4 kinematika

Physics 111: Lecture 2, Pg 1

Physics 111: Lecture 2Physics 111: Lecture 2

Today’s AgendaToday’s Agenda

� Recap of 1-D motion with constant acceleration

� 1-D free fall� example

� Review of Vectors

� 3-D Kinematics� Shoot the monkey� Baseball� Independence of x and y components


Review: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:Recall what you saw:

12

22

32

42

200 at

2

1tvxx ++=


1-D Free-Fall1-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

0 -y yv v gt=2

y00 tg2

1tvyy −+=

y

a

v

t

t

t

gay −=


Gravity facts: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: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: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: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, Lecture 2, Act 1Act 11D free fall1D free fall

� Alice and Bill are standing at the top of a cliff of heightAlice and Bill are standing at the top of a cliff of height HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, Alice straight, Alice straight downdown and Bill straightand Bill straight upup. The speed of the balls when . The speed of the balls when they hit the ground arethey hit the ground are vvAA andand vvBB respectively.respectively. Which of Which of the following is true:the following is true:

(a)(a) vvAA < < vvBB (b) (b) vvAA = = vvBB (c) (c) vvAA > > vvBB

vv00

vv00

BillBillAliceAlice

HH

vvAA vvBB


Lecture 2, Lecture 2, Act 1Act 11D Free fall1D 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:

vv00

BillBill

HH

vv = v= v00

( ) 0HHg2vv 20

2 =−−=− )(Equation:Equation:

This looks just like Bill threw This looks just like Bill threw the ball down with speed the ball down with speed vv00, so, sothe speed at the bottom shouldthe speed at the bottom shouldbe the same as Alice’s ball.be the same as Alice’s ball.

y = 0y = 0


Lecture 2, Act 1Lecture 2, Act 11D Free fall1D Free fall

� We can also just use the equation directly:We can also just use the equation directly:

( )H0g2vv 20

2 −−=− )(Alice:Alice:

vv00

vv00

AliceAlice BillBill

y = 0y = 0

( )H0g2vv 20

2 −−=− )(Bill:Bill:same !!same !!


Vectors (review):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 rr in 2 dimensions.

ExampleExample: Where is Chicago?� Choose origin at UrbanaUrbana� Choose coordinates of distance (miles), and direction (N,S,E,W)� In this case rr is a vector that points 120 miles north.

Chicago

Urbana

r r


Vectors...Vectors...

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

� Boldface notation: AA

� “Arrow” notation:

A A =A

A



� The components of rr are its (x,y,z) coordinates

� rr = (rx ,ry ,rz ) = (x,y,z)

� Consider this in 2-D (since it’s easier to draw):

� rx = x = r cos θ � ry = y = r sin θ

y

x

(x,y)

θ

where r = |rr |

rr θ = arctan( y / x )



� The magnitude (length) of rr is found using the Pythagorean theorem:

r = = +r x y2 2rr y

x

� The length of a vector clearly does notnot depend on its direction.


Unit Vectors:Unit Vectors:

� A Unit Vector Unit Vector is a vector having length 1 and no units

� It is used to specify a direction� Unit vector uu points in the direction of UU

� Often denoted with a “hat”: uu = û

� Useful examples are the Cartesian unit vectors [ i, j, ki, j, k ] � point in the direction of the

x, y and z axes

UU

x

y

z

ii

jj

kk

û û


Vector addition:Vector addition:

� Consider the vectors AA and BB. Find AA + BB.

AA

BB

AA BB AA BB

CC = AA + BB

� We can arrange the vectors as we want, as long as we maintain their length and direction!!


Vector addition using components:Vector addition using components:

� Consider CC = AA + BB.

(a) CC = (Ax ii + Ay jj) + (Bx i i + By jj) = (Ax + Bx)ii + (Ay + By)jj

(b) CC = (Cx ii + Cy jj)

� Comparing components of (a) and (b):

� Cx = Ax + Bx

� Cy = Ay + By

CC

BxAA

ByBB

Ax

Ay


Lecture 2, Lecture 2, Act 2Act 2VectorsVectors

� 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)(a) (3,5,-1)(3,5,-1) (b)(b) (4,-2,5)(4,-2,5) (c)(c) (5,-2,4)(5,-2,4)


Lecture 2, Lecture 2, Act 2Act 2SolutionSolution

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 Kinematics3-D Kinematics

� The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as:

rr = x ii + y jj + z kk

vv = vx ii + vy jj + vz kk (ii , jj , kk unit vectors )

aa = ax ii + ay jj + az kk

� We have already seen the 1-D kinematics equations:

adv

dt

d x

dt= =

2

2v

dx

dt=x x(t= )


3-D Kinematics 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:

rr = rr(t) vv = drr / dt aa = d2rr / 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= ( )



� So for constant acceleration we can integrate to get:

� aa = const

� vv = vv0 + a a t

� rr = rr0 + vv0 t + 1/2 aa t2

(where aa, vv, vv0, rr, rr0, are all vectors)



� Most 3-D problems can be reduced to 2-D problems when acceleration is constant:

� Choose y axis to be along direction of acceleration� Choose x axis to be along the “other” direction of motion

� ExampleExample: Throwing a baseball (neglecting air resistance)� Acceleration is constant (gravity)

� Choose y axis up: ay = -g

� Choose x axis along the ground in the direction of the throw

lost marbles


““x” and “y” components of motion are x” and “y” components of motion are independent.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: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 (θ) 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?What time does the ball reach the fence?� Does Mark get a home run?Does Mark get a home run?

θvv

h

D

y0


Problem...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



� Use geometry to figure out v0x and v0y :

y

x

g

θvv

v0x

v0y

Find v0x = |v| cos θ.

and v0y = |v| sin θ.

y0



� 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 mm

� Since the wall is 3 m high, Mark gets the homer!!


Lecture 2, Lecture 2, Act 3Act 3Motion in 2DMotion 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 44 times as far!!

x = v0x t


Shooting the MonkeyShooting the Monkey(tranquilizer gun)(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...Shooting the Monkey...

� If there were no gravity, simply aim at the monkey

r = r0

r =v0t


Shooting the Monkey...Shooting the Monkey...

rr = vv0 t - 1/2 gg t2

� With gravity, still aim at the monkey! rr = r0 - 1/2 gg t2

Dart hits the monkey!


Recap:Recap:Shooting the monkey...Shooting the monkey...

x x = = xx00

yy = -1/2 gg t2

� This may be easier to think about.

It’s exactly the same idea!!

xx = = vv0 0 tt

yy = -1/2 gg t2

4 kinematika

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