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MS101: PhysicsDr. Ahmed Amin Hussein
2013-2014
May 3, 2023 Prepared By: Dr. Ahmed Amin 1
Chapter 4
Motion with Constant Acceleration
May 3, 2023 2Prepared By: Dr. Ahmed Amin
Motion Along a Line Graphical Representation of
Motion Free Fall Projectile Motion Apparent Weight
§4.1 Motion Along a Line
May 3, 2023 3Prepared By: Dr. Ahmed Amin
For constant acceleration the kinematic equations are:
and
xavv
tavvv
tatvxxx
xixfx
xixfxx
xixif
2
21
22
2
2,
,
fxixxav
xav
vvv
tvx
In a previous example, a box sliding across a rough surface was found to have an acceleration of –2.94 m/s2. If the initial speed of the box is 10.0 m/s, how long does it take for the box to come to rest?
Know: a = 2.94 m/s2, vix = 10.0 m/s, vfx = 0.0 m/s
Want: t.
sec 40.3m/s 942m/s 0.10
0
2
.avt
tavv
x
ix
xixx
Example
May 3, 2023 Prepared By: Dr. Ahmed Amin 4
A train of mass 55,200 kg is traveling along a straight, level track at 26.8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking force has magnitude 84.0 kN, can the train be stopped in time?
22
22
m/s 95.12
02
xva
xavv
ixx
xixx
Know: vfx = 0 m/s, vix = 26.8 m/s, x = 184 m
Determine ax and compare to the train’s maximum ax.
Example
May 3, 2023 Prepared By: Dr. Ahmed Amin 5
Example continued:
The train’s maximum acceleration is:
2brakingnetmax, m/s 52.1
mF
mF
ax
The maximum acceleration is not sufficient to stop the train before it hits the stalled truck.
May 3, 2023 Prepared By: Dr. Ahmed Amin 6
§4.2 Visualizing Motion with Constant Acceleration
Motion diagrams for three carts:
May 3, 2023 Prepared By: Dr. Ahmed Amin 7
Graphs of x, vx, ax for each of the three carts
May 3, 2023 8Prepared By: Dr. Ahmed Amin
A trolley car in New Orleans starts from rest at the St. Charles Street stop and has a constant acceleration of 1.20 m/s2 for 12.0 seconds.
0
2
4
6
8
10
12
14
16
0 2 4 6 8 10 12 14
t (sec)
v (m
/sec
)
(a) Draw a graph of vx versus t.
Example
May 3, 2023 9Prepared By: Dr. Ahmed Amin
(b) How far has the train traveled at the end of the 12.0 seconds?
The area between the curve and the time axis represents the distance traveled.
m 4.86s 12m/s 4.1421
tsec 12t21
vx
(c) What is the speed of the train at the end of the 12.0 s?
This can be read directly from the graph, vx = 14.4 m/s.
Example continued:
May 3, 2023 10Prepared By: Dr. Ahmed Amin
§4.3 Free FallA stone is dropped from the edge of a cliff; if air resistance can be ignored, the FBD for the stone is:
x
y
w
Apply Newton’s Second Law
2m/s 9.8
N/kg 8.9
ga
mamgwFy
The stone is in free fall; only the force of gravity acts on the stone.
May 3, 2023 11Prepared By: Dr. Ahmed Amin
A stone is thrown upwards with an initial velocity of 20 m/s. What is the maximum height risen, and how long does it take to reach this height?
vix = 20 m/sΔx= ?vfx
2 = vix2 + 2 ax Δx
0 = 202 – (2 x 9.8 Δx)Δx = 20.4 m vfx =vix
+ ax Δt
0 = 20 + (-9.8) ΔtΔt = 20/9.8 = 2.04 s
Example
May 3, 2023 12Prepared By: Dr. Ahmed Amin
Example: You throw a ball into the air with speed 15.0 m/s; how high does the ball rise?
Given: viy = +15.0 m/s; ay = 9.8 m/s2
2
21 tatvy yiy
x
y viy
ay
tavv yiyfy
To calculate the final height, we need to know the time of flight.
Time of flight from:
Example
May 3, 2023 13Prepared By: Dr. Ahmed Amin
sec 531m/s 89m/s 015
0
2 ...
av
t
tavv
y
iy
yiyfy
The ball rises until vfy= 0.
m 511
s 531m/s 8921s 531m/s 015
21
22
2
.
....
tatvy yiy
The height:
Example continued:
May 3, 2023 14Prepared By: Dr. Ahmed Amin
Example (text problem 4.24): A penny is dropped from the observation deck of the Empire State Building 369 m above the ground. With what velocity does it strike the ground? Ignore air resistance.
369 m
x
yGiven: viy = 0 m/s, ay = 9.8 m/s2, y = 369 m
Unknown: vyf
Use:
yav
ya
yavv
yfy
y
yiyfy
2
2
222
ay
May 3, 2023 15Prepared By: Dr. Ahmed Amin
What is projectile?Projectile -Any object which
projected by some means and continues to move due to its own inertia (mass).
May 3, 2023 16Prepared By: Dr. Ahmed Amin
Projectiles move in TWO dimensions
Since a projectile moves in 2-dimensions, it therefore has 2 components just like a resultant vector.
Horizontal and Vertical
May 3, 2023 17Prepared By: Dr. Ahmed Amin
Horizontal “Velocity” Component
NEVER changes, covers equal displacements in equal time periods. This means the initial horizontal velocity equals the final horizontal velocity
In other words, the horizontal velocity is CONSTANT. BUT WHY?
Gravity DOES NOT work horizontally to increase or decrease the velocity.
May 3, 2023 18Prepared By: Dr. Ahmed Amin
Vertical “Velocity” Component Changes (due to gravity), does NOT cover equal
displacements in equal time periods.
Both the MAGNITUDE and DIRECTION change. As the projectile moves up the MAGNITUDE DECREASES and its direction is UPWARD. As it moves down the MAGNITUDE INCREASES and the direction is DOWNWARD.
May 3, 2023 19Prepared By: Dr. Ahmed Amin
Combining the ComponentsTogether, these
components produce what is called a trajectory or path. This path is parabolic in nature.
Component Magnitude Direction
Horizontal Constant Constant
Vertical Changes Changes
May 3, 2023 20Prepared By: Dr. Ahmed Amin
Horizontally Launched Projectiles
Projectiles which have NO upward trajectory and NO initial VERTICAL velocity.
0 /oyv m s
constantox xv v
May 3, 2023 21Prepared By: Dr. Ahmed Amin
Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2.
212oxx v t at
oxx v t
Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO!
212y gt
Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.
May 3, 2023 22Prepared By: Dr. Ahmed Amin
Horizontally Launched Projectiles
Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?
What do I know?
What I want to know?
vox=100 m/s t = ?y = 500 m x = ?voy= 0 m/s
g = -9.8 m/s/s
2 2
2
1 1500 ( 9.8)2 2102.04
y gt t
t t
10.1 seconds(100)(10.1)oxx v t 1010 m
May 3, 2023 23Prepared By: Dr. Ahmed Amin
ExampleA place kicker kicks
a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
What I know What I want to know
vox=12.04 m/s t = ?voy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s
2 2
2
1 0 (15.97) 4.9215.97 4.9 15.97 4.9
oyy v t gt t t
t t tt
3.26 s
May 3, 2023 24Prepared By: Dr. Ahmed Amin
ExampleA place kicker kicks
a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(b) How far away does it land?
What I know What I want to know
vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s
(12.04)(3.26)oxx v t 39.24 m
May 3, 2023 25Prepared By: Dr. Ahmed Amin
ExampleA place kicker kicks a
football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(c) How high does it travel?
What I know What I want to know
vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = 39.24 my = 0 ymax=?g = - 9.8 m/s/s2
2
12
(15.97)(1.63) 4.9(1.63)
oyy v t gt
yy
13.01 m
May 3, 2023 26Prepared By: Dr. Ahmed Amin
TMA Exercises
May 3, 2023 27Prepared By: Dr. Ahmed Amin
Problems :
Questions # # 5, 13, 27, 35
Additional questions
Chapter 4, Questions # 1 - 16 & 16 - 20