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Chapter 10: Problem 23

A flywheel with a diameter of \(1.20 \mathrm{~m}\) is rotating at an angular speed of 200 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute- squared) will increase the wheel's angular speed to 1000 rev/min in \(60.0 \mathrm{~s}\) ? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s}\) ?

Short Answer

Expert verified
(a) \(\frac{20\pi}{3} \text{ rad/s}\); (b) \(12.57 \text{ m/s}\); (c) \(800 \text{ rev/min}^2\); (d) 600 revolutions.

Step by step solution

01

Convert Angular Speed to Radians per Second

The angular speed is given as 200 rev/min. To convert this to radians per second, first convert revolutions to radians. Since one revolution is equal to \(2\pi\) radians, multiply the angular speed by \(2\pi\): \[200 \times 2\pi = 400\pi \text{ radians per minute}.\]Next, convert minutes to seconds by dividing by 60:\[\omega = \frac{400\pi}{60} = \frac{20\pi}{3} \text{ radians per second}.\]
02

Calculate Linear Speed at the Rim

The linear speed \(v\) can be calculated using the formula \(v = r\omega\), where \(r\) is the radius of the flywheel. The radius is half the diameter, so \(r = 0.60 \text{ m}\). Using the angular speed \(\omega = \frac{20\pi}{3} \text{ radians/second}\), we find:\[v = 0.60 \times \frac{20\pi}{3} = 4\pi \approx 12.57 \text{ m/s}.\]
03

Find Constant Angular Acceleration

The initial angular speed is 200 rev/min, and the final speed should be 1000 rev/min. First, find the change in angular speed:\[\Delta \omega = 1000 - 200 = 800 \text{ rev/min}.\]The time duration is 60 seconds, which corresponds to 1 minute. Thus, the angular acceleration \(\alpha\) in rev/min虏 is:\[\alpha = \frac{\Delta \omega}{\Delta t} = \frac{800}{1} = 800 \text{ rev/min}^2.\]
04

Calculate Total Revolutions During Acceleration

The angular displacement \(\theta\) can be found using the equation:\[\theta = \omega_0 t + \frac{1}{2} \alpha t^2,\]where \(\omega_0 = 200 \text{ rev/min}\), \(t = 60 \text{ seconds}\) or 1 minute, and \(\alpha = 800 \text{ rev/min}^2\). Substituting these into the equation gives:\[\theta = 200 \times 1 + \frac{1}{2} \times 800 \times 1^2 = 200 + 400 = 600 \text{ revolutions}.\]

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Key Concepts

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

Angular Speed
Angular speed refers to how quickly a rotating object spins, measured in terms of revolutions over time. In this context, it's important to convert the given angular speed from revolutions per minute (rev/min) to radians per second (rad/s) as it offers a more standardized measure in physics. This conversion uses the fact that one full revolution is equivalent to \( 2\pi \) radians.
Here's a simple approach to remember:
  • Multiply the angular speed in rev/min by \( 2\pi \) to convert it to radians per minute.
  • Then, divide by 60 to switch from minutes to seconds.
This step ensures the angular speed is in the correct format for further calculations involving linear speed and angular acceleration.
Linear Speed
Linear speed measures how fast a point on a rotating object is moving along its circular path. It is dependent on both the angular speed and the radius of the object's rotation. The relationship between linear speed \( v \) and angular speed \( \omega \) is given by the formula:\[ v = r\omega \]where \( r \) represents the radius of the circular path.
To make sure you're using these formulas correctly:
  • Remember that radius is half the diameter of the circular path. In this exercise, the diameter is 1.20 m, making the radius 0.60 m.
  • Multiply the given angular speed (in rad/s) by the radius to determine the linear speed.
Understanding this relationship is key to calculating how fast a point on the perimeter of a spinning object, like a flywheel, is moving linearly.
Angular Acceleration
Angular acceleration is an indicator of how quickly an object changes its angular speed over time. It's crucial in understanding how rotational speed increases or decreases. In the exercise, the constant angular acceleration is needed to increase the wheel's angular speed to a higher level.
Here's a practical way to find angular acceleration:
  • Start by identifying the change in angular speed, \( \Delta \omega \), which is the final speed minus the initial speed.
  • Next, divide this change by the time span over which the speed is increasing. This gives the angular acceleration expressed in rev/min虏.
This calculation helps you determine how swiftly an object like a wheel ramps up its rotation speed, which is important in engineering and dynamics analyses.
Angular Displacement
Angular displacement explains how far an object rotates during a given time period, usually expressed in terms of revolutions. It's especially useful for finding out how many spins or turns an object completes in a set time while subjected to angular acceleration.
The core equation for calculating angular displacement \( \theta \) is:\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]In this equation:
  • \( \omega_0 \): initial angular speed given in rev/min.
  • \( t \): time duration of interest, in the exercise it's 60 seconds or 1 minute.
  • \( \alpha \): angular acceleration derived previously.
Apply these values into the equation to calculate the total number of revolutions made. This concept is significant for visualizing rotational movement in mechanical parts and systems.

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Most popular questions from this chapter

A disk, with a radius of \(0.25 \mathrm{~m}\), is to be rotated like a merry-go- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first \(400 \mathrm{rad}\) and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

(a) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} . $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of \(33 \frac{1}{3}\) rev/min, the groove being played is at a radius of \(10.0 \mathrm{~cm}\), and the bumps in the groove are uniformly separated by \(1.75 \mathrm{~mm}\). At what rate (hits per second) do the bumps hit the stylus?

Between 1911 and 1990 , the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of \(1.2 \mathrm{~mm} / \mathrm{y}\). The tower is \(55 \mathrm{~m}\) tall. In radians per second, what is the average angular speed of the tower's top about its base?

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s} ?(\mathrm{~b})\) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?
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