Figure 11-28 gives the angular momentum magnitude Lof a wheel versustime t. Rank the four-lettered time intervals according to the magnitude

of the torque acting on the wheel, greatest first.

Short Answer

Expert verified

The ranking of time intervals according to the magnitude of torque is

.τD>τB>τA=τC

Step by step solution

01

Step 1: Given data

The graph L vs t is given.

02

Understanding the concept of torque

The torque acting on the object is equal to the moment of force. The toque is also equal to the product of the moment of inertia and angular acceleration of the object. It can be written as the time rate of change of angular momentum of the object.

Use the concept of torque and from the given graph find angular momentum at given time intervals.

The formula is as follows:

Ï„=dLdt

WhereL is angular momentum,τis torque andtis time.

03

Determining therank of time intervals according to the magnitude of the torque

From the graph, see that for time intervals A and C the slopes are 0.Therefore, torques are 0 Nm.

For time interval B the slope is positive and for D slope is negative, but here, magnitudes want to be found.So, the slope of time interval D is greater than B.

From this, rank torques from the time intervals,

τD>τB>τA=τC

Therefore, torques can be ranked from the given time intervals using the concept of torque. The rank is,

τD>τB>τA=τC

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform disk of mass 10m and radius 3.0rcan rotate freely about its fixed centre like a merry-go-round. A smaller uniform disk of mass mand radius rlies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of 20rad/s.Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding).

(a) What then is their angular velocity about the centre of the larger disk?

(b) What is the ratio ofK/K0 the new kinetic energy of the two-disk system to the system’s initial kinetic energy?

Question: In Figure, a solid brass ball of mass 0.280 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radiusr=14.o cm and the ball has radius r<<R(a) What is hif the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height, h = 6.00R (b) What is the magnitude of the horizontal force component acting on the ball at point Q? (c) What is the direction of the horizontal force component acting on the ball at point Q?

The rotor of an electric motor has rotational inertia Im=2.0×10−3kgm2about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertiarole="math" localid="1660985808865" Ip=12kg.m2about this axis. Calculate the number of revolutions of the rotor required to turn the probe through30°about its central axis.

Figure 11-29 shows a particle moving at constant velocity and five points with their xycoordinates. Rank the points according to the magnitude of the angular momentum of the particle measured about them, greatest first.

A ballerina begins a tour jet (Figure a) with angular speed ωiand a rotational inertia consisting of two parts : role="math" localid="1661005078220" Ileg= 1.44 kg.m2 for her leg extended outward at angle θ= 90.0°to her body and Itrunk= 0.660 kg.m2 for the rest of her body (primarily her trunk). Near her maximum height she holds both legs at angle30.0°to her body and has angular speedωf(Figure b). Assuming that Ihas not changed, what is the ratioωfωi ?

(a) Initial phase of a tour jet: large rotational inertia and small angular speed. (b) Later phase: smaller rotational inertia and larger angular speed.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free