Transcript University Physics - Erwin Sitompul

Lecture 3
Ch2. STRAIGHT LINE MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
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18.30–20.15
: Class
: Quiz 1
: Class & Make-up Quiz 1
: Class & Discuss Quiz 1
: Mid-term Examination
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15.10.10
22.06.10
29.06.10
Erwin Sitompul
17.30–19.30
19.30–20.45
17.30–18.30
19.00–20.45
18.30–20.15
: Class
: Quiz 1
: Class & Make-up Quiz 1
: Discussion of Quiz 1
: Mid-term Examination
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Solution of Homework 2: Aprilia vs. Kawasaki
An Aprilia and a Kawasaki are separated by 200 m when they
start to move towards each other at t = 0. The Aprilia moves
with initial velocity 5 m/s and acceleration 4 m/s2. The Kawasaki
runs with initial velocity 10 m/s and acceleration 6 m/s2.
200 m
v 0,apr  5 m s
a ap r  4 m s
v 0,kw sk   10 m s
a kw sk   6 m s
2
2
(a) Determine the point where the two motorcycles meet
each other.
x apr  x kw sk
x 0,apr  v 0,apr t 
1
2
a apr t  x 0,kw sk  v 0,kw sk t 
2
1
2
a kw sk t
0  (5) t  2 (4) t  200  (  10) t  2 (  6) t
1
2
1
2
2
5 t  15 t  200  0
2
Erwin Sitompul
University Physics: Mechanics
3/3
Solution of Homework 2: Aprilia vs. Kawasaki
t1, 2 
5 t  15 t  200  0
2
 15 
t1  5 s
t2  8 s
x apr  x 0,apr  v 0,apr t 
1
2
a apr t
 0  (5)(5)  2 (4)(5)
1
 75 m
2
2
(15)  4(5)(  200)
2
2(5)
• Possible answer
• Both motorcycles meet
after 5 seconds
• Where?
x kwsk  x 0,kwsk  v 0,kwsk t 
 200  (  10)(5) 
1
2
a kwsk t
2
(  6)(5)
2
1
2
 75 m
Thus, both motorcycles meet at a point 75 m from the original
position of Aprilia or 125 m from the original position of
Kawasaki.
Erwin Sitompul
University Physics: Mechanics
3/4
Solution of Homework 2: Aprilia vs. Kawasaki
(b) Determine the velocity of Aprilia and Kawasaki by the
time they meet each other.
v apr  v 0 ,apr  a apr t
v kw sk  v 0 , kw sk  a kw sk t
 (5)  (4)(5)
 (  10)  (  6)(5)
 25 m s
  40 m s
• What is the meaning of
negative velocity?
Erwin Sitompul
University Physics: Mechanics
3/5
Free Fall Acceleration
 A free falling object accelerates downward constantly (air
friction is neglected).
 The acceleration is a = –g = –9.8 m/s2, which is due to the
gravitational force near the earth surface.
 This value is the same for all masses,
densities and shapes.
 With a = –g = –9.8 m/s2.
v  v0 
 at
gt
xy  xy00  vv00tt  22 at
gt
11
22
v  v0 
 2 ag ((xy  xy00))
2
2
xy  xy00  22 ((vv00  vv))tt
11
xy  xy00  vt
vt  22 at
gt
11
Erwin Sitompul
22
Accelerating feathers and
apples in a vacuum
University Physics: Mechanics
3/6
Example: Niagara Free Fall
In 1993, Dave Munday went over the
Niagara Falls in a steel ball equipped
with an air hole and then fell 48 m to
the water. Assume his initial velocity
was zero, and neglect the effect of
the air on the ball during the fall.
(a) How long did Munday fall to
reach the water surface?
y  y 0  v0t 
1
2
gt
2
 48  0  0  t  2  9.8  t
1
t 
2
(48)(2)
 9 .8 0
9.8
t
9.80   3.13 s
Erwin Sitompul
2
Niagara Falls
v  v 0  gt
y  y0  v0t 
1
2
gt
2
v  v0  2 g ( y  y 0 )
2
2
y  y0 
1
2
( v0  v )t
y  y 0  vt 
1
2
gt
University Physics: Mechanics
2
3/7
Example: Niagara Free Fall
In 1993, Dave Munday went over the Niagara Falls in a steel
ball equipped with an air hole and then fell 48 m to the water.
Assume his initial velocity was zero, and neglect the effect of
the air on the ball during the fall.
(b) What was Munday’s velocity as
he reached the water surface?
v  v 0  gt
v  0  (9.8)(3.13)   30.67 m /s
v  v 0  gt
or
y  y0  v0t 
1
2
gt
2
v  v0  2 g ( y  y 0 )
v  v0  2 g ( y  y 0 )
2
2
v  0  2(9.8)(  48  0)  9 4 0 .8
y  y0 
v
y  y 0  vt 
2
2
2
940.8   30.67 m/s
2
1
2
( v0  v )t
1
2
gt
2
• Positive (+) or negative (–) ?
Erwin Sitompul
University Physics: Mechanics
3/8
Example: Niagara Free Fall
In 1993, Dave Munday went over the Niagara Falls in a steel
ball equipped with an air hole and then fell 48 m to the water.
Assume his initial velocity was zero, and neglect the effect of
the air on the ball during the fall.
(c) Determine Munday’s position and
velocity at each full second.
y  y0  v0t 
1
2
gt
2
v  v 0  gt
t  1  y  0  0 1 
(9.8)(1)   4.9
2
1
2
v  0  (9.8)(1)   9.8
t  2  y  0  02 
(9.8)(2)   19.6
2
1
2
v  0  (9.8)(2)   19.6
t 3
Erwin Sitompul
University Physics: Mechanics
3/9
Example: Baseball Pitcher
A pitcher tosses a baseball up along a
y axis, with an initial speed of 12 m/s.
(a) How long does the ball take to
reach its maximum height?
v  v 0  gt
0  12  (9.8)t
t
12
 1.22 s
9.8
v  v 0  gt
y  y0  v0t 
1
2
gt
2
v  v0  2 g ( y  y 0 )
2
2
y  y0 
1
2
( v0  v )t
y  y 0  vt 
Erwin Sitompul
1
2
gt
2
University Physics: Mechanics
3/10
Example: Baseball Pitcher
A pitcher tosses a baseball up along a
y axis, with an initial speed of 12 m/s.
(b) What is the ball’s maximum
height above its release point?
y  y 0  v0t 
1
2
gt
2
y  0  (12)(1.22)  2 (9.8)(1.22)
1
2
 7.35 m
v  v 0  gt
y  y0  v0t 
1
2
gt
2
v  v0  2 g ( y  y 0 )
2
2
y  y0 
1
2
( v0  v )t
y  y 0  vt 
Erwin Sitompul
1
2
gt
2
University Physics: Mechanics
3/11
Example: Baseball Pitcher
A pitcher tosses a baseball up along a
y axis, with an initial speed of 12 m/s.
(c) How long does the ball take to
reach a point 5.0 m above its
release point?
y  y 0  v0t 
1
2
gt
2
5  0  (12) t  2 (9.8) t
1
2
4.9 t  12 t  5  0
2
t1  0.53,
t 2  1.92
• Both t1 and t2 are correct
• t1 when the ball goes up,
t2 when the ball goes down
Erwin Sitompul
University Physics: Mechanics
3/12
Lecture 3
Ch3. VECTOR QUANTITIES
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Vectors and Scalars
 A vector quantity has magnitude as well as direction.
 Some physical quantities that are vector quantities are
displacement, velocity, and acceleration.
 A scalar quantity has only magnitude.
 Some physical quantities that are scalar quantities are
temperature, pressure, energy, mass, and time.
 A displacement vector
represents the change of
position.
 Both magnitude and direction
are needed to fully specify a
displacement vector.
 The displacement vector,
however, tells nothing about
the actual path.
Erwin Sitompul
University Physics: Mechanics
3/14
Adding Vectors Geometrically
 Suppose that a particle moves from
A to B and then later from B to C.
 The overall displacement (no matter
what its actual path) can be
represented with two successive
displacement vectors, AB and BC.
 The net displacement of these two
displacements is a single
displacement from A to C.
 AC is called the vector sum (or
resultant) of the vectors AB and BC.
 The vectors are now redrawn and
relabeled, with an italic symbol and an
arrow over it.
 For handwriting, a plain symbol
and
→ → →
an arrow is enough, such as a, b, s.
Erwin Sitompul
s  ab
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3/15
Adding Vectors Geometrically
 Two vectors can be added in either order.
ab b a
 Three vectors can be grouped in any way as they are added.
a  (b  c )  ( a  b )  c
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University Physics: Mechanics
3/16
Adding Vectors Geometrically
→
→
 The vector –b is a vector with the same magnitude as b but
the opposite direction. Adding the two vectors would yield
b  (b )  0
→
→
 Subtracting b can be done by adding –b.
d  a  b  a  (b )
 d  b  a or a  d  b
Erwin Sitompul
University Physics: Mechanics
3/17
Components of Vectors
 Adding vectors geometrically can be tedious.
 An easier technique involves algebra and requires that the
vectors be placed on a rectangular coordinate system.
 For two-dimensional vectors, the x and y axes are usually
drawn in the drawing plane.
a x  a cos 
a y  a sin 
a
ax  a y
ay
tan  
ax
 A component of a vector is the
projection of the vector on an axis.
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2
2
a | a |
  a
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3/18
Components of Vectors
 Note, that the angle θ is the angle that the vector makes with
the positive direction of the x axis.
 The calculation of θ is done in counterclockwise directions.
What is the value of θ on
the figure above?
θ = 324.46° or –35.54°
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University Physics: Mechanics
3/19
Components of Vectors
 Vectors with various sign of components are shown below:
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3/20
Components of Vectors
Which of the indicated →methods for combining the x and y
components of vector a are proper to determine that vector?
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University Physics: Mechanics
3/21
Components of Vectors
Example:
A small airplane leaves an airport on an overcast day and it is
later sighted 215 km away, in a direction making an angle of 22°
east of due north. How far east and north is the airplane from
the airport when sighted?
Solution:
  90  22  68
d x  d cos   (215 km)(cos68  )
 80.54 km
d y  d sin   (215 km )(sin 68  )
 199.34 km
Thus, the airplane is 80.54 km east and
199.34 km north of the airport.
Erwin Sitompul
University Physics: Mechanics
3/22
Trigonometric Functions
Sine
Cosine
Tangent
Erwin Sitompul
University Physics: Mechanics
3/23
Unit Vectors
 A unit vector is a vector that has a
magnitude of exactly 1 and points in
a particular direction.
 The unit vectors in the positive
directions of the x, y, and z axes
are labeled ^i, ^j, and ^k.
 Unit vectors are very useful for expressing other vectors:
a  a x ˆi  a y ˆj
Erwin Sitompul
b  b x ˆi  b y ˆj
University Physics: Mechanics
3/24
Adding Vectors Geometrically
 Vectors can be added geometrically by subsequently linking
the head of one vector with the tail of the other vector.
 The addition result is the vector that goes from the tail of the
first vector to the head of the last vector.
Erwin Sitompul
University Physics: Mechanics
3/25
Adding Vectors by Components
 Another way to add vectors is to combine their
components axis by axis.
rx  a x  b x
ry  a y  b y
r  ab
r
a
ax
Erwin Sitompul
b
by
r  rx ˆi  ry ˆj
by
r
r
ry
a y bx
ay
ax
bx
rx
University Physics: Mechanics
3/26
Adding Vectors by Components
r
ry
rx
r  rx ˆi  ry ˆj
Component Notation
r  r 
Magnitude-angle Notation
where r 
rx  ry
ry
tan  
rx
2
2
In the figure below, find out the
→ → → →
components of d1, d2, d1+d2.
Erwin Sitompul
University Physics: Mechanics
3/27
Adding Vectors by Components
Desert Ant
Erwin Sitompul
University Physics: Mechanics
3/28
Trivia
A bear walks 5 km to the south. Resting a while, it continues
walking 5 km east. Then, it walks again 5 km to the north and
the bear reaches its initial position.
What is the color of the bear?
5 km,
south
5 km,
north
5 km
east
Polar Bear with White Fur
Erwin Sitompul
University Physics: Mechanics
3/29
Homework 3: The Beetles
Two beetles run across flat sand, starting at the same point. The
red beetle runs 0.5 m due east, then 0.8 m at 30° north of due
east.
The green beetle also makes two runs; the first is 1.6 m at 40°
east of due north.
What must be (a) the magnitude and (b) the direction of its
second run if it is to end up at the new location of red beetle?
2nd run?
1st run
2nd run
1st run
Erwin Sitompul
University Physics: Mechanics
3/30