Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 13, Problem 68PQ

(a)

To determine

The total angular momentum of the initial wheel-clay system using estimated values of masses of clay and wheel, the radius of the wheel, and the density of the clay.

(a)

Expert Solution
Check Mark

Answer to Problem 68PQ

The total angular momentum of the initial wheel-clay system is 5.05kgm2/s .

Explanation of Solution

Let the radius of pottery wheel is 7in , the  approximate mass of the pottery wheel is 25.0kg , and density of clay is 1500kg/m3 . Estimated mass of a clay vase is 2.50kg.

It is given that clay is in the approximate shape of a sphere.

Write the expression for the density of the sphere.

  ρ=MsphereVsphere

Here, ρ is the density of the sphere, Msphere is the mass of the sphere and Vsphere is the volume of the sphere.

Rearrange above equation to get expression of volume of sphere.

  Vsphere=Msphereρ                                                                                                          (I)

Write the expression for the volume of sphere.

  Vsphere=43πRsphere3

Here, Rsphere is the radius of sphere.

Substitute 43πRsphere3 for Vsphere in equation (I) to modify equation (I).

  43πRsphere3=MsphereρRsphere=3Msphere4ρπ3                                                                                              (II)

The wheel is in form of a disk. Thus, consider wheel as disk to find rotational inertia.

Write the expression for the total rotational inertia of wheel-clay system.

  Itotal,i=Idisk+Isphere                                                                                                   (III)

Here, Itotal,i is the initial rotational inertia of wheel-clay system, Idisk is the rotational inertia of the disk and Isphere is the rotational inertia of disk.

Write the expression for the rotational inertia of disk.

  Idisk=12MdiskRdisk2                                                                                                  (IV)

Here, Mdisk is the mass of the disk and Rdisk is the radius of the disk.

Write the expression for the rotational inertia of the sphere.

  Isphere=25MsphereRsphere2                                                                                              (V)

Substitute (IV) and (V) in equation (III) to get Itotal,i .

  Itotal,i=12MdiskRdisk2+25MsphereRsphere2                                                                        (VI)

Write the expression for the total angular momentum of the initial wheel-clay system.

  Li=Itotal,iωi                                                                                                            (VII)

Here, Li is the total angular momentum of the initial wheel-clay system and ωi is the initial angular momentum of the system.

Conclusion:

It is given that the angular velocity of the system is 120rpm.

Convert radius of disk from 7in to meter.

  Rdisk=(7in.)(1m39.37in.)=0.178m

Convert angular momentum 120rpm into rads .

  ωi=(120revmin)(2πrad1rev)(1min60s)=12.56rads

Substitute 2.50kg for Msphere and 1500kg/m3 for ρ in equation (II) to get Rsphere .

  Rsphere=3(2.50kg)4(1500kg/m3)(3.14)3=0.074m

Substitute 0.178m for Rdisk, 25.0kg for Mdisk, 2.50kg for Msphere and 0.074m for Rsphere in equation (VI) to get Itotal,i .

  Itotal,i=12(25.0kg)(0.178m)2+25(2.50kg)(0.074m)2=0.401kgm2

Substitute 0.401kgm2 for Itotal,i and 12.56rads for ωi in equation (VII) to get Li .

  Li=(0.401kgm2)(12.56rads)=5.05kgm2/s

Therefore, the total angular momentum of the initial wheel-clay system is 5.05kgm2/s .

(b)

To determine

Whether the system need to be sped up or slowed down, or should nothing be done to maintain a constant angular speed, if shape of the clay changed into a tall, cylindrical vase.

(b)

Expert Solution
Check Mark

Answer to Problem 68PQ

According to conservation of angular momentum, system would speed up due to the change of shape of the clay. The speed increases by 1% .

Explanation of Solution

The shape of clay changed from sphere to cylinder. It is given that radius of the cylindrical vase is one fourth of radius of sphere.

Since shape of clay is cylinder, calculate rotational inertia of cylinder to find rotational inertia of clay.

Write the equation for the total rotational inertia of the wheel-clay system after the shape is changed.

  Itotal,f=Idisk+Icylinder                                                                                             (VIII)

Here, Itotal, f is the rotational inertia of the system after shape of sphere clay changed to cylindrical shape and Icylidner is the rotational inertia of cylinder about its own axis.

Write he expression for the rotational inertia of cylinder.

  Icylinder=12McylidnerRcylinder2                                                                                         (IX)

Here, Mcylinder is the mass of clay and Rcylinder is the radius of cylindrical vase.

Substitute (IX) and (IV) in (VIII) and to get Itotal,f .

  Itotal,f=12MdiskRdisk2+12McylinderRcylinder2                                                                       (X)

Write the expression for the percentage change in rotational inertia of system.

  %change=Itotal,fItotal,iItotal,i×100                                                                                (XI)

Write the expression for final angular momentum of the system.

  Lf=Itotal,fωf                                                                                                         (XII)

Here, Lf is the final angular momentum and ωf is the final angular velocity.

Write the expression for the conservation of angular momentum.

  Li=Lf

Substitute (VII) and (XII) in above equation to get ωf .

  Itotal,iωi=Itotal,fωf                                                                                                 (XIII)

Conclusion:

Radius of cylinder is 14th of sphere.

Calculate radius of cylinder.

  Rcylinder=14(0.074m)=0.018m

Substitute 0.178m for Rdisk, 25.0kg for Mdisk, 2.50kg for Mcylinder and 0.018m for Rcylinder in equation (X) to get Itotal,f .

  Itotal,f=12(25.0kg)(0.178m)2+25(2.50kg)(0.018m)2=0.397kgm2

Substitute 0.397kgm2 for Itotal,f and 0.401kgm2 for Itotal,i in (XI) to get %change.

  %change=0.397kgm20.401kgm20.401kgm2×100=1.0%

Thus, rotational inertia of the system is decreases by 1.0% so that from conservation of angular momentum given in equation (XIII), angular momentum of the system increases by 1.0% .

Therefore, according to conservation of angular momentum, system would speed up due to the change of the clay. The speed increases by 1% .

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Chapter 13 Solutions

Physics for Scientists and Engineers: Foundations and Connections

Ch. 13 - Rotational Inertia Problems 5 and 6 are paired. 5....Ch. 13 - A 12.0-kg solid sphere of radius 1.50 m is being...Ch. 13 - A figure skater clasps her hands above her head as...Ch. 13 - A solid sphere of mass M and radius Ris rotating...Ch. 13 - Suppose a disk having massMtot and radius R is...Ch. 13 - Problems 11 and 12 are paired. A thin disk of...Ch. 13 - Given the disk and density in Problem 11, derive...Ch. 13 - A large stone disk is viewed from above and is...Ch. 13 - Prob. 14PQCh. 13 - A uniform disk of mass M = 3.00 kg and radius r =...Ch. 13 - Prob. 16PQCh. 13 - Prob. 17PQCh. 13 - The system shown in Figure P13.18 consisting of...Ch. 13 - A 10.0-kg disk of radius 2.0 m rotates from rest...Ch. 13 - Prob. 20PQCh. 13 - Prob. 21PQCh. 13 - In Problem 21, what fraction of the kinetic energy...Ch. 13 - Prob. 23PQCh. 13 - Prob. 24PQCh. 13 - Prob. 25PQCh. 13 - A student amuses herself byspinning her pen around...Ch. 13 - The motion of spinning a hula hoop around one's...Ch. 13 - Prob. 28PQCh. 13 - Prob. 29PQCh. 13 - Prob. 30PQCh. 13 - Sophia is playing with a set of wooden toys,...Ch. 13 - Prob. 32PQCh. 13 - A spring with spring constant 25 N/m is compressed...Ch. 13 - Prob. 34PQCh. 13 - Prob. 35PQCh. 13 - Prob. 36PQCh. 13 - Prob. 37PQCh. 13 - Prob. 38PQCh. 13 - A parent exerts a torque on a merry-go-round at a...Ch. 13 - Prob. 40PQCh. 13 - Today, waterwheels are not often used to grind...Ch. 13 - Prob. 42PQCh. 13 - A buzzard (m = 9.29 kg) is flying in circular...Ch. 13 - An object of mass M isthrown with a velocity v0 at...Ch. 13 - A thin rod of length 2.65 m and mass 13.7 kg is...Ch. 13 - A thin rod of length 2.65 m and mass 13.7 kg is...Ch. 13 - Prob. 47PQCh. 13 - Two particles of mass m1 = 2.00 kgand m2 = 5.00 kg...Ch. 13 - A turntable (disk) of radius r = 26.0 cm and...Ch. 13 - CHECK and THINK Our results give us a way to think...Ch. 13 - Prob. 51PQCh. 13 - Prob. 52PQCh. 13 - Two children (m = 30.0 kg each) stand opposite...Ch. 13 - A disk of mass m1 is rotating freely with constant...Ch. 13 - Prob. 55PQCh. 13 - Prob. 56PQCh. 13 - The angular momentum of a sphere is given by...Ch. 13 - Prob. 58PQCh. 13 - Prob. 59PQCh. 13 - Prob. 60PQCh. 13 - Prob. 61PQCh. 13 - Prob. 62PQCh. 13 - A uniform cylinder of radius r = 10.0 cm and mass...Ch. 13 - Prob. 64PQCh. 13 - A thin, spherical shell of mass m and radius R...Ch. 13 - To give a pet hamster exercise, some people put...Ch. 13 - Prob. 67PQCh. 13 - Prob. 68PQCh. 13 - The velocity of a particle of mass m = 2.00 kg is...Ch. 13 - A ball of mass M = 5.00 kg and radius r = 5.00 cm...Ch. 13 - A long, thin rod of mass m = 5.00 kg and length =...Ch. 13 - A solid sphere and a hollow cylinder of the same...Ch. 13 - A uniform disk of mass m = 10.0 kg and radius r =...Ch. 13 - When a person jumps off a diving platform, she...Ch. 13 - One end of a massless rigid rod of length is...Ch. 13 - A uniform solid sphere of mass m and radius r is...Ch. 13 - Prob. 77PQCh. 13 - A cam of mass M is in the shape of a circular disk...Ch. 13 - Prob. 79PQCh. 13 - Consider the downhill race in Example 13.9 (page...Ch. 13 - Prob. 81PQ
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