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Chapter 10: Problem 19

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 \(29000 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
Angular velocity: 0.0025 rad/s; Radial acceleration: 20.16 m/s虏; Tangential acceleration: 0 m/s虏.

Step by step solution

01

Convert Units

First, convert the speed from km/h to meters per second. Given speed is \(29000\, \text{km/h}\). We know, \(1\, \text{km} = 1000\, \text{m}\) and \(1\, \text{h} = 3600\, \text{s}\). Thus, \[ v = 29000 \times \frac{1000}{3600} \approx 8055.56\, \text{m/s} \] The radius is already given in kilometers, convert it to meters: \(3220\, \text{km} = 3220\, 000\, \text{m}\).
02

Calculate Angular Velocity

Use the formula for angular velocity \(\omega\), where \(\omega = \frac{v}{r}\). Substitute the values: \[ \omega = \frac{8055.56}{3220\times 1000} \approx 0.0025\, \text{rad/s} \].
03

Calculate Radial Acceleration

The radial acceleration (also known as centripetal acceleration) \(a_r\) is given by \(a_r = \frac{v^2}{r}\). Substitute the known values: \[ a_r = \frac{(8055.56)^2}{3220\times 1000} \approx 20.16\, \text{m/s}^2 \].
04

Calculate Tangential Acceleration

Since the spaceship is moving at a constant speed around a circular path, its tangential acceleration \(a_t\) is zero because there is no change in the magnitude of the velocity. So, \(a_t = 0\).

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Key Concepts

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

Angular Velocity
Angular velocity refers to the rate at which an object rotates around a central point. Imagine the hands of a clock, swirling at a constant pace around its center, and that's a simple way of grasping angular velocity.
In terms of mathematics, angular velocity \( \omega \) is the measure of the angle an object sweeps per unit time. For a point on a rotating object, it is given by the formula: \( \omega = \frac{v}{r} \). Here, \( v \) is the linear speed of the object, and \( r \) is the radius of the circular path. Think of it as how quickly you're looping around in a circle.
Some important points to remember about angular velocity include:
  • Measured in radians per second ( ad/s").
  • Higher values indicate faster rotation.
  • It is a vector quantity, having both magnitude and direction.
For the spaceship taking a circular turn in our problem, the angular velocity comes out to approximately \( 0.0025 \, ext{rad/s} \). This tells us the spaceship is turning slowly around its path.
Radial Acceleration
Radial acceleration, often known as centripetal acceleration, describes acceleration towards the center of a circular path. It's an essential concept in circular motion as it keeps the object moving in a circle rather than shooting off in a straight line.
Mathematically, radial acceleration \( a_{r} \) can be computed using the formula: \( a_{r} = \frac{v^2}{r} \). Here, \( v \) is the object鈥檚 linear speed, and \( r \) is the radius of the circle. The essence of radial acceleration is the same force you feel when a car takes a sharp turn, pushing you towards the outside of your seat.
Key characteristics of radial acceleration include:
  • Measured in meters per second squared (\
Tangential Acceleration
Tangential acceleration comes into play when an object speeds up or slows down along its circular path. Unlike radial acceleration, which points towards the circle's center, tangential acceleration acts along the path's tangent.
In our exercise, the spaceship maintains a steady speed, indicating that its tangential acceleration \( a_{t} \) is zero. Tangential acceleration is concerned with the change in speed along the circular path, not with turning itself. There's no change in magnitude, therefore, there's no tangential acceleration in this scenario.
Here's what you need to know about tangential acceleration:
  • Measured in meters per second squared (\

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Most popular questions from this chapter

Starting from rest at \(t=0,\) a wheel undergoes a constant angular acceleration. When \(t=2.0 \mathrm{~s},\) the angular velocity of the wheel is \(5.0 \mathrm{rad} / \mathrm{s} .\) The acceleration continues until \(t=20 \mathrm{~s},\) when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=40 \mathrm{~s} ?\)

A uniform helicopter rotor blade is \(7.80 \mathrm{~m}\) long, has a mass of \(110 \mathrm{~kg},\) and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 320 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in \(6.70 \mathrm{~s}\). Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade in order for the blade to reach a speed of 320 rev/min?

If an airplane propeller rotates at 2000 rev \(/\) min while the airplane flies at a speed of \(480 \mathrm{~km} / \mathrm{h}\) relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius \(1.5 \mathrm{~m},\) as seen by (a) the pilot and (b) an observer on the ground? The plane's velocity is parallel to the propeller's axis of rotation.

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a \(30 \mathrm{~kg}\) box as shown in Fig. \(10-59 .\) The outer radius \(R\) of the device is \(0.50 \mathrm{~m},\) and the radius \(r\) of the hub is \(0.20 \mathrm{~m} .\) When a constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(140 \mathrm{~N}\) is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude \(0.80 \mathrm{~m} / \mathrm{s}^{2} .\) What is the rotational inertia of the device about its axis of rotation?

Starting from rest, a wheel has constant \(\alpha=3.0 \mathrm{rad} / \mathrm{s}^{2} .\) During a certain \(4.0 \mathrm{~s}\) interval, it turns through 120 rad. How much time did it take to reach that \(4.0 \mathrm{~s}\) interval?
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