Chapter 18 MC Explanations by TA Ajay Kailas
1.) Current, by default, goes in the opposite direction of electron movement. Current flows in the direction of positive charge. Therefore, if Chlorine is moving left (it's negative) current is moving right. Si
a) The initial angular momentum with respect to the pivot is mvr, and the final total moment of inertia is I mr 2 , so the final angular velocity is mvr mr 2 I . b) The kinetic energy after the collision is 1 2 K mr 2 I M m gh, or 2 2 M m gh . mr 2 I c) S
10.96: The initial angular momentum is 1 mRv1 , with the minus sign indicating that runner's motion is opposite the motion of the part of the turntable under his feet. The final angular momentum is 2 ( mR 2 ), so 1 mRv1 2 mR 2
(80 kg m 2 )( 0.200 rad s) (
10.97: From Eq. (10.36),
r
(50.0 kg)(9.80 m s2 )(0.040 m) 12.7 rad s, (0.085 kg m2 )(6.0 m s) (0.33 m)
or 13 rad s to two figures, which is quite large.
10.98: The velocity of the center of mass will change by velocity will change by
vcm
J m
, and the angular
The change is velocity of the end of the bat will J J ( x xcm ) xcm then be vend Setting vcm _ xcm m vend 0 allows cancellation of J , and gives ( x
10.99: In Fig. (10.34(a), if the vector r , and hence the vector L are not horizontal but make an angle with the horizontal, the torque will still be horizontal (the torque must be perpendicular to the vertical weight). The magnitude of the torque will be
10.100:
a)
The distance from the center of the ball to the midpoint of the line joining the points where the ball is in contact with the rails is R 2 d 2 , so vcm R 2 d 2 4. R, the same as rolling on a flat surface. When when d 0, this reduces to vcm 0 fo
10.101:
a)
The friction force is f acceleration is gives
t
fR I k MgR 1 2 MR 2
k n
2 k g R
k Mg , so a
at
k g. The magnitude of the angular
R
0
. b) Setting v
t R and solving for t
R0 a R
R0 k g 2 k g
R0 , 3 k g
and
d 1 2 at 2 1 R0 k g 2 3 k g
2 2 R 20 .
10.102: Denoting the upward forces that the hands exert as FL and FR , the conditions that FL and FR must satisfy are FL FR w
I , r where the second equation is L, divided by r. These two equations can be solved for the forces by first adding and then sub
10.94:
a), g)
b) Using the vector product form for the angular momentum, v1
v2 and r1
r2 , so
mr1 v1 , ^. Then, j so the angular momenta are the same. c) Let ^ ^ v1 r1 zi xk , and ^ L1 mr1 v1 m xR i x 2 y 2 ^ j
mr2 v2
^ xR k .
With x 2
y2
m 2R2 ( 2 m
10.93: The train's speed relative to the earth is 0.600 m s 0.475 m , so the total angular momentum is
0.600 m s
0.475 m 1.20 kg 0.475 m
1.00 m 1 2 7.00 kg 2
2
0,
from which 0.298 rad s ,with the minus sign indicating that the turntable moves clockwise,
10.85: The speed of the ball just before it hits the bar is v 2gy 15.34 m s. Use conservation of angular momentum to find the angular velocity of the bar just after the collision. Take the axis at the center of the bar. L1 mvr 5.00 kg 15.34 m s 2.00 m 153
10.86: a) The rings and the rod exert forces on each other, but there is no net force or torque on the system, and so the angular momentum will be constant. As the rings slide toward the ends, the moment of inertia changes, and the final angular velocity
10.87: The intial angular momentum of the bullet is m 4 v L 2 , and the final moment of intertia of the rod and bullet is m 3 L2 angular moment equal to b)
m 4 L 2
2
I and solving for
gives
2
19 48 mL2 . Setting the initial mvL 8 6 v L. 2 19 19 48 mL
1 2
10.88: Assuming the blow to be concentrated at a point (or using a suitably chosen rFave, and L rFave t rJ . "average" point) at a distance r from the hinge, ave The angular velocity is then L rFave t l 2 Fave t 3 Fave t , 1 I I ml 2 2 ml 3 Where l is the
10.89: is I
a) The initial angular momentum is L
mv l 2 and the final moment of inertia
I0
m l 2 , so
mv l 2 M 3 l2 m l 2 5.46 rad s.
2
2
b) M m gh 1 2 2 I , and after solving for h and substitution of numerical values, will be h 3.16 10 2 m. c) Rather t
10.90: Angular momentum is conserved, so I 0 0 I 2 2 , or, using the fact that for a common mass the moment of inertia is proportional to the square of the radius, 2 2 2 2 R020 R2 2 , or R0 0 R0 R2 0 ~ R0 0 2R0 R 0 R0 , where the terms in R and 2 have bee
10.91: The initial angular momentum is L1
K1
0 I A and the initial kinetic energy is
2 I A0 2. The final total moment of inertia is 4 I A , so the final angular velocity
is 1 4 0 and the final kinetic energy is 1 2 4I A 0 4 2 obtained more directly from K
10.92: The tension is related to the block's mass and speed, and the radius of the circle, v2 by T m . The block's angular momentum with respect to the hole is L mvr , so in r terms of the angular momentum,
T 1 mv r
2
m 2v 2 r 2 m r3
mvr mr 3
2
L2 . mr 3
3/10/2011
Chapter 22: Electromagnetic
Waves
A magnetic dipole antenna
works by Faradays law:
changing magnetic flux
induces a current in the
loop.
Nature of EM waves
Antennae
The EM Spectrum
EM
Speed of EM Waves
Energy Transport
Polarization
Doppler Effec
2/24/2011
Unit 13, Chapter 20 Overview:
Electromagnetic Induction
Motional EMF
The motional EMF is
= vBL
where L is the separation between the rails.
The current in the rod is
Electric Generators
Faradays Law
I=
V vBL
=
R
R
R
where R is the resistance in
Chapter 20 ELECTROMAGNETIC INDUCTION
Conceptual Questions
4. The farmer could place wire loops around the edge of his field near the high voltage wires. The alternating current would produce a changing magnetic flux through the wire loops, and therefore,
Chapter 21 ALTERNATING CURRENT
Conceptual Questions
2. The average power transmitted is proportional to the product of the rms current and the rms voltage. The power delivered by a high voltage and low current combination is the same as that delivered by
Chapter 22 ELECTROMAGNETIC WAVES
Conceptual Questions
2. The magnetic field of the waves must be vertical-parallel to the axis of the antenna. If you are due south of the antenna, the electric field must lie in the east-west direction, perpendicular to bo
Chapter 23 REFLECTION AND REFRACTION OF LIGHT
Conceptual Questions
2. In a virtual image, light rays do not actually come from a point on an image but instead only appear to diverge from a single point. In a real image, the rays actually do pass through t
Chapter 16: Electric Forces and Field 1.) This is a great question! Because we are talking about acceleration, have a mass, and are finding force, the key here is to realize we use the F=ma equation. Since there are 2 objects acting on each other that are
1.) Electric field is defined as Force per unit charge. The SI unit for force is a Newton while the unit for charge is C. Therefore, the E field unit is N/C. It is also V/M from the equation delta V=E times distance. The answer is F. 2.) Because the dista
REVIEW AND SYNTHESIS: CHAPTERS 1618
Review Exercises
1. Strategy and Solution Since the spheres are identical, the charge will be shared evenly by the two spheres, so the spheres will have (18.0 C 6.0 C) 2 12.0 C of charge each. 5. Strategy The force on c
Chapter 16 multiple choice answers
1 j
2 c
3 e
4 a
5 c
6 b
7 d
8 c
9 b
10 b
Chapter 17 multiple choice answers
1 f
2 a
3 e
4 c
5 d
6 b
7 f
8 e
9 b
10 b
11 b
12 d
Chapter 19 MAGNETIC FORCES AND FIELDS
Conceptual Questions
1. The magnetic field cannot be described as the magnetic force per unit charge because unlike the electric force, the magnetic force depends upon the velocity of the charge. 2. (a) It is possible