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A computer disk drive is turned on starting from rest and has constant angular acceleration. If took 0.750 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 \(\operatorname{rad} / s^{2} ?\)

Short Answer

Expert verified
(a) 0.53 seconds, (b) 44.63 rad/s².

Step by step solution

01

Understanding the Problem

The computer disk drive starts from rest with a constant angular acceleration. We know the time taken for it to complete the second revolution and need to find the time for the first revolution and the angular acceleration.
02

Use Angular Kinematics Equation for Second Revolution

For the disk drive starting from rest with constant angular acceleration, after completing the second revolution, the total angular displacement is \(4\pi\) radians. We use the equation \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\). Here, \(\theta = 4\pi\), \(\omega_0 = 0\), and \(t = 0.750 \) s for the second revolution. Substituting gives: \[ 4 \pi = \frac{1}{2} \alpha (0.750)^2 \] Solve for \(\alpha\).
03

Solve for Angular Acceleration

After substituting values into the equation from Step 2: \[ 4\pi = \frac{1}{2} \alpha (0.750)^2 \] Calculating, we have \[ \alpha = \frac{4 \pi}{0.28125} \approx 44.63 \text{ rad/s}^2 \]
04

Use Angular Kinematics Equation for First Revolution

For the first revolution, \(\theta = 2\pi\) radians. Using the same kinematic equation: \[ 2\pi = \frac{1}{2} \alpha t_1^2 \] Substituting \(\alpha = 44.63 \), and solving for \(t_1\), we get: \[ 2\pi = \frac{1}{2} \times 44.63 \times t_1^2 \] Solve for \(t_1\).
05

Solve for Time of First Revolution

Rearrange the equation from Step 4 and calculate: \[ t_1^2 = \frac{2 \times 2 \pi}{44.63} \approx 0.2815 \] \[ t_1 = \sqrt{0.2815} \approx 0.53 \text{ seconds} \]

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Key Concepts

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

Constant Angular Acceleration
In angular kinematics, angular acceleration represents how quickly an object's rotational velocity changes over time. Constant angular acceleration means that this change is steady and uniform, much like how gravity provides constant acceleration to a freely falling object. When a rotating system begins from rest, this concept helps in predicting how fast it will turn at any given point.

With constant angular acceleration, the rotation speed increases linearly with time. This property simplifies calculations because you don't need to worry about changes in the rate of increase. You can apply kinematic equations similar to those used in linear motion but adapted for rotational scenarios. These include equations for angular displacement, angular velocity, and time.
  • Useful for predicting motion in circular or rotary systems like wheels, disks, and motors.
  • Equations used: \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\), where \(\theta\) is the angular displacement, \(\omega_0\) is initial angular velocity, and \(\alpha\) is angular acceleration.
Angular Displacement
Angular displacement refers to the angle through which a point or line has been rotated in a specified sense about a specified axis. It's analogous to linear displacement, measuring how far an object has traveled, but in the context of rotational motion.

In this exercise, the total angular displacement is given for each complete revolution. One complete revolution corresponds to an angular displacement of \(2\pi\) radians. Therefore, for the second revolution, the disk underwent a total of \(4\pi\) radians. This displacement can be determined using the kinematic equation, assuming constant angular acceleration.
  • For the first complete revolution: \(\theta = 2\pi\) radians.
  • For the second complete revolution: \(\theta = 4\pi\) radians.
Angular Acceleration Calculation
To find the angular acceleration, we used details from the second complete revolution: the time taken (0.750 seconds) and the total angular displacement \(4\pi\) radians. Knowing that the system starts from rest simplifies the equation, as the initial angular velocity \(\omega_0\) is zero.

Substitute these known values into the equation:
\[4\pi = \frac{1}{2}\alpha (0.750^2)\]
Rearranging gives us:
\[\alpha = \frac{4\pi}{0.28125} \approx 44.63 \, \text{rad/s}^2\]
This calculation shows the uniform rate of increase in angular velocity from zero for this system.
  • It's crucial to plug values correctly to avoid errors in finding \(\alpha\).
  • Understanding the formula allows you to solve similar problems by just changing known variables.
Revolution Time Calculation
Once the angular acceleration is calculated, you can determine how long different revolutions take. For this problem, once \(\alpha\) is known, finding the time for the first revolution is straightforward. We know from the problem that \(\theta = 2\pi\) radians.

Using the equation:
\[2\pi = \frac{1}{2}\alpha t_1^2\]
Substitute \(\alpha = 44.63\) into the equation:
\[t_1^2 = \frac{4\pi}{44.63}\approx 0.2815\]
Taking the square root provides the time for the first revolution:
\[t_1 \approx 0.53 \, \text{seconds}\]
These kinds of calculations help in understanding performance and timing in systems relying on rotational motion, such as disk drives or engines.
  • Careful calculation ensures accurate results and understanding of rotational timelines.
  • This principle can be reused across different rotational scenarios by adjusting figures.

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Most popular questions from this chapter

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 turms, 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 safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s}\) . What angular velocity is required to achieve these values?

A straight piece of reflecting tape extends from the center of a wheel to its rim. You darken the room and use a camera and strobe unit that flashes once every 0.050 s to take pictures of the wheel as it rotates counterclockwise. You trigger the strobe so that the first flash \((t=0)\) occurs when the tape is horizontal to the right at an angular displacement of zero. For the following situations draw a sketch of the photo you will get for the time exposure over five flashes (at \(t=0,0.050 \mathrm{s}, 0.100 \mathrm{s}, 0.150 \mathrm{s},\) and 0.200 \(\mathrm{s} )\) and graph \(\theta\) versus \(t\) and \(\omega\) versus \(t\) for \(t=0\) to \(t=0.200 \mathrm{s}\) (a) The angular velocity is constant at 10.0 rev \(/ \mathrm{s}\) . (b) The wheel starts from rest with a constant angular acceleration of 25.0 rev \(/ \mathrm{s}^{2}\) . (c) The wheel is rotating at 10.0 rev \(/ \mathrm{s}\) at \(t=0\) and changes angular velocity at a constant rate of \(-50.0 \mathrm{rev} / \mathrm{s}^{2}\) .

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 one-fourth of the length from one end.

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