Physics 111: Lecture 2, Pg 1
Physics 111: Lecture 2
Today’s Agenda
Recap of 1-D motion with constant acceleration
1-D free fallexample
Review of Vectors
3-D KinematicsShoot the monkeyBaseballIndependence of x and y components
Physics 111: Lecture 2, Pg 2
Review: For constant acceleration we found:
atvv 0
200 at
2
1tvxx
consta
x
a
vt
t
t
v)(v2
1v
)x2a(xvv
0av
020
2
From which we derived:
Physics 111: Lecture 2, Pg 3
Recall what you saw:
12
22
32
42
200 at
2
1tvxx
Physics 111: Lecture 2, Pg 4
1-D Free-Fall This is a nice example of constant acceleration (gravity): In this case, acceleration is caused by the force of gravity:
Usually pick y-axis “upward”Acceleration of gravity is “down”:
y
ay = g
gtvvy
0y
2y00 tg
2
1tvyy
y
a
v
t
t
t
gay
Physics 111: Lecture 2, Pg 5
Gravity facts:
g does not depend on the nature of the material!Galileo (1564-1642) figured this out without fancy clocks
& rulers!
demo - feather & penny in vacuum
Nominally, g = 9.81 m/s2 At the equator g = 9.78 m/s2
At the North pole g = 9.83 m/s2
More on gravity in a few lectures!
Penny & feather
Ballw/ cup
Physics 111: Lecture 2, Pg 6
Problem: The pilot of a hovering helicopter
drops a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance)
1000 m
Physics 111: Lecture 2, Pg 7
Problem: First choose coordinate system.
Origin and y-direction.
Next write down position equation:
Realize that v0y = 0.
20y0 gt
2
1tvyy
20 gt
2
1yy
1000 m
y = 0
y
Physics 111: Lecture 2, Pg 8
Problem: Solve for time t when y = 0 given
that y0 = 1000 m.
Recall:
Solve for vy:
y0 = 1000 m
y
s314sm819
m10002
g
y2t
20 .
.
20 gt
2
1yy -
y = 0
)( 02
y02y yya2vv --
sm140
gy2v 0y
/
Physics 111: Lecture 2, Pg 9
Lecture 2, Act 11D free fall
Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are vA and vB respectively. Which of the following is true:
(a) vA < vB (b) vA = vB (c) vA > vB
v0
v0
BillAlice
H
vA vB
Physics 111: Lecture 2, Pg 10
Lecture 2, Act 11D Free fall
Since the motion up and back down is symmetric, intuition should tell you that v = v0
We can prove that your intuition is correct:
v0
Bill
H
v = v0
0HHg2vv 20
2 )(Equation:
This looks just like Bill threw the ball down with speed v0, sothe speed at the bottom shouldbe the same as Alice’s ball.
y = 0
Physics 111: Lecture 2, Pg 11
Lecture 2, Act 11D Free fall
We can also just use the equation directly:
H0g2vv 20
2 )(Alice:
v0
v0
Alice Bill
y = 0
H0g2vv 20
2 )(Bill:same !!
Physics 111: Lecture 2, Pg 12
Vectors (review): In 1 dimension, we could specify direction with a + or - sign.
For example, in the previous problem ay = -g etc.
In 2 or 3 dimensions, we need more than a sign to specify the direction of something:
To illustrate this, consider the position vector r in 2 dimensions.
Example: Where is Chicago? Choose origin at Urbana Choose coordinates of
distance (miles), and direction (N,S,E,W)
In this case r is a vector that points 120 miles north.
Chicago
Urbana
r
Physics 111: Lecture 2, Pg 13
Vectors...
There are two common ways of indicating that something is a vector quantity:
Boldface notation: A
“Arrow” notation:
A =A
A
Physics 111: Lecture 2, Pg 14
Vectors...
The components of r are its (x,y,z) coordinates r = (rx ,ry ,rz ) = (x,y,z)
Consider this in 2-D (since it’s easier to draw):rx = x = r cos ry = y = r sin
y
x
(x,y)
where r = |r |
r arctan( y / x )
Physics 111: Lecture 2, Pg 15
Vectors...
The magnitude (length) of r is found using the Pythagorean theorem:
r r x y2 2r
y
x
The length of a vector clearly does not depend on its direction.
Physics 111: Lecture 2, Pg 16
Unit Vectors:
A Unit Vector is a vector having length 1 and no units
It is used to specify a direction Unit vector u points in the direction of U
Often denoted with a “hat”: u = û
Useful examples are the Cartesian unit vectors [ i, j, k ] point in the direction of the
x, y and z axes
U
x
y
z
i
j
k
û
Physics 111: Lecture 2, Pg 17
Vector addition:
Consider the vectors A and B. Find A + B.
A
B
A B A B
C = A + B
We can arrange the vectors as we want, as long as we maintain their length and direction!!
Physics 111: Lecture 2, Pg 18
Vector addition using components:
Consider C = A + B.
(a) C = (Ax i + Ay j) + (Bx i + By j) = (Ax + Bx)i + (Ay + By)j
(b) C = (Cx i + Cy j)
Comparing components of (a) and (b):
Cx = Ax + Bx
Cy = Ay + By
C
BxA
ByB
Ax
Ay
Physics 111: Lecture 2, Pg 19
Lecture 2, Act 2Vectors
Vector A = {0,2,1} Vector B = {3,0,2} Vector C = {1,-4,2}
What is the resultant vector, D, from adding A+B+C?
(a) {3,5,-1} (b) {4,-2,5} (c) {5,-2,4}
Physics 111: Lecture 2, Pg 20
Lecture 2, Act 2Solution
D = (AXi + AYj + AZk) + (BXi + BYj + BZk) + (CXi + CYj + CZk)
= (AX + BX + CX)i + (AY + BY+ CY)j + (AZ + BZ + CZ)k
= (0 + 3 + 1)i + (2 + 0 - 4)j + (1 + 2 + 2)k
= {4,-2,5}
Physics 111: Lecture 2, Pg 21
3-D Kinematics
The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as:
r = x i + y j + z k
v = vx i + vy j + vz k (i , j , k unit vectors )
a = ax i + ay j + az k
We have already seen the 1-D kinematics equations:
adv
dt
d x
dt
2
2v
dx
dtx x(t )
Physics 111: Lecture 2, Pg 22
3-D Kinematics
For 3-D, we simply apply the 1-D equations to each of the component equations.
Which can be combined into the vector equations:
r = r(t) v = dr / dt a = d2r / dt2
ad x
dtx
2
2
vdx
dtx
x x(t )
ad y
dty
2
2
vdy
dty
y y t ( )
ad z
dtz
2
2
vdz
dtz
z z t ( )
Physics 111: Lecture 2, Pg 23
3-D Kinematics
So for constant acceleration we can integrate to get:
a = constv = v0 + a tr = r0 + v0 t + 1/2 a t2
(where a, v, v0, r, r0, are all vectors)
Physics 111: Lecture 2, Pg 24
2-D Kinematics
Most 3-D problems can be reduced to 2-D problems when acceleration is constant:Choose y axis to be along direction of accelerationChoose x axis to be along the “other” direction of motion
Example: Throwing a baseball (neglecting air resistance)Acceleration is constant (gravity)Choose y axis up: ay = -gChoose x axis along the ground in the direction of the
throw
lost marbles
Physics 111: Lecture 2, Pg 25
“x” and “y” components of motion are independent.
A man on a train tosses a ball straight up in the air.View this from two reference frames:
Reference frame on the moving train.
Reference frame on the ground.
Cart
Physics 111: Lecture 2, Pg 26
Problem: Mark McGwire clobbers a fastball toward center-field. The
ball is hit 1 m (yo ) above the plate, and its initial velocity is 36.5 m/s (v ) at an angle of 30o () above horizontal. The center-field wall is 113 m (D) from the plate and is 3 m (h) high.
What time does the ball reach the fence?Does Mark get a home run?
v
h
D
y0
Physics 111: Lecture 2, Pg 27
Problem...
Choose y axis up. Choose x axis along the ground in the direction of the hit. Choose the origin (0,0) to be at the plate. Say that the ball is hit at t = 0, x = x0 = 0
Equations of motion are:
vx = v0x vy = v0y - gt
x = vxt y = y0 + v0y t - 1/ 2 gt2
Physics 111: Lecture 2, Pg 28
Problem...
Use geometry to figure out v0x and v0y :
y
x
g
v
v0x
v0y
Find v0x = |v| cos .and v0y = |v| sin .
y0
Physics 111: Lecture 2, Pg 29
Problem...
The time to reach the wall is: t = D / vx (easy!) We have an equation that tell us y(t) = y0 + v0y t + a t2/ 2 So, we’re done....now we just plug in the numbers:
Find: vx = 36.5 cos(30) m/s = 31.6 m/s vy = 36.5 sin(30) m/s = 18.25 m/s t = (113 m) / (31.6 m/s) = 3.58 s y(t) = (1.0 m) + (18.25 m/s)(3.58 s) -
(0.5)(9.8 m/s2)(3.58 s)2
= (1.0 + 65.3 - 62.8) m = 3.5 m
Since the wall is 3 m high, Mark gets the homer!!
Physics 111: Lecture 2, Pg 30
Lecture 2, Act 3Motion in 2D
Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it?
(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1
Physics 111: Lecture 2, Pg 31
Lecture 2, Act 3Solution
The distance a ball will go is simply x = (horizontal speed) x (time in air) = v0x t
2y00 tg
2
1tvyy
To figure out “time in air”, consider the equation for the height of the ball:
0tg2
1tv 2
y0 When the ball is caught, y = y0
0tg2
1vt y0
g
v2t y0
t 0 (time of throw)
(time of catch)
two solutions
Physics 111: Lecture 2, Pg 32
Lecture 2, Act 3Solution
g
v2t y0 So the time spent in the air is proportional to v0y :
Since the angles are the same, both v0y and v0x for ball 2are twice those of ball 1.
ball 1ball 2
v0y ,1
v0x ,1
v0y ,2
v0x ,2
v0,1
v0,2
Ball 2 is in the air twice as long as ball 1, but it also has twice the horizontal speed, so it will go 4 times as far!!
x = v0x t
Physics 111: Lecture 2, Pg 33
Shooting the Monkey(tranquilizer gun)
Where does the zookeeper aim if he wants to hit the monkey?( He knows the monkey willlet go as soon as he shoots ! )
Physics 111: Lecture 2, Pg 34
Shooting the Monkey...
If there were no gravity, simply aim at the monkey
r = r0
r =v0t
Physics 111: Lecture 2, Pg 35
Shooting the Monkey...
r = v0 t - 1/2 g t2
With gravity, still aim at the monkey! r = r0 - 1/2 g t2
Dart hits the monkey!
Physics 111: Lecture 2, Pg 36
Recap:Shooting the monkey...
x = x0
y = -1/2 g t2
This may be easier to think about. It’s exactly the same idea!!
x = v0 t
y = -1/2 g t2
Physics 111: Lecture 2, Pg 37
Recap of Lecture 2
Recap of 1-D motion with constant acceleration. (Text: 2-3)
1-D Free-Fall (Text: 2-3)example
Review of Vectors (Text: 3-1 & 3-2)
3-D Kinematics (Text: 3-3 & 3-4)Shoot the monkey (Ex. 3-11)Baseball problemIndependence of x and y components
Look at textbook problems Chapter 3: # 35, 39, 69, 97