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A spherical asteroid with radius \(r = 123 m\) and mass \(M = 2.25 \times 10^{10} kg\) rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the asteroid's south pole (as defined by the axis of rotation) and fires its engine, applying a force \(F\) tangentially to the asteroid's surface as shown in Fig. 8-65. If \(F = 285 N\), how long will it take the tug to rotate the asteroid's axis of rotation through an angle of 5.0\(^{\circ}\) by this method?

Short Answer

Expert verified
It will take approximately 5.27 hours.

Step by step solution

01

Calculate the Asteroid's Moment of Inertia

The moment of inertia for a solid sphere rotating about an axis through its center is given by \( I = \frac{2}{5} M r^2 \). Given \( M = 2.25 \times 10^{10} \) kg and \( r = 123 \) m, substitute these values to find \( I \):\[I = \frac{2}{5} \times 2.25 \times 10^{10} \times (123)^2.\]
02

Calculate Initial Angular Velocity

The asteroid completes 4 revolutions per day. First, convert this to radians per second. One revolution is \( 2\pi \) radians.\[\omega_i = \frac{4 \times 2\pi}{24 \times 3600} \, \text{rad/s}.\]
03

Calculate Torque Applied by Force

Torque due to the force \( F \) is given by \( \tau = F \cdot r \). Given \( F = 285 \) N and \( r = 123 \) m, compute the torque:\[\tau = 285 \times 123.\]
04

Calculate the Angular Acceleration

The angular acceleration \( \alpha \) can be calculated using the equation \( \tau = I \cdot \alpha \). Solve for \( \alpha \):\[\alpha = \frac{\tau}{I}.\]
05

Calculate the Angle in Radians

The desired angle of rotation is 5.0 degrees. Convert this angle to radians:\[\theta = \frac{5.0 \times \pi}{180} \text{ radians}.\]
06

Calculate Time to Rotate the Asteroid

Using the rotational kinematic equation \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \), solve for time \( t \). This is a quadratic equation in \( t \), which can be simplified and solved:\[0 = \frac{1}{2} \alpha t^2 + \omega_i t - \theta.\]Using the quadratic formula, solve for the positive value of \( t \):\[t = \frac{-\omega_i + \sqrt{\omega_i^2 + 2\alpha\theta}}{\alpha}.\]
07

Final Calculation

Substitute the values from Steps 2, 3, 4, and 5 into the equation derived in Step 6, and compute the time \( t \) it takes to rotate the asteroid's axis by 5.0 degrees.

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Key Concepts

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

Moment of Inertia
Moment of inertia is a fundamental concept in rotational motion. It is essentially the rotational equivalent of mass in linear motion. It describes how difficult it is to change an object's rotational speed. The larger the moment of inertia, the harder it is to rotate the object.

For different shapes, the formula for moment of inertia, denoted as \( I \), varies. For a solid sphere like the asteroid in the problem, rotating about an axis through its center, the formula is:
  • \( I = \frac{2}{5} M r^2 \)
where:
  • \( M \) is the mass of the object in kilograms
  • \( r \) is the radius in meters
This formula is derived from integrating all the infinitesimal mass elements over the volume of the sphere, considering how far each element is from the axis of rotation.Knowing the moment of inertia is crucial for understanding how much torque is needed to achieve a desired angular acceleration, setting the stage for calculating the angular velocity, and ultimately solving the given problem.
Angular Velocity
Angular velocity is the rate at which an object rotates or revolves. It tells us how fast the rotational motion is happening. The concept is akin to linear velocity in translational motion but in a rotational sense.

Measured in radians per second (rad/s), angular velocity is typically denoted by \( \omega \). In the problem, the asteroid rotates four times a day, so we first need to convert this to a standard unit of radians per second:
  • One complete revolution equals \( 2\pi \) radians.
  • With four revolutions per day, the angular velocity \( \omega_i \) is calculated as:\[ \omega_i = \frac{4 \times 2\pi}{24 \times 3600} \] rad/s.
This conversion is vital because rotational kinematic equations and other calculations typically require angular velocity in rad/s.Understanding angular velocity is key to solving problems involving rotational motion, helping determine how long and how much force is needed to cause a change in the rotational state.
Torque
Torque is the rotational equivalent of force. It measures how much a force acting on an object causes it to rotate. When a force is applied tangentially, like in the case of the astronaut's "tug" on the asteroid, it creates torque.

The formula for torque, symbolized by \( \tau \), is given by:
  • \( \tau = F \cdot r \)
where:
  • \( F \) is the force applied in Newtons
  • \( r \) is the distance from the axis of rotation, or the radius
In this example, a force of 285 N applied at a radius of 123 m results in:
  • \( \tau = 285 \times 123 \)
This calculated torque is then used to find the angular acceleration using the formula \( \tau = I \cdot \alpha \).By understanding torque, we can better comprehend how rotational motion is controlled and altered, making it possible to predict the rotation of objects under various forces.

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Most popular questions from this chapter

(I) Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.0 min to turn through 23 complete revolutions before reaching its final speed. \((a)\) What was its angular acceleration (assumed constant), and \((b)\) what was its final angular speed in rpm?

(II) A sphere of radius \(r = 34.5 cm\) and mass \(m = 1.80 kg\) starts from rest and rolls without slipping down a 30.0\(^{\circ}\) incline that is 10.0 m long. \((a)\) Calculate its translational and rotational speeds when it reaches the bottom. \((b)\) What is the ratio of translational to rotational kinetic energy at the bottom? Avoid putting in numbers until the end so you can answer: \((c)\) do your answers in \((a)\) and \((b)\) depend on the radius of the sphere or its mass?

(II) A potter is shaping a bowl on a potter's wheel rotating at constant angular velocity of 1.6 rev/s (Fig. 8-48). The friction force between her hands and the clay is 1.5 N total. \((a)\) How large is her torque on the wheel, if the diameter of the bowl is 9.0 cm? \((b)\) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hands? The moment of inertia of the wheel and the bowl is \(0.11 kg\cdot m^2\).

A small mass \(m\) attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table (Fig. 8-62). Initially, the mass revolves with a speed \(\upsilon_1 = 2.4 m/s\) in a circle of radius \(r_1 = 0.80 m\). The string is then pulled slowly through the hole so that the radius is reduced to \(r_2 = 0.48 m\). What is the speed, \(\upsilon_2\) , of the mass now?

The Moon orbits the Earth such that the same side always faces the Earth. Determine the ratio of the Moon's spin angular momentum (about its own axis) to its orbital angular momentum. (In the latter case, treat the Moon as a particle orbiting the Earth.)

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