Transcript PHYSICS 218 (517-520) recitation 1

PHYSICS 218 (517-520)
recitation 1
• In recitation,
– We shall discuss exercises and problems. They may include
homework problems.
– Quiz: you are supposed to solve a quiz at each recitation.
– Make a group so that each group contains 3 (or 4) students.
Take a seat as in your Lab.
– You should collaborate with your partners to solve quiz problems.
You work as teams on the quiz problem. One group has one
answer paper. The answer paper should have names and UIN of
all group members. The group members will have the same credit
for recitation quiz. One of you may serve as a focused peer
mentor.
• There is no Lab today (Sep.1).
• There is no quiz today (Sep.1).
Unites are treated just like algebraic symbols.
DO NOT forget to give your answers
with correct units.
•
Unit
– An equation must always be dimensionally consistent.
– You cannot add length and time.
•
•
Unit conversion
Eg. The official world land speed record is 1228.0 km/h. Express this
speed in m/s.
1km = 103 m,
1 h = 3600 s
ж
ж
km ц
1000 m ц
ч
ч
з
1228.0 km/h = зз1228.0
=
1228.0
ґ
= 341.11m/s
ч
ч
з
ч
ч
зи
з
h ш и
3600 s ш
ж
ж
km ц
km 1000 m
1h ц
ч
ч
з
= зз1228.0
=
1228.0
ґ
ґ
ч
ч
з
ч
ч
зи
з
h ш и
h
km
3600 s ш
=1
=1
Ex. The inhabitants of a small island begin exporting beautiful cloth made from a rare
plant that grows only on their island. Seeing how popular the small quantity that they
export has been, they steadily raise their prices. A clothing maker from New York,
thinking that he can save money by "cutting out the middleman," decides to travel to
the small island and buy the cloth himself. Ignorant of the local custom of offering
strangers outrageous prices and then negotiating down, the clothing maker accepts
(much to everyone's surprise) the initial price of 400 tepizes/m2. The price of this
cloth in New York is 120 dollars/yard2 .
If he bought 500 m2 of this fabric, how much money did he lose?
(1 tepiz = 0.625 dollar and 1 yard = 0.9144 m.)
Strategy
How much money he spent?
How much he would spent in NY?
400 tepizes 0.625dollar
ґ
= 125, 000dollars
2
1m
1 tepiz
2
ж
ц
120
dollars
1
yard
ч
in NY, 500 m 2 ґ
ґ зз
» 71,800 dollars
ч
2
ч
з
и
ш
1 yard
0.9144 m
in the island, 500 m2 ґ
\ He lost 125, 000 - 71,800 = 53, 200 dollars
Problem 1.67(a)
• Approximately how many atoms make up our planet? For simplicity,
assume the average atomic mass of the atoms is 14 g/mol. Avogadro's
number gives the number of atoms in a mole.
mass of the earth = 5.97 ґ 10 27 g (Appendix F of the textbook)
5.97 ґ 10 27 g 6.02ґ 10 23 atoms
\ number of atoms in the earth =
ґ
14 g/mol
1mol
= 2.6ґ 1050 atoms
y
r
A
In Cartesian coordinate system,
r
A is represented by Ax and Ay .
In polar coordinate system,
r
r
A is represented by A (= A) and q.
q
y
q
Ax = A cos q,
A=
Ay = A sin q,
x
жAy ч
ц
з
A + A , q = tan зз ч
.
ч
ч
иA ш
2
x
2
y
- 1
x
r
A
• q is defined from the positive x-axis to the direction of a vector toward the positive yaxis, i.e., counterclockwise.
• arctan function has two solutions in the range of (0, 2p).
Vector calculus
r
A = ( Ax , Ay , Az ) or
z
k̂
ĵ
iˆ
x
iˆ Чiˆ =
iˆ Чˆj =
y
ˆj Чˆj =
iˆ Чkˆ =
r
A = Ax iˆ + Ay ˆj + Az kˆ
kˆ Чkˆ = 1
ˆj Чkˆ = 0
r r
A ЧB = ( Axiˆ + Ay ˆj + Az kˆ) Ч( Bxiˆ + By ˆj + Bz kˆ)
= Ax Bxiˆ Чiˆ + Ax By iˆ Чˆj + Ax Bz iˆ Чkˆ
+ Ay Bx ˆj Чiˆ + Ay By ˆj Чˆj + Ay Bz ˆj Чkˆ
+ Az Bx kˆ Чiˆ + Az By kˆ Чˆj + Az Bz kˆ Чkˆ
= Ax Bx + Ay By + Az Bz
r r
r2
2
2
2
A ЧA = Ax + Ay + Az = A
Scalar product (dot product, inner product)
r r
r r
A ЧB = A B cos q
q : angle between the two vectors (0 Ј q Ј p )
r r
p
if q < , then A ЧB > 0
2
r r
p
if q > , then A ЧB < 0
2
r
Ex. A = (1, 2,3),
r
B = (3, 2,1). Calculate the angle between the two vectors.
A = 12 + 22 + 32 = 14
B = 32 + 22 + 12 = 14
r r
A ЧB = Ax Bx + Ay By + Az Bz = 1ґ 3 + 2ґ 2 + 3ґ 1 = 10
= A B cos q = 14 cos q
\ cos q =
10 5
= ,
14 7
q = 44.4o
Exercise
A particle is moving from the point A tor the rpointr B.
r
Define the displacement of the particle R = B - A, where A = 360
r
and B = 880.
r
r
We write A = ( Ax , Az ) and B = ( Bx , Bz ).
Then A = 360, q=40o and B = 880, f =163o
r
R
Ax = A cos q = 360ґ cos 40o = 276
Az = A sin q = 360ґ sin 40o = 231
Bx = B cos f = 880ґ cos163o = - 842
Bz = B sin f = 880ґ sin163o = 257
r
R = ( Rx , Rz ),
Rx = Bx - Ax = - 842 - 276 = - 1118,
Rz = Bz - Az = 257 - 231 = 26
Problem 1.71
r
You are to program a robotic arm on an assembly line to move in the xy-plane. Its first displacement is A;
r
its second displacement is B, of magnitude 6.42 cm and direction 63.2o measured in the sense from the
r
r r
+ x-axis toward the - y -axis. The resultant C = A + B of the two displacements should also have
a magnitude of 6.42 cm, but a direction 22.2o measured in the sense from the +x-axis toward the +y-axis.
r
What is A?
Bx = B cos b , By = B sin b ,
B = 6.42, b = 360o - 63.2o = 296.8o
Cx = C cos g , C y = C sin g ,
r
r r
A = C - B,
C = B,
g = 22.2o
Ax = B (cos g - cos b ) = 6.42ґ (cos 22.2o - cos 296.8o ) = 3.05
Ay = B (sin g - sin b ) = 6.42ґ (sin 22.2o - sin 296.8o ) = 8.16
A = (3.05, 8.16)
Problem 1.99
At Enormous State University (ESU), the football team records its plays using vector displacements,
with the origin taken to be the position of the ball before the play starts. In a certain pass play,
the receiver starts at + 2.0 iˆ - 4.5 ˆj, where the units are yards, iˆ is to the right, and ˆj is downfield.
Subsequent displacements of the receiver are +8.5 iˆ (in motion before the snap), +11 ˆj (breaks downfield),
- 6.0 iˆ + 4.0 ˆj (zigs), and +12.0 iˆ + 18.0 ˆj (zags). Meanwhile, the quarterback has dropped straight
back to a position - 7.0 ˆj. How far and in which direction must the quaterback throw the ball?
r
receiver's final position R
r
R = (+ 2.0 iˆ - 4.5 ˆj ) + (+8.5 iˆ) + (+11 ˆj ) + (- 6.0 iˆ + 4.0 ˆj) + (+12.0 iˆ + 18.0 ˆj )
= 16.5 iˆ + 28.5 ˆj
r
quarterback's position Q
downfield
r
Q = - 7.0 ˆj
r r
The relative distance between the two: R - Q = 16.5 iˆ + 35.5 ˆj
How far?
r r
R - Q = 16.52 + 35.52 » 38 yards
Which direction?
ж35.5 ч
ц
o
tan - 1 зз
»
65
(Both x and y components are positive.)
ч
зи16.5 ч
ш
65o from the right line, 25o to the right of downfield
25o
65o