/*! 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} Q10Q Figure 11-29 shows a particle mo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified

Rank of the points according to the magnitude of the angular momentum of the particle measured about them is,

.Lb>Lc=Ld>La=Le

Step by step solution

01

Step 1: Given

Figure is given to rank the points according to the magnitude of the angular momentum of the particle measured about them.

02

Understanding the concept of angular momentum

Using the equation of angular momentum, rank the points according to the magnitude of the angular momentum of the particle measured about them.

The formula is as follows:

L=r×P=rPsinθ

Where,θ is the angle between Pand the position vectorr.

03

Step 3: Ranking the points according to the magnitude of the angular momentum of the particle measured about them

Now,

L=r×P=rPsinθ

For points a and e, the angle between the momentum vector and the distance from the moving particle to points a and e is zero. So,

From figure,r=4for point a andforr=8point b,

La=(4)(P)sin0=0

Le=(8)(P)sin0=0

For point c, the angle between the momentum vector and the position vector is900.So,r=13

From figure,r=2for point c,

Lc=(2)(P)sin900=2P

For point b, the angle between the momentum vector and the position vector is 1080.So,

From figure r=13, for point b,

Lb=13(P)sin1080=3.42P

For point d, the angle between the momentum vector and the position vector isSo,

From figure,r=13for point d,

Ld=13(P)sin33.690=2P

Therefore,

Lb>Lc=Ld>La=Le

Hence, using the equation of angular momentum, the points can be ranked according to the magnitude of the angular momentum of the particle measured about them and the rank is,

.Lb>Lc=Ld>La=Le

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 91Ó°ÊÓ!

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?

A wheel of radius 0.250 m, which is moving initially at 43.0 m/s, rolls to a stop in 225 m. Calculate the magnitudes of its (a) linear acceleration and (b) angular acceleration. (c) Its rotational inertia is 0.155 kg.m2 about its central axis. Find the magnitude of the torque about the central axis due to friction on the wheel.

Figure 11-27 shows an overheadview of a rectangular slab that can

spin like a merry-go-round about its center at O. Also shown are seven

paths along which wads of bubble gum can be thrown (all with the

same speed and mass) to stick onto the stationary slab. (a) Rank the paths according to the angular speed that the slab (and gum) will have after the gum sticks, greatest first. (b) For which paths will the angular momentum of the slab(and gum) about Obe negative from the view of Fig. 11-27?

Question: A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector r→=(2.00m)i^-(3.00m)j^+(2.00m)k^,the force is given by F→=Fxi^+(7.00N)j^-(6.00N)k^and the corresponding torque about the origin isτ→=(4.00Nm)i^+(2.00Nm)j^-(1.00Nm)k^. Determine.

A uniform solid sphere rolls down in an incline (a) what must be the incline angle if the linear acceleration of the center of the sphere is tohave a magnitude of the 0.10g? (b) If the frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to 0.10g? why?

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.