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Chapter 11: Problem 22

Flywheel Rotating A flywheel with a diameter of \(1.20 \mathrm{~m}\) has a rotational speed of 200 rev/min. (a) What is the rotational speed of the flywheel in radians per second? (b) What is the translational speed of a point on the rim of the flywheel? (c) What constant rotational acceleration (in revolutions per minute-squared) will increase the wheel's rotational speed to 1000 rev/min in 60 s? (d) How many revolutions does the wheel make during that \(60 \mathrm{~s}\) ?

Short Answer

Expert verified
a) 20.94 rad/s b) 12.56 m/s c) 13.33 rev/min^2 d) 6,000 rev

Step by step solution

01

Convert Rotational Speed to Radians per Second

First, we need to convert the given rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). There are 2蟺 radians in one revolution and 60 seconds in one minute.So, oRotational speed in rad/s = \(\omega = 200 \text{ rev/min} \times \frac{2\pi \text{ radians}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}\).Calculate the value: \(\omega = \frac{200 \times 2\pi}{60} = \frac{400\pi}{60} = \frac{40\pi}{6} \cong 20.94 \text{ rad/s} \).
02

Find the Translational Speed

The translational speed of a point on the rim of the flywheel can be determined by the formula: \(v = r \omega \), where \( r \) is the radius of the flywheel and \( \omega \) is the angular velocity. Given the diameter of the flywheel is 1.20 meters, the radius (\( r \)) is \( 0.60 \text{ m}\). Substitute the values into the equation: \(v = 0.60 \text{ m} \times 20.94 \text{ rad/s} \cong 12.56 \text{ m/s} \).
03

Calculate the Constant Rotational Acceleration

To find the constant rotational acceleration, we need the initial and final speeds and the time taken to change between these speeds. Given the initial speed is 200 rev/min, the final speed is 1000 rev/min, and the time taken is 60 s. Convert these speeds to revolutions per minute-squared. The formula for angular acceleration is: \( \alpha = \frac{\Delta \omega}{\Delta t}\) First, calculate the change in rotational speed: \( \Delta \omega = 1000 \text{ rev/min} - 200 \text{ rev/min} = 800 \text{ rev/min}\) Now, divide this change by the time to get the acceleration: \( \alpha = \frac{800\text{ rev/min}}{60 \text{ s}}\equiv 13.33 \text{ rev/min}^2\)
04

Determine the Number of Revolutions in 60 s

To determine the total number of revolutions the wheel makes in 60 seconds, use the kinematic equation for rotational motion: \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2\) Here, \( \theta \) is the total number of revolutions, \( \omega_0 \) is the initial rotational speed, \( \alpha \) is the angular acceleration, and \( t \) is the time. Convert \( \omega_0 \) to revolutions per second: \( \omega_0 = 200 \text{ rev/min} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{200}{60} \text{ rev/s} \equiv 3.33 \text{ rev/s}\) And \( \alpha \) remains 13.33 rev/min^2. Given: \(\theta = 3.33 \text{ rev/s} \times 60 \text{ s} + \frac{1}{2} \times 13.33 \text{ rev/min}^2 \times 60 \text{ s} \equiv 200 + 6,000)\equiv 6000 \text{ rev}\)

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Key Concepts

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

Rotational Speed
Rotational speed is how fast an object rotates or revolves relative to another point, often the center of rotation. It's measured in revolutions per minute (rev/min) or radians per second (rad/s). Imagine a flywheel spinning. If it completes 200 revolutions in one minute, its rotational speed is 200 rev/min. Think of it like the speedometer of a car, but instead of miles per hour, we measure in rotations per minute.
Radians per Second
Radians per second (rad/s) is a unit that measures angular velocity. To convert from rev/min to rad/s:
  • Know that 1 revolution is equal to \(2\pi\) radians.
  • There are 60 seconds in a minute.
So, for a flywheel spinning at 200 rev/min: \[\text{Rotational speed in rad/s} = 200 \times \frac{2 \pi}{60}\to\frac{400\pi}{60} \approx 20.94 \text{ rad/s}\ruby\]Radians per second gives us a more precise idea of the rotational speed in terms of angles.
Translational Speed
Translational speed refers to the linear speed of a point on the rim of the rotating object. It's related to the rotational speed through the formula: \(v = r \omega\). Here:\r
  • \r v is the translational speed,
  • r is the radius of the flywheel,
  • \omega is the angular velocity in rad/s.
For the flywheel (where diameter = 1.20 m, so radius r = 0.60 m), and with \omega \approx 20.94 \text{ rad/s},\r \[\text{v} = 0.60 \times 20.94 \approx 12.56 \text{ m/s}\r.\]This means that a point on the edge of the flywheel moves at a speed of 12.56 meters per second.
Rotational Acceleration
Rotational acceleration is how quickly the rotational speed changes over time. It's similar to how a car accelerates in speed. For rotational motion, the formula is: \( \alpha = \frac{\Delta \omega}{\Delta t}\),
where:
\r
    \r
  • \alpha is the rotational acceleration,
    • \r
  • \Delta \omega is the change in rotational speed,
    • \r
  • \Delta t is the change in time.
\rFor our flywheel:
\r
    \r
  • initial speed \omega_i = 200 rev/min,
    • \r
  • final speed \omega_f = 1000 rev/min,
    • \r
  • time \Delta t = 60 s.
Change in speed \Delta \omega = 800 rev/min.
Therefore,
\r \alpha = \frac{800 \text{ rev/min}}{60 \text{ s}}\approx 13.33 \text{ rev/min}^2.
This means the rotation speed increases by 13.33 rev/min every second.
Revolutions per Minute-squared
Revolutions per minute-squared (rev/min^2) is a unit of rotational acceleration. It tells how the rotational speed changes over time. For example, in our exercise:
\r
    \r
  • \r The flywheel starts at 200 rev/min.
    • \r
  • \r It accelerates to 1000 rev/min.
    • \r
  • \rhover 60 seconds.
This acceleration \approx 13.33 rev/min^2.
To find the total number of revolutions during this period, use \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2\).
\r
    \r \r
  • \theta is the total revolutions,
    • \r \r
  • \omega_0 = initial speed = 200 rev/min = \approx 3.33 rev/s,
    • \r \r
  • \alpha = 13.33 rev/min^2,
    • \r\r
  • t = 60 s.
    • \r
So:
\r\theta = 3.33 \times 60 + \frac{1}{2} \times 13.33 \times 60^2\r\approx 6000 \text{ rev}.\rThis means the flywheel makes about 6000 revolutions in 60 seconds.

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Most popular questions from this chapter

A Disk A disk, initially rotating at \(120 \mathrm{rad} / \mathrm{s}\), is slowed down with a constant rotational 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?

Bicycle Pedal Arm The length of a bicycle pedal arm is \(0.152 \mathrm{~m}\), and a downward force of \(111 \mathrm{~N}\) is applied to the pedal by the rider's foot. What is the magnitude of the torque about the pedal arm's pivot point when the arm makes an angle of (a) \(30^{\circ}\), (b) \(90^{\circ}\). and (c) \(180^{\circ}\) with the vertical?

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Rotational Vs. Translational Energy of Motion (a) Describe how a solid ball can move so that i. Its total kinetic energy is just the energy of motion of its center of mass ii. Its total kinetic energy is the energy of its motion relative to its center of mass (b) Two bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just slides down the alley, while the other rolls down the alley. Which ball has more energy? Explain your reasoning.
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