one dimensional motion and gravity

Week 2 Physics 214Spring 2009

1

One dimensional motion and gravityWhen we drop an object it’s velocity continues to increase. That means there is an acceleration.Near the earths surface the value of this acceleration is g = 9.8m/s2.

This is due to the attractive force of gravity. g varies slightly over the earth because the earth is not a perfect sphere and because of the rotation of the earth.On the moon g is much smaller (~1/6)

For CONSTANT acceleration,The equations of motion are:v = v0 + at , d = v0t + ½(at2)v2 = v0

2 + 2ad , d = ½(v +v0)t(starting at d =0)

at is the change in v for time interval t

+

For this case all quantities are +And g is rounded up to 10.0 m/swhich is 2% off and might fail CHIP


2

One dimensional motion and gravity

For CONSTANT acceleration,The equations of motion are:v = v0 + at does not involve d

d = v0t + ½(at2) does not involve v

v2 = v02 + 2ad does not involve t

d = ½(v0 + v)t does not involve a

(starting at d =0)

d, vo, v, a, t

For constant acceleration, if you are told the values of three of these quantities, you can solve for a fourth quantity --- and not have to solve for the fifth quantity if you don’t need to, by picking the proper equation.

Displacement = avg. velocity * time

Combine 1st and 4th eqns. 2d eq.Combine eqs 1 & 4 differently, get 3d eqn., ‘cuz (v-v0)/a = t, plug into eq.4

d = ½(v0+v)(v-v0)/a = (v2-v02)/2a is same as eqn. 3


3

Throwing verticallyIn the example shown a ball is thrown vertically.The acceleration is minus 9.8m/s2 and the motion is symmetric.

v = v0 + at at t=0 v = v0 At the top v = 0 and t = v0/9.8 At the bottom t = 2v0/9.8 and v = - v0

time up and down = 2 x time to top

d = v0t + 1/2at2 so d=0 when t = 0 ort = -2v0/a or 2v0/9.8 since a = -9.8

WHY? d = v0t+1/2at2 = t(v0+at/2)So d = 0 if t=0 or if ( ) = 0

g = -9.8m/s2

+


4

Throwing verticallyIn the example shown a ball is thrown vertically.The acceleration is minus 9.8m/s2 and the motion is symmetric.

v = v0 + at at t=0 v = v0 At the top v = 0 and t = v0/9.8 At the bottom t = 2v0/9.8 and v = - v0

time up and down = 2 x time to top

d = v0t + 1/2at2 so d=0 when t = 0 ort = -2v0/a or 2v0/9.8 since a = -9.8

v2 = v02 + 2ad

At d = 0 v = +v0 (t = 0)Or v = -v0 ( t = 2v0/9.8)

g = -9.8m/s2

+


5

QUIZ Ch 3 E6Ball is droppedWhat is the change in velocitybetween 1 and 4 seconds?

After 1 sec v = 9.8 t = 9.8 m/sAfter 4 sec v = 9.8 x 4 = 39.2 m/sChange in velocity is 29.4 m/s

g +v

.

A. 9.8 m/s B. 19.6 m/sC 29.4 m/s D. 39.2 m/s.


6

Projectile MotionIf we throw an object so that it’s initial velocity is horizontal then ignoring friction it will continue to move with this velocity in the horizontal direction but it will also start to fall so that it’s trajectory will be curved.


7

Motion in two dimensionsIf we take two axes at right angles we can analyze the motion along each axis separately and determine properties of the wholemotion. We will only deal with cases where there is a constant velocity along one axis and a constant acceleration along the other axis. This means that for

axis 1 v1 = constant and d1 = v1t note: a1=0axis 2 v2 = v02 + at d = v02t + ½ at2 v2

2 = v022 + 2ad

Axis 1 usually x

Axis 2 usually yup

down


8

Projectile MotionWe will focus on projectiles with a = 9.8m/s2

v1

v0v

At the highest point the vertical velocity is zerovv = v0v + at so t = v0v/9.8 since a = -gh = v0vt + ½ at2 = vertical height above launch

At impact t = 2v0v/9.8 and R = v1 x 2v0v/9.8 And the vertical velocity is minus v0v

Maximum range R for 45o

degree upward launch, easiest to see with trig.

9.8

R = rangeθ


9

Projectile MotionWe will focus on projectiles with a = 9.8m/s2

v1

v0v

At impact t = 2v0v/9.8 and R = v1 x 2v0v/9.8

v1=v0cos(θ) v0v=v0sin(θ) R=2c.s.v02/9.8m/s2

2sin.cos = sin(2θ) , max when 2θ = 90o

Maximum range R for 45o

degree upward launch, easiest to see with trig.

9.8

R = rangeθ

v2=vv


10

Trajectories

http://faculty.tcc.fl.edu/scma/carrj/Java/baseball4.html

http://www.mhhe.com/physsci/physical/giambattista/proj/projectile.html

With no friction there are always two angles which give the same range for the same starting velocity 450 + X and 450 – X

R varies as sin(2θ)

http://www.physics.purdue.edu/class/applets/phe/projectile.htm

http://www.physics.purdue.edu/academic_programs/courses/phys214/movies.php (anim0002.mov)


11

Throwing a ball horizontally from a building

Use + down sog is positive and h is positiveh = v02t + ½ at2

v = v02 + atv2 = v02

2 + 2ah

v02 = 0, t2 = 2h/a

R = v1t = v1sqrt(2h/a)

NOTE h is measured down

v1

g

h

v2

R


12

Summary Chapter 3Any motion in a plane can analyzed using two axes at right angles. The motion along each axis is independent of the motion along the other axis so the two dimensional motion can be analyzed as two one dimensional motions linked by time.Special case axis 1 v1 = constant and d1 = v1taxis 2 v2 = v02 + at , h = v02t + 1/2at2 , v2 = v02

2 + 2ah

For projectile motion a = 9.8m/s2

v1

v1

v0v

v0v

v1

9.8m/s2


13

1D-20Independence of Vertical & Horizontal Motions (Drop-Kick)

THE VERTICAL & HORIZONTAL MOTIONS ARE INDEPENDENT. THE HORIZONTAL VELOCITY DOES NOT AFFECT THE VERTICAL MOTION.THE VERTICAL FALL TIME IS THE SAME AS LONG AS THE BALLS DROP SIMULTANEOUSLY FROM THE SAME HEIGHT.

Listen to the SOUND when they hit the ground and when they bounce.

One ball drops from rest. The other ball is simultaneously projected horizontally

Which ball will hit the ground first ?


14

1D-21Independence of Vertical and Horizontal Motions

THE HORIZONTAL MOTION OF THE BALL IS UNAFFECTED BY ITS VERTICAL MOTION.

The trajectory

in the room frame

A ball is projected vertically from a cart traveling horizontally

The trajectory in the cart

frame


15

1G-03 Measurement of g

d = ½ gt2 t = sqrt ( 2d/g )g = 2d/t2

Hand timing would not be accurate because of the short fall time

Measuring g by dropping an object

What difficulties might be

encountered in measuring h & t of

fall ?

Let d = 1 m,

t = sqrt(2/9.8) = 0.45 s


16

1K-11 Coin and Feather

IN AIR, WEIGHT AND SURFACE AREA MAKE OBJECTS FALL AT DIFFERENT SPEEDS BECAUSE OF AIR FRICTION.

DROPPING A COIN AND A FEATHER ?

DO ALL OBJECTS HAVE THE SAMEACCELERATION

WHEN DROPPED ?


17

1D-22 Water Jets & Projectile Motion

NEGLECTING FRICTION THE RANGE IS A MAXIMUM AT 450. TWO DIFFERENT ANGLES CAN GIVE THE SAME RANGE (ANGLES SYMMETRIC ABOUT 45°). A LARGER ANGLE MEANS A LONGER TIME OF FLIGHT, BUT LESS HORIZONTAL VELOCITY. A SMALLER ANGLE MEANS A LARGER HORIZONTAL VELOCITY, AND LESS FLIGHT TIME. THE TRAJECTORY IS SYMMETRIC.

g

PROJECTILE MOTION OF A WATER JETWhat angle gives the

maximum range?


18

1D-23 Shoot the Monkey

THE VERTICAL MOTION IS INDEPENDENT OF THE HORIZONTAL MOTIONTHE EFFECT OF FRICTION IS MINIMIZED BY USING A HEAVY TARGET

WHERE SHOULD ONE AIM, ABOVE,

BELOW OR AT?

Ignoring friction y = v0yt – 1/2gt2 , t = d/v0x , v0yt = v0yd/v0x = hSo at x = d y = h – ½ gt2 In the same time the monkey falls ½ gt2

So the bullet always hits the monkey no matter what the value of v0, Unless range R is less than d!

The monkey falls out of the tree at the instant the gun is fired

|--------------d-----------------|

Turn off g, bullet hits monkey at height h.

Turn on g, both bullet and monkey fall same amount

v0y/v0x = h/d

y=0.


19

Questions Chapter 3Q1 A small piece of paper is dropped and flutters to the floor. Isthe piece of paper accelerating at any time during this motion? Explain?

Q4 A lead ball and an aluminum ball, each 1 in. in diameter, are released simultaneously and allowed to fall to the ground. Due to its greater density, the lead ball has a substantially larger mass than the aluminum ball. Which of these balls, if either, has the greater acceleration due to gravity? Explain.

Yes at the start

They both have the same gravitational acceleration. Any difference in how they fall is due to friction. If the balls have identical shapes size and surface polish the friction is the same but the weight is different, so the two motions are .not quite identical


20

Q8 A rock is dropped from the top of a diving platform into the swimming pool below. Will the distance traveled by the rock in a 0.1-second interval near the top of its flight be the same as the distance covered in a 0.1-second interval just before it hits the water? Explain.

Q10 A ball is thrown downward with a large starting velocity.

A. Will this ball reach the ground sooner than one that is dropped from rest at the same time from the same height? Explain.

B. Will this ball accelerate more rapidly than one that is dropped with no initial velocity? Explain.

No because the velocity is increasing

Yes because it will have a higher average velocity

No the acceleration is the same

.

.

.

QUIZZES


21

Q14 A ball is thrown straight upward. At the very top of its flight, the velocity of the ball is zero. Is its acceleration at this point also zero? Explain.

Q15 A ball rolls up an inclined plane, slows to a stop, and then rolls back down. Do you expect the acceleration to be constant during this process? Is the velocity constant?

Q19 Is it possible for an object to have a horizontal component of velocity that is constant at the same time that the object is accelerating in the vertical direction? Explain by giving an example, if possible.

No the acceleration is 9.8m/s2 down

The acceleration is constant the velocity is not

Yes a projectile


22

Ch 3 E4

Heart beat = 75 beats/minutea) What is the time between pulses?b)How far does an object fall in this time?(from rest)

a) t = 60/75 = 0.8 s

b) d = v0t + ½ 9.8t2 = 3.136 m9.8m/s2


23

Ch 3 E8Ball thrown up at 15 m/sa) How high after 1 second?b)How high after 2 seconds?

After 1 sec d = v0t + ½ at2 = 15 – 4.9 = 10.1 mAfter 2 sec d = 15 x 2 – ½ 9.8 x 22 = 10.4 mTime to top v = v0 + at t = 15/9.8 = 1.53 s

Height at top d = 11.48m

+

g

15 m/s

t = 1.53 s

t = 2 s


24

Ch 3 E10

V0 = 18 m/s a = - 2 m/s2

a) What is v after 4 seconds?b)What is time to top?

a) v = v0 + at

b) v = 0 t = 18/2 = 9s

+

a=2m/s2

18 m/s

= 18 – 2 x 4 = 10m/s


25

Ch 3 E16

V0v = 30 m/s V0H = 30 m/sg = - 9.8m/s2

a) What is time to top?b)What is the range?

a) v = v0 + at t = 30/9.8 = 3.06s tR = 6.12s

b) d = 30 x tR = 183.6m

+30 m/s

30 m/s

g


26

Ch 3 CP2V01 = 0 m/s V02 = 12 m/sa) What are the velocities after 1.5s?b) How far has each ball dropped in 1.5s?c) Does the velocity difference change?

a) v1 = at = 9.8 x 1.5 = 14.7m/s

v2 = 12 + 9.8 x 1.5 = 26.7m/sb)d1 = ½at2 = 11.03m

d2 = v2t + ½at2 = 29.03mc) No

12 m/s

1 2


27

Ch 3 CP4

a) v = v0 + at time to top = 200/9.8 = 20.4s

time to range = 400/9.8 = 40.8s

b) d = 346 x 40.8 = 14120m

c) ↑346 →200 time = 692/9.8 = 70.6s

d = 200 x 70.6 = 14120

V0v = 200m/s v0H = 346m/sa) How long in the air?b) How far?c) v0v = 346 v0H = 200

200m/s

346m/s

g


28

Extinction of the dinosaursThere are accurate ways to look at events in geologic

time back to the formation of the earth about 4.5 billion years ago.Radioactive elements are very accurate clocks and sedimentary layers reveal geologic events as a function of time. 65 million years ago the extinction of the dinosaurs 250 million years ago over 90% of all species became extinct.

Experimental measurements show that the 65 million extinction occurred about when a large asteroid impacted in the gulf of Mexico near the Yucatan peninsula, enveloping the Earth in a cloud of dust.


29

Sedimentary layersLooking back in time

Iridium Hill Montana


30

Sedimentary layers


31

Asteroid impact simulation

Less than a minute after impact, the dissipation of the asteroid kinetic energy produces a stupendous explosion that melts, vaporizes, and ejects a substantial volume of calcite, granite, and water. The dominant feature here is the conical curtain of hot debris that has been ejected and is now falling back to Earth. The turbulent material inside this curtain is still being accelerated by the explosion from the crater excavation.

one dimensional motion and gravity

Documents