Physics 111: Lecture 2 Today’s Agenda

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


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:


Recall what you saw:

12

22

32

42

200 at

2

1tvxx


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


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


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


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


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

/


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


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


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 !!


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


Vectors...

There are two common ways of indicating that something is a vector quantity:

Boldface notation: A

“Arrow” notation:

A =A

A


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 )


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.


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


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!!


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


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}


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}


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 )


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 ( )


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)


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


“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


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


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


Problem...

Use geometry to figure out v0x and v0y :

y

x

g

v

v0x

v0y

Find v0x = |v| cos .and v0y = |v| sin .

y0


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!!


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



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



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


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 ! )


Shooting the Monkey...

If there were no gravity, simply aim at the monkey

r = r0

r =v0t


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!


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


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

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Physics 111: Lecture 2 Today’s Agenda