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Chapter 11: Problem 68

A spherical asteroid with radius \(r=123 \mathrm{m}\) and mass \(M=2.25 \times 10^{10} \mathrm{kg}\) rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the pole (as defined by the axis of rotation) and fires its engine, applying a force \(F \quad\) tangentially to the asteroid's surface as shown in Fig. \(44 .\) If \(F=265 \mathrm{N},\) how long will it take the tug to rotate the asteroid's axis of rotation through an angle of \(10.0^{a}\) by this method?

Short Answer

Expert verified
It takes approximately 10.61 hours for the tug to rotate the asteroid by 10 degrees.

Step by step solution

01

Calculate the Initial Angular Velocity

The asteroid rotates four times per day. First, convert revolutions per day to radians per second.A full revolution is equal to \(2\pi\) radians. Thus, \(4\) revolutions per day means \(4 \times 2\pi \) radians per day. Conversion from days to seconds (1 day = 86400 seconds):\[\omega_i = \frac{4 \times 2\pi}{86400}\, \text{radians/second} \approx 2.91 \times 10^{-4} \,\text{rad/s}\]
02

Calculate the Moment of Inertia

For a sphere, the moment of inertia \(I\) is given by the formula:\[I = \frac{2}{5} Mr^2 \]Substituting the given mass \(M = 2.25 \times 10^{10} \,\text{kg}\) and radius \(r = 123 \,\text{m}\):\[I = \frac{2}{5} (2.25 \times 10^{10}) (123)^2 = 1.36 \times 10^{14} \,\text{kg} \, \text{m}^2\]
03

Calculate the Torque Applied by the Tug

Torque \(\tau\) is given by \(\tau = rF\) where \(r\) is the radius and \(F\) is the force.\[\tau = 123 \,\text{m} \times 265 \,\text{N} = 32595 \,\text{N} \cdot \text{m}\]
04

Determine the Angular Acceleration

Angular acceleration \(\alpha\) is related to torque \(\tau\) and moment of inertia \(I\) by \(\alpha = \frac{\tau}{I}\).\[\alpha = \frac{32595}{1.36 \times 10^{14}} \approx 2.40 \times 10^{-10} \,\text{rad/s}^2\]
05

Calculate the Time to Rotate the Axis by 10 Degrees

Convert \(10.0^\circ\) to radians: \[\theta = 10.0 \times \left(\frac{\pi}{180}\right) \approx 0.175 \,\text{rad}\]The rotational equation is \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\).Assuming the initial rotation contributes minimally over \(t\) (due to small \(\omega_i\)), solve approximately for \(t:\)\[0.175 = \frac{1}{2} \times 2.40 \times 10^{-10} \times t^2\]\[t^2 \approx \frac{0.175 \times 2}{2.40 \times 10^{-10}}\]\[t^2 \approx 1.46\times 10^{9}\]\[t \approx \sqrt{1.46 \times 10^{9}} \approx 38219 \,\text{s} \approx 10.61 \,\text{hours}\]

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Key Concepts

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

Torque
Torque is a measure of the rotational force applied to an object. It is pivotal in rotating bodies like asteroids and planets. Mathematically, torque \( \tau \) is defined as the product of the force applied and the distance from the axis of rotation to where the force is applied, or \( \tau = rF \). In our exercise, the tug spaceship applies a force tangentially to the asteroid's surface, creating a torque that influences the asteroid's angular motion.
  • Formula: \( \tau = r \times F \)
  • Units: Newton-meters (N路m)
  • Effect: Higher torques cause quicker changes in rotational speed.
When the spaceship exerts a force of 265 N at a distance (radius) of 123 m from the rotation axis, it generates a torque of 32595 N路m. This rotational influence is vital for altering the asteroid's angular momentum, especially when you aim to change its rotational axis or speed. By understanding torque, you can control the rotational behavior of an object effectively.
Moment of Inertia
The moment of inertia \( I \) is likened to the rotational equivalent of mass. It describes an object's resistance to changes in its rotational state. For a spherical object like an asteroid, this property is determined by both its mass and the distribution of that mass around the axis of rotation.
  • Formula for a Sphere: \( I = \frac{2}{5} Mr^2 \)
  • Units: Kilogram meter squared (kg路m虏)
  • Significance: A higher moment of inertia means the object is harder to spin or stop spinning.
In the given problem, substituting the asteroid's mass \( M = 2.25 \times 10^{10} \, \text{kg} \) and its radius \( r = 123 \, \text{m} \) into the formula gives a moment of inertia of \( 1.36 \times 10^{14} \, \text{kg} \, \text{m}^2 \). This tells us how the mass of the asteroid is spread out and how it will respond when the torque is applied. Effectively, the larger the moment of inertia, the more torque is needed to achieve the same amount of rotation.
Angular Acceleration
Angular acceleration \( \alpha \) signifies the rate of change of angular velocity. It is the rotational counterpart of linear acceleration and plays a crucial role when rotational changes are studied. This value depends on the applied torque and the moment of inertia.
  • Formula: \( \alpha = \frac{\tau}{I} \)
  • Units: Radians per second squared (rad/s虏)
  • Function: Determines how quickly an object speeds up or slows down its rotation.
In the exercise, the torque applied by the spaceship is \( 32595 \, \text{N} \cdot \text{m} \), and the moment of inertia of the asteroid is \( 1.36 \times 10^{14} \, \text{kg} \, \text{m}^2 \). The resulting angular acceleration is calculated to be approximately \( 2.40 \times 10^{-10} \, \text{rad/s}^2 \). Although small, this acceleration will gradually alter the asteroid's rotation, allowing it to eventually rotate the desired 10 degrees. This illustrates how even a modest force, over time, can produce significant changes in the rotation of massive objects.

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Most popular questions from this chapter

Determine the angular momentum of the Earth (a) about its rotation axis (assume the Earth is a uniform sphere), and \((b)\) in its orbit around the Sun (treat the Earth as a particle orbiting the Sun). The Earth has mass \(=6.0 \times 10^{24} \mathrm{~kg}\) and radius \(=6.4 \times 10^{6} \mathrm{~m},\) and is \(1.5 \times 10^{8} \mathrm{~km}\) from the Sun.

Show that the kinetic energy \(K\) of a particle of mass \(m\), moving in a circular path, is \(K=L^{2} / 2 I,\) where \(L\) is its angular momentum and \(I\) is its moment of inertia about the center of the circle.

A particle of mass \(m\) uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius \(R\) : $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta+\hat{\mathbf{j}} R \sin \theta $$ with \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2},\) where the constants \(\omega_{0}\) and \(\alpha\) are the initial angular velocity and angular acceleration, respectively. Determine the object's tangential acceleration \(\overrightarrow{\mathbf{a}}_{\tan }\) and determine the torque acting on the object using \((a) \vec{\tau}=\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\) (b) \(\vec{\tau}=I \vec{\alpha}\)

Water drives a waterwheel (or turbine) of radius \(R=3.0 \mathrm{~m}\) as shown in Fig. \(11-47 .\) The water enters at a speed \(v_{1}=7.0 \mathrm{~m} / \mathrm{s}\) and exits from the waterwheel at a speed \(v_{2}=3.8 \mathrm{~m} / \mathrm{s} . \quad(a)\) If \(85 \mathrm{~kg}\) of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? \((b)\) What is the torque the water applies to the waterwheel? \((c)\) If the water causes the waterwheel to make one revolution every \(5.5 \mathrm{~s}\), how much power is delivered to the wheel?

A toy gyroscope consists of a 170 -g disk with a radius of \(5.5 \mathrm{~cm}\) mounted at the center of a thin axle \(21 \mathrm{~cm}\) long (Fig. 11-41). The gyroscope spins at \(45 \mathrm{rev} / \mathrm{s}\). One end of its axle rests on a stand and the other end precesses horizontally about the stand. (a) How long does it take the gyroscope to precess once around? \((b)\) If all the dimensions of the gyroscope were doubled (radius \(=11 \mathrm{~cm},\) axle \(=42 \mathrm{~cm})\) how long would it take to precess once?
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