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Chapter 9: Problem 31

Consult Multiple-Concept Example 10 to review an approach to problems such as this. A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angular velocity of 21 rad/s in 0.80 s. Assuming the CD is a uniform solid disk, determine the net torque acting on it.

Short Answer

Expert verified
The net torque acting on the CD is approximately \(8.03 \times 10^{-4} \text{ N} \cdot \text{m}\).

Step by step solution

01

Convert Units

First, convert the mass and radius of the CD to standard SI units. The mass is given in grams, so we convert it from 17 g to kilograms:\[17 \text{ g} = 0.017 \text{ kg}\]The radius is given in centimeters, so we convert it from 6.0 cm to meters:\[6.0 \text{ cm} = 0.06 \text{ m}\]
02

Calculate the Moment of Inertia

The moment of inertia \( I \) for a uniform solid disk is given by:\[I = \frac{1}{2} m r^2\]Substitute the values of mass \( m = 0.017 \text{ kg} \) and radius \( r = 0.06 \text{ m} \):\[I = \frac{1}{2} (0.017) (0.06)^2 = 3.06 \times 10^{-5} \text{ kg} \cdot \text{m}^2\]
03

Determine Angular Acceleration

The angular acceleration \( \alpha \) can be found using the formula:\[\alpha = \frac{\Delta \omega}{\Delta t}\]where \( \Delta \omega = 21 \text{ rad/s} - 0 \text{ rad/s} = 21 \text{ rad/s} \) and \( \Delta t = 0.80 \text{ s} \):\[\alpha = \frac{21 \text{ rad/s}}{0.80 \text{ s}} = 26.25 \text{ rad/s}^2\]
04

Calculate the Net Torque

Use the relation between torque \( \tau \), moment of inertia \( I \), and angular acceleration \( \alpha \):\[\tau = I \alpha\]Substitute the values for \( I = 3.06 \times 10^{-5} \text{ kg} \cdot \text{m}^2 \) and \( \alpha = 26.25 \text{ rad/s}^2 \):\[\tau = (3.06 \times 10^{-5}) (26.25) = 8.03 \times 10^{-4} \text{ N} \cdot \text{m}\]

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Key Concepts

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

Uniform Solid Disk
In physics, a uniform solid disk is a common model used to represent objects with rotational symmetry and uniform mass distribution. A CD, as mentioned in the exercise, can be considered a uniform solid disk due to its consistent thickness and homogeneous material. This assumption simplifies calculations when exploring rotational dynamics.

In the context of this exercise:
  • The mass of the CD is evenly distributed across its area.
  • This makes it possible to use specific formulas that are designed for uniform solid disks, like the moment of inertia formula.
Understanding the nature of the object helps in estimating factors such as angular motion and resultant torques during rotational motion.
Angular Acceleration
Angular acceleration (\(\alpha\)) refers to the rate at which an object's angular velocity changes over time. It plays a crucial role in rotational motion, similar to linear acceleration in translational motion. In this exercise, the CD starts from rest and reaches a specific angular velocity, allowing us to calculate its angular acceleration.

Angular acceleration can be calculated using the formula:
  • \[\alpha = \frac{\Delta \omega}{\Delta t}\]
Where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time taken for this change. Here, the CD's angular velocity changes from 0 rad/s to 21 rad/s in 0.80 seconds. By substituting these values, you find the angular acceleration to be 26.25 rad/s虏.

Angular acceleration is essential for understanding the dynamics of rotating bodies, as it directly influences the net torque and energy aspects in rotational systems.
Moment of Inertia
The moment of inertia (\(I\)) is a measure of an object's resistance to changes in its rotational motion. It depends on both the mass of the object and the distribution of that mass relative to the axis of rotation. For a uniform solid disk such as the CD in question, the moment of inertia can be calculated using the formula:
  • \[I = \frac{1}{2} m r^2\]
Where \(m\) is the mass, and \(r\) is the radius of the disk. For the CD, with a mass of 0.017 kg and a radius of 0.06 m, the moment of inertia is computed to be \(3.06 \times 10^{-5}\, \text{kg} \cdot \text{m}^2\).

Understanding the moment of inertia is critical, as it affects the angular acceleration and torque. With a higher moment of inertia, the object requires more torque to achieve the same angular acceleration compared to an object with a lower moment of inertia. Thus, it acts as a rotational analogue to mass in linear motion and is fundamental in solving problems related to rotational dynamics.

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Most popular questions from this chapter

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300-mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of 0.30 \(\mathrm{m}\) have to rotate to store this much energy? Give your answer in rev/min.

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

A solid sphere is rolling on a surface. What fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass?

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, M is the total mass of the object, and h is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius R relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A square, 0.40 m on a side, is mounted so that it can rotate about an axis that passes through the center of the square. The axis is perpendicular to the plane of the square. A force of 15 N lies in this plane and is applied to the square. What is the magnitude of the maximum torque that such a force could produce?
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