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Chapter 11: Problem 38

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Short Answer

Expert verified
(a) Angular momentum is 0.528 Nm路s, (b) Angular velocity is 440 rad/s.

Step by step solution

01

Identify given values

First, we identify the parameters given in the problem: rotational inertia \( I = 1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2} \), torque \( \tau = 16 \mathrm{~N} \cdot \mathrm{m} \), and time \( t = 33 \mathrm{~ms} = 0.033 \mathrm{~s} \).
02

Determine angular momentum using torque-time relation

Angular momentum \( L \) can be determined from the relation between torque and angular momentum: \( \tau = \frac{\Delta L}{\Delta t} \). Rearranging, we have \( \Delta L = \tau \cdot \Delta t \). Substituting given values, \( \Delta L = 16 \mathrm{~N} \cdot \mathrm{m} \times 0.033 \mathrm{~s} = 0.528 \mathrm{~N} \cdot \mathrm{m} \cdot \mathrm{s} \).
03

Use angular momentum to find angular velocity

The change in angular momentum \( \Delta L \) is also related to the change in angular velocity \( \Delta \omega \) as \( \Delta L = I \cdot \Delta \omega \). Solving for angular velocity, we get \( \Delta \omega = \frac{\Delta L}{I} \). Substitute the values \( \Delta L = 0.528 \mathrm{~N} \cdot \mathrm{m} \cdot \mathrm{s} \) and \( I = 1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2} \). Thus, \( \Delta \omega = \frac{0.528}{1.2 \times 10^{-3}} = 440 \mathrm{~rad/s} \).

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Key Concepts

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

Torque
Torque is a twisting force that can cause an object to rotate around an axis. It is a crucial factor in understanding how objects move in a circular path. Think of torque as a measure of how much a force acting on an object causes that object to rotate.
  • The equation for torque \( \tau \) is \( \tau = F \times r \times \sin(\theta) \), where:
    • \( F \) is the force applied,
    • \( r \) is the distance from the rotation point to where the force is applied, and
    • \( \theta \) is the angle between the force and the lever arm.
  • In our exercise, a torque of 16 N路m is applied to the sanding disk.
The greater the torque, the more effectively it can rotate the object. In this scenario, applying torque results in changes in angular momentum and velocity for the disk. By applying this concept, students can visualize how forces cause rotational effects.
Rotational Inertia
Rotational inertia, or moment of inertia, is the resistance of an object to changes in its rotational motion. It depends on the mass of the object and how that mass is distributed relative to the axis of rotation.
  • For a given torque, an object with high rotational inertia will accelerate more slowly in rotation compared to one with low rotational inertia.
  • The formula is \( I = \sum m_i r_i^2 \), where:
    • \( m_i \) is the mass of each point in the object, and
    • \( r_i \) is the distance from the rotation axis.
Understanding rotational inertia helps in predicting how easily an object can change its state of rotation. In the exercise, the disk鈥檚 rotational inertia \( 1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \) determines how the disk reacts to the applied torque.
Angular Velocity
Angular velocity describes the speed of rotation and the direction in which an object rotates. It is the rate of change of an object鈥檚 angular position concerning time.
  • The formula for angular velocity \( \omega \) can be represented as \( \omega = \frac{\Delta \theta}{\Delta t} \), where:
    • \( \Delta \theta \) is the change in angular position, and
    • \( \Delta t \) is the change in time.
  • The angular velocity can also be linked to angular momentum as shown in our exercise. The formula \( \Delta \omega = \frac{\Delta L}{I} \) illustrates this relationship.
In the exercise example, applying torque over a short period results in a change in angular velocity. The calculated angular velocity of the disk becomes 440 rad/s, reflecting the speed at which it rotates. This connection between torque, inertia, and angular velocity exemplifies how these concepts integrate to explain rotational dynamics.

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Most popular questions from this chapter

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. 11-62). The radius of the axle is \(0.200 \mathrm{~m}\), and the rotational inertia of the wheel- axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m}\), what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

\- 52 (co A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 m\) that can rotate freely about its center like a merrygo-round. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach- disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the ki- \(\quad\) netic energy?

Two particles, each of mass \(2.90 \times 10^{-4} \mathrm{~kg}\) and speed \(5.46 \mathrm{~m} / \mathrm{s}\), travel in opposite directions along parallel lines separated by \(4.20 \mathrm{~cm} .\) (a) What is the magnitude \(L\) of the angular momentum of the two-particle system around a point midway between the two lines? (b) Does the value of \(L\) change if the point about which it is calculated is not midway between the lines? If the direction of travel for one of the particles is reversed, what would be (c) the answer to part (a) and (d) the answer to part (b)?

Shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800\) \(\mathrm{m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2} .\) By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1} ?\)

A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost?
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