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Chapter 10: Problem 30

A gyroscope flywheel of radius \(2.83 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is \(2760 \mathrm{rev} / \mathrm{min} .\) (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

Short Answer

Expert verified
(a) 0.40 m/s虏, (b) 2364.94 m/s虏, (c) 83.12 m.

Step by step solution

01

Convert Angular Speed to Standard Units

First, we need to convert the final angular speed from revolutions per minute to radians per second. Since there are \(2\pi\) radians in one revolution, and 60 seconds in a minute, the conversion is:\[\omega = 2760 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 289 \text{ rad/s}\]
02

Calculate Tangential Acceleration

The tangential acceleration \(a_t\) is found using the relation between linear acceleration and angular acceleration:\[a_t = r \alpha\]Where \(r = 0.0283 \text{ m}\) (converted from \(2.83 \text{ cm}\)) and \(\alpha = 14.2 \text{ rad/s}^{2}\):\[a_t = 0.0283 \text{ m} \times 14.2 \text{ rad/s}^{2} = 0.40226 \text{ m/s}^{2}\]
03

Calculate Radial (Centripetal) Acceleration

The radial or centripetal acceleration \(a_r\) is given by the formula:\[a_r = \omega^2 r\]Substituting \(\omega = 289 \text{ rad/s}\) and \(r = 0.0283 \text{ m}\):\[a_r = (289 \text{ rad/s})^2 \times 0.0283 \text{ m} = 2364.94 \text{ m/s}^{2}\]
04

Calculate Distance Traveled During Spin-Up

The distance \(s\) traveled by a point on the rim is given by the angular distance \(\theta\) converted to linear distance, \(s = r\theta\), where \(\theta = \frac{\omega^2 - \omega_0^2}{2\alpha}\). Since the initial angular speed \(\omega_0 = 0\),\[\theta = \frac{289^2}{2 \times 14.2} = 2939.92 \text{ rad}\]Thus, the linear distance is:\[s = 0.0283 \text{ m} \times 2939.92 \text{ rad} = 83.12 \text{ m}\]
05

Final Answers to Parts a, b, and c

Combining all the calculations:(a) The tangential acceleration of a point on the rim is approximately \(0.40 \text{ m/s}^{2}\).(b) The radial acceleration at full speed is approximately \(2364.94 \text{ m/s}^{2}\).(c) The distance through which a point on the rim moves during spin-up is approximately \(83.12 \text{ m}\).

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Key Concepts

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

Tangential Acceleration
Imagine you're riding on the edge of a spinning wheel. How quickly you're picking up speed as the wheel spins is your tangential acceleration. Tangential acceleration is the change in your speed along the circle per unit of time.

It's calculated by multiplying the radius of the wheel by the angular acceleration, which tells you how fast the wheel is speeding up its spin. The formula you'll use is:
  • \( a_t = r \alpha \)
Where:
  • \(a_t\) is the tangential acceleration,
  • \(r\) is the radius of the circle,
  • \(\alpha\) is the angular acceleration.
To use this formula correctly, remember to convert any measurements to standard units like meters and radians per second squared. This ensures you're calculating the tangential acceleration right and understanding how quickly things speed up along the edge as they start to move.
Radial Acceleration
When you're moving in a circle, your body feels a tug towards the center. This push keeps you moving in a curved path rather than a straight line. It's known as radial or centripetal acceleration.

The formula for radial acceleration brings together how fast the circle is spinning (angular speed), and how far you are from the center (the radius):
  • \( a_r = \omega^2 r \)
Here:
  • \(a_r\) represents radial acceleration,
  • \(\omega\) is the angular speed of the object,
  • \(r\) defines the radius of the circle from its center to where you are.
This lets you find out how strongly you're getting pulled inward, which is crucial for understanding how an object sticks to a circular path. The faster you go or the smaller the circle's radius, the stronger that inward pull is.
Angular Speed Conversion
Sometimes, rotational problems require you to switch between units, like going from revolutions per minute to radians per second. This conversion is crucial for solving rotational motion problems accurately.

To convert angular speed from one unit to another, you can use the relationship
  • 1 revolution = \(2\pi\) radians and
  • 1 minute = 60 seconds.
Therefore, the formula to convert revolutions per minute (rpm) to radians per second is:
  • \( \omega = \text{rpm} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \)
This conversion ensures that everything lines up correctly with other calculations in the physics of rotation, keeping the math consistent and your equations accurate. Remember, working in standardized units like radians and seconds helps maintain a clear path to the solution of rotational problems.

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Most popular questions from this chapter

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 s? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s} ?\)

Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius \(3220 \mathrm{~km}\) at a speed of \(29000 \mathrm{~km} / \mathrm{h} ?\)

During the launch from a board, a diver's angular speed about her center of mass changes from zero to \(6.20 \mathrm{rad} / \mathrm{s}\) in \(220 \mathrm{~ms}\). Her rotational inertia about her center of mass is \(12.0 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?

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 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 \(\alpha_{1} ?\)
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