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Chapter 10: Problem 88

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 an angular 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?

Short Answer

Expert verified
(a) Rotational inertia is approximately \(154.84\ \text{kg}\cdot \text{m}^2\); (b) mass is approximately \(64.48\ \text{kg}\).

Step by step solution

01

Understand the Problem

We need to find the rotational inertia and the mass of a spherical shell given its radius, applied torque, and resulting angular acceleration. The formula for torque \( \tau \) is \( \tau = I \alpha \), where \( I \) is the rotational inertia and \( \alpha \) is the angular acceleration.
02

Find Rotational Inertia (Part a)

To find the rotational inertia \( I \), use the formula \( \tau = I \alpha \). \[ I = \frac{\tau}{\alpha} \] Substituting the given values: \[ I = \frac{960 \ \text{N} \cdot \text{m}}{6.20 \ \text{rad/s}^2} \approx 154.84 \ \text{kg} \cdot \text{m}^2 \].
03

Use Rotational Inertia Formula for Shell

For a thin spherical shell, the rotational inertia \( I \) around its center is \( I = \frac{2}{3} m r^2 \). We already computed \( I \) and know \( r = 1.90 \text{ m} \).
04

Solve for Mass (Part b)

Rearrange the formula \( I = \frac{2}{3} m r^2 \) to solve for mass \( m \): \[ m = \frac{3I}{2r^2} \] Substitute \( I = 154.84 \ \text{kg} \cdot \text{m}^2 \) and \( r = 1.90 \ \text{m} \): \[ m = \frac{3 \times 154.84 \ \text{kg} \cdot \text{m}^2}{2 \times (1.90 \ \text{m})^2} \approx 64.48 \ \text{kg} \].

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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, making it rotate around an axis. It is like the twist you apply when you turn a screwdriver.
Torque is represented by the Greek letter \( \tau \) and is calculated using the formula:
  • \( \tau = I \alpha \)
where \( I \) is the rotational inertia, and \( \alpha \) is the angular acceleration.
In the exercise, an applied torque of 960 N 路 m leads to the shell spinning faster. The ability of the shell to resist changes in its rotational state is influenced by this torque. Thus, torque highlights the relationship between the force applied, the effort to rotate, and the resulting speed of rotation.
Angular Acceleration
Angular acceleration is the rate at which an object's angular velocity changes with time. It tells us how quickly something speeds up or slows down as it spins.
This is similar to how in a car, acceleration describes how quickly you can speed up. Measured in radians per second squared (\( \text{rad/s}^2 \)), angular acceleration is generated by applying torque to an object. In our problem, the angular acceleration is 6.20 \( \text{rad/s}^2 \), which represents how swiftly the spherical shell's rotation speed changes due to the applied torque.
Knowing angular acceleration helps in understanding how effectively a rotating force (torque) impacts the motion of an object over time.
Spherical Shell
A spherical shell is a hollow three-dimensional object where its mass is concentrated at a certain distance from the center, similar to a basketball. It can rotate about an axis through its center.
When talking about rotational inertia, spherical shells have a unique property. Their rotational inertia can be expressed by the formula:
  • \( I = \frac{2}{3} m r^2 \)
This formula indicates how the mass \( m \) and radius \( r \) of the shell determine its resistance to changes in rotational motion.
The spherical shell in our problem has a radius of 1.90 m, which influences how it behaves when the torque is applied, and aids in calculating its mass from the resulting rotational inertia.
Mass Calculation
Mass calculation involves finding the amount of matter within an object based on its rotational properties. For objects like the spherical shell, we can use rotational inertia to uncover its mass.
Given the rotational inertia \( I \) calculated from torque and angular acceleration, the mass \( m \) of a spherical shell is determined by rearranging its rotational inertia equation.
  • The mass formula becomes: \( m = \frac{3I}{2r^2} \)
This allows us to use known values of \( I \) and radius \( r \) to find the shell's mass.
This approach shows how rotational movement principles directly link to finding physical attributes like mass, vital for understanding dynamics in rotational systems such as wheels or planets.

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Most popular questions from this chapter

A golf ball is launched at an angle of \(20^{\circ}\) to the horizontal, with a speed of \(60 \mathrm{~m} / \mathrm{s}\) and a rotation rate of \(90 \mathrm{rad} / \mathrm{s}\). Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

A pulley, with a rotational inertia of \(1.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its axle and a radius of \(10 \mathrm{~cm}\), is acted on by a force applied tangentially at its rim. The force magnitude varies in time as \(F=0.50 t+0.30 t^{2}\), with \(F\) in newtons and \(t\) in seconds. The pulley is initially at rest. At \(t=3.0 \mathrm{~s}\) what are its (a) angular acceleration and (b) angular speed?

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 30 \(\mathrm{kg}\) box as shown in Fig. \(10-56\). 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?

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.

Our Sun is \(2.3 \times 10^{4}\) ly (light-years) from the center of our Milky Way galaxy and is moving in a circle around that center at a speed of \(250 \mathrm{~km} / \mathrm{s}\). (a) How long does it take the Sun to make one revolution about the galactic center? (b) How many revolutions has the Sun completed since it was formed about \(4.5 \times 10^{9}\) years ago?
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