This preview shows page 1. Sign up to view the full content.
Unformatted text preview: NG Let the two disks constitute the system. Since there are no
external torques acting on the system, the principle of conservation of angular momentum
applies. Therefore we have Linitial = L final , or I A ω A + I Bω B = ( I A + I B ) ω final
This expression can be solved for the moment of inertia of disk B.
SOLUTION Solving the above expression for I B , we obtain IB = IA F
ω
G
ω
H final
B –ωA – ω final I = ( 3.4 kg ⋅ m ) L –2.4 rad / s – 7.2 rad / s O 4.4 kg ⋅ m
=
J
M rad / s – (–2.4 rad / s) P
–9.8
K
N
Q
2 2 Chapter 9 Problems 479 60. REASONING Before any sand strikes the disk, only the disk is rotating. After the sand has
landed on the disk, both the sand and the disk are rotating. If the sand and disk are taken to
be the system of objects under consideration, we note that there are no external torques
acting on the system. As the sand strikes the disk, each exerts a torque on the other.
However, these torques are exerted by members of the system, and, as such, are internal
torques. The conservation of angular momentum states that the total angular momentum of a
system remains constant (is conserved) if the net average external torque acting on the
system is zero. We will use this principle to find the final angular velocity of the system.
SOLUTION The angular momentum of the system (sand plus disk) is given by
Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or
L = Iω. The conservation of angular momentum can be written as
I3
1ω
2
Final angular
momentum = I 0ω0
13
2
Initial angular
momentum where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are final
and initial moments of inertia. The initial moment of inertia is given, while the final moment
of inertia is the sum of the values for the rotating sand and disk, I = Isand + I0. We note that
the sand forms a thin ring, so its moment of inertia is given by (see Table 9.1)
2
Isand = M sand Rsand , where Msand is the mass of the sand and Rsand is the radius of the ring.
Thus, the final angular velocity of the system is, then, I0
I0 I0 = ω0 = ω0 2
I M + I0 Rsand + I 0 I sand sand ω = ω0 0.10 kg ⋅ m 2 = 0.037 rad/s
= ( 0.067 rad/s ) ( 0.50 kg )( 0.40 m )2 + 0.10 kg ⋅ m 2 61. REASONING We consider a system consisting of the person and the carousel. Since the
carousel rotates on frictionless bearings, no net external torque acts on this system.
Therefore, its angular momentum is conserved, and we can set the total angular momentum
of the system before the person hops on equal to the total angular momentum afterwards.
SOLUTION The initial angular momentum of the carousel is I carousel ω0 (Equation 9.10) , where Icarousel is the moment of inertia of the carousel and ω0 is its initial angular velocity.
After the person climbs aboard, the total angular momentum of the carousel and person is
I carouselωf + I personωf , where ωf is the final angular velocity. According to Equation 9.6, the 480 ROTATIONAL DYNAMICS person’s moment of inertial is I person = MR 2 , since he is at the outer edge of the carousel,
which has a radius R. Applying the conservation of angular momentum, we have
I carouselωf + I personωf
144424443 = I carouselω
1 24 0
43
Initial total
angular momentum Final total
angular momentum ωf = = I carouselω0
I carouselω0
=
I carousel + I person I carousel + MR 2 (125 kg ⋅ m2 ) ( 3.14 rad/s) 125 kg ⋅ m2 + ( 40.0 kg ) (1.50 m ) 2 = 1.83 rad/s 62. REASONING Once the motorcycle is in the air, it is subject to no net external torque, since
gravity and air resistance are being ignored. Therefore, its total angular momentum is
conserved: Lf = L0 . The total angular momentum of the motorcycle is the sum of the
angular momentum LE = I EωE (Equation 9.10) of the engine and the angular momentum LM = I MωM of the rest of the motorcycle (including the rider): I EωEf + I ω = I EωE0 + I ω
144 M Mf 144 M M0
244
3
244
3
Lf (1) L0 We will use Equation (1) to find the ratio IE/IM of the moments of inertia of the engine and
the rest of the motorcycle. We note that, so long as all angular velocities are expressed in
rev/min, there is no need to convert to SI units (rad/s). SOLUTION Initially, only the engine is rotating, so the rest of the motorcycle has no
angular velocity: ωM0 = 0 rad/s. Solving Equation (1) for the ratio IE/IM, we obtain I EωEf + I MωMf = I EωE0 + I M ( 0) or I E (ωEf − ωE0 ) = − I MωMf or −ωMf
IE
=
I M ωEf − ωE0 As usual, clockwise rotation is negative, and counterclockwise is positive. The ratio of the
moments of inertia is, then, IE
IM = −ωMf ωEf − ωE0 = − ( −3.8 rev/min ) ( +12 500 rev/min ) − ( +7700 rev/min ) = 7.9 ×10−4 Chapter 9 Problems 481 63. REASONING The rod and bug are taken to be the system of objects under consideration,
and we note that there are no external torques acting on the system. As the bug crawls out to
the end of the rod, each exerts a torque on the other. However, these torques are internal
torques. The conservation of angular momentum states that the total angular momentum of a
system re...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details