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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: 2.60 m/s虏; Tangential acceleration: 0 m/s虏.

Step by step solution

01

Understanding Angular Velocity

Angular velocity \( \omega \) can be found using the formula \( \omega = \frac{v}{r} \), where \(v\) is the linear velocity, and \(r\) is the radius of the circular path. Convert \(29000 \mathrm{~km/h}\) into \(\mathrm{km/s}\) to get 29000 / 3600 km/s.
02

Calculate Angular Velocity

Substitute the known values into the formula: \( \omega = \frac{29000 / 3600 \text{ km/s}}{3220 \text{ km}} \). Simplify the expression to find \( \omega \approx 0.0025 \text{ rad/s} \).
03

Understanding Radial Acceleration

Radial acceleration \( a_{r} \) is given by \( a_{r} = \frac{v^2}{r} \). Use the linear velocity in \(\text{km/s}\) that we calculated earlier as part of this solution.
04

Calculate Radial Acceleration

Plug in the converted speed and the radius into the formula: \( a_{r} = \frac{(29000 / 3600)^2}{3220} \). Simplify to find \( a_{r} \approx 2.60 \mathrm{~m/s^2} \) (after converting units as needed).
05

Understanding Tangential Acceleration

Tangential acceleration \( a_{t} \) is defined as \( a_{t} = r \cdot \alpha \), where \( \alpha \) is the angular acceleration. Here, no angular acceleration is mentioned, implying \( a_{t} = 0 \).
06

Conclusion

The magnitude of the angular velocity is approximately \( 0.0025 \text{ rad/s} \), the radial acceleration is approximately \( 2.60 \text{ m/s}^2 \), and the tangential acceleration is \( 0 \text{ m/s}^2 \).

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Key Concepts

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

Radial Acceleration
Radial acceleration, also known as centripetal acceleration, keeps an object moving in a circular path. It acts towards the center of the circle. In the case of the spaceship taking a circular turn, radial acceleration ensures it stays on its curved path rather than flying off in a straight line. It can be calculated using the formula:\[a_{r} = \frac{v^2}{r} \]where \(v\) is the velocity of the object and \(r\) is the radius of the circular path.

For our specific example, the spaceship has a velocity of \(29000\) km/h. Before using the formula, we convert this speed into km/s by dividing by \(3600\) (the number of seconds in an hour). Substituting into the formula, we find that the radial acceleration is approximately \(2.60\; \mathrm{m/s^2}\).

Key points to remember:
  • Radial acceleration is responsible for maintaining circular motion.
  • It always points towards the center of the circle.
  • The magnitude depends on both the speed of the object and the radius of the path.
Tangential Acceleration
Tangential acceleration refers to the rate of change of speed along a circular path. Unlike radial acceleration, which changes the direction, tangential acceleration changes the speed of an object moving in a circle. It is calculated using:\[ a_{t} = r \cdot \alpha \]where \(r\) is the radius of the path and \(\alpha\) is the angular acceleration.

In many scenarios involving uniform circular motion, such as our spaceship example, there is no angular acceleration mentioned. This means no change in the speed over time, i.e., the speed is constant, and hence the tangential acceleration is zero.

Important aspects to consider:
  • Tangential acceleration affects how fast or slowly the speed of an object changes.
  • A zero tangential acceleration implies the object's speed is constant during the motion.
Circular Motion
Circular motion is the movement of an object following a circular path. It can be either uniform or non-uniform. Uniform circular motion implies that the object travels at a constant speed through the path, while non-uniform involves changing speeds.

Circular motion involves two main components:
  • Angular velocity, which measures how fast the object spins around the circle.
  • Radial and tangential accelerations, which influence the direction and speed of the motion respectively.
In our spaceship example, it follows a circular turn with a given radius of \(3220\) km at a steady speed of \(29000\) km/h. The calculations confirm it's moving with a small angular velocity, indicating a slow and steady spin. Additionally, there is no tangential acceleration, further reinforcing the notion of a consistent speed.

Key takeaways:
  • In circular motion, speed and direction are controlled by different accelerations.
  • Uniform circular motion is characterized by zero tangential acceleration.
  • Knowledge of angular velocity helps in understanding the rate of rotation.

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Most popular questions from this chapter

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?

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?

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. 10-44). A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm}\), and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at \(10 \mathrm{rev} / \mathrm{s} ; 60\) revolutions later, its angular speed is \(15 \mathrm{rev} / \mathrm{s}\). Calculate (a) the angular acceleration, (b) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second.
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