/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Communications \(\quad\) Satelli... [FREE SOLUTION] | 91Ó°ÊÓ

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Communications \(\quad\) Satellite \(A\) communications satellite is a solid cylinder with mass \(1210 \mathrm{~kg}\), diameter \(1.21 \mathrm{~m}\), and length \(1.75 \mathrm{~m}\). Prior to launching from the shuttle cargo bay, it is set spinning at \(1.52 \mathrm{rev} / \mathrm{s}\) about the cylinder axis (Fig. \(11-27)\). Calculate the satellite's (a) rotational inertia about the rotation axis and (b) rotational kinetic energy.

Short Answer

Expert verified
(a) Rotational inertia is 221.45 kg·m². (b) Rotational kinetic energy is 10096.6 J.

Step by step solution

01

Determine the Moment of Inertia

The moment of inertia for a solid cylinder rotating about its central axis is given by the formula: \[ I = \frac{1}{2} m r^2 \] where \( m \) is the mass of the cylinder and \( r \) is the radius. Given: \[ m = 1210 \, \text{kg}, \, \text{diameter} = 1.21 \, \text{m} \] Therefore, the radius is: \[ r = \frac{1.21}{2} = 0.605 \, \text{m} \] Plug in the values: \[ I = \frac{1}{2} \times 1210 \, \text{kg} \times (0.605 \, \text{m})^2 \] Calculate the moment of inertia.
02

Evaluate the formula for the Moment of Inertia

Substitute the values and compute: \[ I = \frac{1}{2} \times 1210 \times (0.605)^2 \] First, compute the square of the radius: \[ (0.605)^2 = 0.366025 \] Now, multiply to find the moment of inertia: \[ I = \frac{1}{2} \times 1210 \times 0.366025 = 221.45 \, \text{kg} \cdot \text{m}^2 \]
03

Find the Rotational Kinetic Energy

The formula for rotational kinetic energy is: \[ K = \frac{1}{2} I \omega^2 \] where \( \omega \) is the angular velocity in radians per second. First, convert the given angular velocity from revolutions per second to radians per second. Given: \( 1.52 \, \text{rev/s} \) and \( 1 \, \text{rev} = 2 \pi \, \text{radians} \) \[ \omega = 1.52 \times 2 \pi = 1.52 \times 6.2832 = 9.5483 \, \text{rad/s} \] Substitute the values of \( I \) and \( \omega \) into the kinetic energy formula: \[ K = \frac{1}{2} \times 221.45 \times (9.5483)^2 \] Calculate the rotational kinetic energy.
04

Evaluate the Kinetic Energy Formula

Substitute and compute: \[ K = \frac{1}{2} \times 221.45 \times (9.5483)^2 \] First, compute the square of the angular velocity: \[ (9.5483)^2 = 91.15 \] Now multiply to find the kinetic energy: \[ K = \frac{1}{2} \times 221.45 \times 91.15 = 10096.6 \, \text{J} \]

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Key Concepts

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

moment of inertia
The moment of inertia, often symbolized as \( I \), denotes an object's resistance to rotational motion. For a solid cylinder rotating about its central axis, the moment of inertia is calculated using the formula: \( I = \frac{1}{2} m r^2 \). This formula considers the cylinder’s mass \( m \) and radius \( r \). For our satellite example, given that the mass is 1210 kg and the diameter is 1.21 meters, the radius is therefore half of the diameter, which is 0.605 meters. Plugging these values into the formula, we obtain: \( I = \frac{1}{2} \times 1210 \text{kg} \times (0.605 \text{m})^2 \) Calculating inside the parentheses first: \( (0.605 \text{m})^2 = 0.366025 \text{m}^2 \) Now multiply by the other values: \( I = \frac{1}{2} \times 1210 \times 0.366025 = 221.45 \text{kg} \cdot \text{m}^2 \) This tells us the amount of resistance the satellite has to changes in its rotational motion.
rotational kinetic energy
Rotational kinetic energy is the energy an object possesses due to its rotation. It’s given by the formula: \[ K = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. To solve for the rotational kinetic energy of our communication satellite, we first need to convert its rotational speed from revolutions per second to radians per second. Given the rotation rate of 1.52 revolutions per second: \(1\text{ rev} = 2 \pi \text{ radians} \), thus \( \omega = 1.52 \times 2 \pi = 1.52 \times 6.2832 = 9.5483 \text{ rad/s} \) Now, plug \( I \) and \( \omega \) into the energy formula: \( K = \frac{1}{2} \times 221.45 \times (9.5483)^2 \)First, calculate the square of \( \omega \): \( (9.5483)^2 = 91.15 \text{rad}^2/\text{s}^2 \) Then multiply through: \( K = \frac{1}{2} \times 221.45 \times 91.15 = 10096.6 \text{ J} \) Thus, the satellite's rotational kinetic energy is approximately 10096.6 Joules.
solid cylinder physics
Understanding the physics behind a solid cylinder is crucial when dealing with various physical scenarios, such as our satellite. Here’s a brief breakdown: **Properties** - A solid cylinder has uniform mass distribution throughout its volume. - Its rotational inertia is a measure of how the mass is distributed in relation to its axis of rotation. **Key Equations** - Moment of inertia: \( I = \frac{1}{2} m r^2 \) - Rotational kinetic energy: \( K = \frac{1}{2} I \omega^2 \)For our scenario, given the mass, radius, and rotational speed of the cylinder, we used these key concepts to determine both the moment of inertia and the rotational kinetic energy. This exemplifies how fundamental principles in physics can directly apply to real-world objects like communication satellites. In nutshell, understanding these concepts not only helps in calculating values effectively but also in comprehending the underlying principles and their broader applications in physics.

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Most popular questions from this chapter

Starting from Rest A wheel, starting from rest, rotates with a constant rotational acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2} .\) During a certain \(3.00 \mathrm{~s}\) interval, it turns through \(90.0\) rad. (a) How long is the wheel turning before the start of the \(3.00 \mathrm{~s}\) interval? (b) What is the rotational velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval?

Tall Cylinder-Shaped Chimney A tall, cylinder-shaped chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length \(H\), and let \(\theta\) be the angle the chimney makes with the vertical. In terms of these symbols and \(g\), express the following: (a) the rotational speed of the chimney, (b) the radial acceleration of the chimney's top, and (c) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque. In part (c) recall that \(\alpha=d \omega / d t .)(\mathrm{d})\) At what angle \(\theta\) does the tangential acceleration equal \(g\) ?

Flywheel Rotating A flywheel with a diameter of \(1.20 \mathrm{~m}\) has a rotational speed of 200 rev/min. (a) What is the rotational speed of the flywheel in radians per second? (b) What is the translational speed of a point on the rim of the flywheel? (c) What constant rotational acceleration (in revolutions per minute-squared) will increase the wheel's rotational speed to 1000 rev/min in 60 s? (d) How many revolutions does the wheel make during that \(60 \mathrm{~s}\) ?

Turntable A record turntable rotating at \(33 \frac{1}{3}\) rev/min slows down and stops in \(30 \mathrm{~s}\) after the motor is turned off. (a) Find its (constant) rotational acceleration in revolutions per minutesquared. (b) How many revolutions does it make in this time?

Thin Spherical Shell A thin spherical shell has a radius of \(1.90 \mathrm{~m}\). An applied torque of \(960 \mathrm{~N} \cdot \mathrm{m}\) gives the shell a rotational acceleration of \(6.20 \mathrm{rad} / \mathrm{s}^{2}\) about an axis through the center of the shell. What are (a) the rotational inertia of the shell about that axis and (b) the mass of the shell?

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