_i\): Initial angular velocity (\(0\) in this problem as it starts from rest)
  • \(\alpha\): Angular acceleration
  • \(t\): Time
This formula is essential to understanding how the drive completes its revolutions, beginning with zero speed and gradually speeding up.
Revolution Timing
Revolution timing refers to the time taken for the disk drive to complete one full spin or revolution. In this scenario, we鈥檙e examining the timing of both the first and second revolutions.
To find out how long it takes to reach one complete rotation, we need to consider its angular kinematics. The first complete revolution timing calculation uses \(2\pi = \frac{1}{2} \alpha t_1^2\) to find that it took about 0.0611 seconds.
  • The second revolution takes 0.0865 seconds, as given.
  • The difference between the two provides insights into how acceleration impacts timing.
By comparing these times, we can learn about the disk鈥檚 increasing speed during startup.
Angular Displacement
Angular displacement is the change in position of a point moving along a circular path. It's similar to how linear displacement measures change in position along a straight line, but in a circle, this change is expressed in radians.
For circular revolutions:
  • One complete revolution equals \(2\pi\) radians.
  • Two complete revolutions equal \(4\pi\) radians.
As seen in the exercise, these values are integral to determining the time required and calculating angular acceleration. Angular displacement, observed as \(2\pi\) or \(4\pi\) radians, helps quantify how much the disk has rotated over time. It's the foundation for exploring motion in circular paths.

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Most popular questions from this chapter

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius 20.0 cm. A 1.50-kg stone is attached to a very light wire that is wrapped around the rim of the pulley (\(\textbf{Fig. E9.43}\)), and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

A wheel is rotating about an axis that is in the \(z\)-direction.The angular velocity \(\omega_z\) is \(-\)6.00 rad/s at \(t =\) 0, increases linearly with time, and is \(+\)4.00 rad/s at \(t =\) 7.00 s. We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t =\) 7.00 s?

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.45. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 m and moment of inertia for rotation about the axle of \(I =\) 0.480 kg \(\cdot\) m\(^2\). A small block with mass 0.340 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-\)9.00 J of work as the block descends 3.00 m. What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 m?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 25.0 cm. Starting from rest at \(t =\) 0, the flywheel rotates with constant angular acceleration 3.00 rad/s\(^2\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of 8600 kg/m\(^3\), what thickness must it have to store 800 J of kinetic energy at \(t =\) 8.00 s?
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