Chap. 3: Kinematics (2D) - Physics and Astronomy at TAMUpeople.physics.tamu.edu/kamon/teaching/phys218/slide… ·  · 2013-01-29Kinematics (2D) Chap. 3: Kinematics (2D) ... Projectile

Kinematics (2D)

Chap. 3: Kinematics

(2D)

Recap: Kinematics (1D)

1. Vector Kinematics

2. Projectile Motion

3. Uniform Circular Motion

4. Relative Velocity

1

Kinematics (2D)

Last, This and Next Weeks

• [Last Week]

– Chap. 1 and Chap. 2

• [This Week]

– Chap. 3 Chap. 4

• [Next Week]

– Chap. 5

Question?

2

Kinematics (2D)

Critical Thinker

Kinematics (2D)

Laws, Principles

(so-called formulae)

Solution A Solution B Solution C

Problem

Answer

Critical ThinkerCritical Thinker

One would just plug in the numbers and if it didn't come out to be a

correct answer then he/she would just change the positive to

negative and so on.

What’s wrong with this? This is a typical practice of memorizing

the problems. In the exam, you wouldn’t necessarily expect to see

the same problems. What will be the safest bet? First, see

“concept”. This should take you to a proper setting.

ISEE

5

Kinematics (2D)

Common Steps for Thinker

Step 1: Draw a diagram (or picture) of the

situation, with coordinate axes.

Step 2: Think about which principle(s) of physics

apply in this problem.

Step 3: Write down principle(s).

Step 4: Solve them.

ISEE

ISEE

ISEE

6

Kinematics (2D)

Where should the professor be

when you release the egg?

Egg Drop

Chap.2: Problem II-2

x = x0 + v0x t + ½ ax t 2

y = y0 + v0y t + ½ ay t 2

7

A set of 1-D kinematic eqs. for

motion of each of TWO bodies

with constant acceleration

[Chap.3] a set of 2-D kinematic

eqs. for motion of ONE body

Kinematics (2D)

Vector Kinematics

x = x0 + v0x t + ½ ax t 2 (1)

vx = v0x + ax t (2)

vx2 = v0x

2 + 2ax (x – x0) (3)

y = y0 + v0y t + ½ ay t 2 (4)

vy = v0y + ay t (5)

vy2 = v0y

2 + 2ay (y – y0) (6)

One Set of 2-D Kinematic Eqs. for Motion of

One Body with Constant Acceleration

2

00

2

002

00

2

1

2

1

2

1

tavyy

tavxx

tatvrr

yy

xx

x component y component

8

Kinematics (2D)

I

III

G

Visual Vector Kinematics

II

9

Kinematics (2D)

How to study Chap. 3 HWs

Key Categories Examples Exercises, Problems

I 1 body, with q = 0 3.6 10, 71

II 1 body, with q 3.7, 3.8, 3.9 19, 21, 47, 52, 54, 60, 62,

63, 64, 65, 67, 71

III 2 bodies with 2 a’s 3.10 11, 51, 56

C Circular Motion 3.11, 3.12 25, 28, 30

R Relative Velocity 3.12, 3.13, 3.14 32, 34

10

Version 2

Kinematics (2D) Direction of Velocity Direction of Position

33

22

32

00

)m/s (0.156

1 m/s) (1.0)(

)m/s 0.5(2

1m) (2.0)(

6

1

2

1

ttty

ttx

tbtatvrr

11

Kinematics (2D)

x(t) = (2.0 m) – (1/2) (0.50 m/s2) t 2

vx(t) = – (0.50 m/s2) t

ax(t) = – 0.50 m/s2

Example 1(a): Motion along with x axis

Kinematic Eqs. are related by derivatives and integrals.

Motion with constant

acceleration.

Initial velocity = zero

Initial position = 2.0 m

[Eq. 1]

12

Kinematics (2D)

y(t) = (1.0 m/s) t + (1/6) (0.150 m/s3) t 3

vy(t) = (1.0 m/s) + (1/2) (0.150 m/s3) t 2

ay(t) = (0.150 m/s3) t

Example 1(b): Motion along with y axis

Kinematic Eqs. are related by derivatives and integrals.

Motion with varying

acceleration.

Initial velocity = 1 m/s

Initial position = zero

[Eq. 2]

13

Kinematics (2D)

ONE 2D Motion (Eqs. 1 & 2)

Study each motion between 0 s and 10 s

using spreadsheet:

• Calculate position every 0.5 s and plot them.

• Calculate displacement

• Calculate average velocity

• Calculate velocity every 0.5 s

• Calculate acceleration every 0.5 s

14

Kinematics (2D)

position displacement

average

velocity velocity acceleration

time (s) x (m) y (m) distance dx dy dx/dt dy/dt vx vy ax ay

0.00 2.00 0.00 2.00 -0.063 0.503 -0.125 1.006 0.000 1.000 -0.500 0.000

0.50 1.94 0.50 2.00 -0.188 0.522 -0.375 1.044 -0.250 1.019 -0.500 0.075

1.00 1.75 1.03 2.03 -0.313 0.559 -0.625 1.119 -0.500 1.075 -0.500 0.150

1.50 1.44 1.58 2.14 -0.438 0.616 -0.875 1.231 -0.750 1.169 -0.500 0.225

2.00 1.00 2.20 2.42 -0.563 0.691 -1.125 1.381 -1.000 1.300 -0.500 0.300

2.50 0.44 2.89 2.92 -0.688 0.784 -1.375 1.569 -1.250 1.469 -0.500 0.375

3.00 -0.25 3.68 3.68 -0.813 0.897 -1.625 1.794 -1.500 1.675 -0.500 0.450

3.50 -1.06 4.57 4.69 -0.938 1.028 -1.875 2.056 -1.750 1.919 -0.500 0.525

4.00 -2.00 5.60 5.95 -1.063 1.178 -2.125 2.356 -2.000 2.200 -0.500 0.600

4.50 -3.06 6.78 7.44 -1.188 1.347 -2.375 2.694 -2.250 2.519 -0.500 0.675

5.00 -4.25 8.13 9.17 -1.313 1.534 -2.625 3.069 -2.500 2.875 -0.500 0.750

5.50 -5.56 9.66 11.15 -1.438 1.741 -2.875 3.481 -2.750 3.269 -0.500 0.825

6.00 -7.00 11.40 13.38 -1.563 1.966 -3.125 3.931 -3.000 3.700 -0.500 0.900

6.50 -8.56 13.37 15.87 -1.688 2.209 -3.375 4.419 -3.250 4.169 -0.500 0.975

7.00 -10.25 15.58 18.65 -1.813 2.472 -3.625 4.944 -3.500 4.675 -0.500 1.050

7.50 -12.06 18.05 21.71 -1.938 2.753 -3.875 5.506 -3.750 5.219 -0.500 1.125

8.00 -14.00 20.80 25.07 -2.063 3.053 -4.125 6.106 -4.000 5.800 -0.500 1.200

8.50 -16.06 23.85 28.76 -2.188 3.372 -4.375 6.744 -4.250 6.419 -0.500 1.275

9.00 -18.25 27.23 32.78 -2.313 3.709 -4.625 7.419 -4.500 7.075 -0.500 1.350

9.50 -20.56 30.93 37.15 -2.438 4.066 -4.875 8.131 -4.750 7.769 -0.500 1.425

10.00 -23.00 35.00 41.88 -2.563 4.441 -5.125 8.881 -5.000 8.500 -0.500 1.500 15

Kinematics (2D)

position displacement

average

velocity velocity acceleration

time (s) x (m) y (m) distance dx dy dx/dt dy/dt vx vy ax ay

0.00 2.00 0.00 2.00 -0.063 0.503 -0.125 1.006 0.000 1.000 -0.500 0.000

0.50 1.94 0.50 2.00 -0.188 0.522 -0.375 1.044 -0.250 1.019 -0.500 0.075

1.00 1.75 1.03 2.03 -0.313 0.559 -0.625 1.119 -0.500 1.075 -0.500 0.150

1.50 1.44 1.58 2.14 -0.438 0.616 -0.875 1.231 -0.750 1.169 -0.500 0.225

2.00 1.00 2.20 2.42 -0.563 0.691 -1.125 1.381 -1.000 1.300 -0.500 0.300

2.50 0.44 2.89 2.92 -0.688 0.784 -1.375 1.569 -1.250 1.469 -0.500 0.375

3.00 -0.25 3.68 3.68 -0.813 0.897 -1.625 1.794 -1.500 1.675 -0.500 0.450

3.50 -1.06 4.57 4.69 -0.938 1.028 -1.875 2.056 -1.750 1.919 -0.500 0.525

4.00 -2.00 5.60 5.95 -1.063 1.178 -2.125 2.356 -2.000 2.200 -0.500 0.600

4.50 -3.06 6.78 7.44 -1.188 1.347 -2.375 2.694 -2.250 2.519 -0.500 0.675

5.00 -4.25 8.13 9.17 -1.313 1.534 -2.625 3.069 -2.500 2.875 -0.500 0.750

5.50 -5.56 9.66 11.15 -1.438 1.741 -2.875 3.481 -2.750 3.269 -0.500 0.825

6.00 -7.00 11.40 13.38 -1.563 1.966 -3.125 3.931 -3.000 3.700 -0.500 0.900

6.50 -8.56 13.37 15.87 -1.688 2.209 -3.375 4.419 -3.250 4.169 -0.500 0.975

7.00 -10.25 15.58 18.65 -1.813 2.472 -3.625 4.944 -3.500 4.675 -0.500 1.050

7.50 -12.06 18.05 21.71 -1.938 2.753 -3.875 5.506 -3.750 5.219 -0.500 1.125

8.00 -14.00 20.80 25.07 -2.063 3.053 -4.125 6.106 -4.000 5.800 -0.500 1.200

8.50 -16.06 23.85 28.76 -2.188 3.372 -4.375 6.744 -4.250 6.419 -0.500 1.275

9.00 -18.25 27.23 32.78 -2.313 3.709 -4.625 7.419 -4.500 7.075 -0.500 1.350

9.50 -20.56 30.93 37.15 -2.438 4.066 -4.875 8.131 -4.750 7.769 -0.500 1.425

10.00 -23.00 35.00 41.88 -2.563 4.441 -5.125 8.881 -5.000 8.500 -0.500 1.500 16

Kinematics (2D)

position displacement

average

velocity velocity acceleration

time (s) x (m) y (m) distance dx dy dx/dt dy/dt vx vy ax ay

0.00 2.00 0.00 2.00 -0.063 0.503 -0.125 1.006 0.000 1.000 -0.500 0.000

0.50 1.94 0.50 2.00 -0.188 0.522 -0.375 1.044 -0.250 1.019 -0.500 0.075

1.00 1.75 1.03 2.03 -0.313 0.559 -0.625 1.119 -0.500 1.075 -0.500 0.150

1.50 1.44 1.58 2.14 -0.438 0.616 -0.875 1.231 -0.750 1.169 -0.500 0.225

2.00 1.00 2.20 2.42 -0.563 0.691 -1.125 1.381 -1.000 1.300 -0.500 0.300

2.50 0.44 2.89 2.92 -0.688 0.784 -1.375 1.569 -1.250 1.469 -0.500 0.375

3.00 -0.25 3.68 3.68 -0.813 0.897 -1.625 1.794 -1.500 1.675 -0.500 0.450

3.50 -1.06 4.57 4.69 -0.938 1.028 -1.875 2.056 -1.750 1.919 -0.500 0.525

4.00 -2.00 5.60 5.95 -1.063 1.178 -2.125 2.356 -2.000 2.200 -0.500 0.600

4.50 -3.06 6.78 7.44 -1.188 1.347 -2.375 2.694 -2.250 2.519 -0.500 0.675

5.00 -4.25 8.13 9.17 -1.313 1.534 -2.625 3.069 -2.500 2.875 -0.500 0.750

5.50 -5.56 9.66 11.15 -1.438 1.741 -2.875 3.481 -2.750 3.269 -0.500 0.825

6.00 -7.00 11.40 13.38 -1.563 1.966 -3.125 3.931 -3.000 3.700 -0.500 0.900

6.50 -8.56 13.37 15.87 -1.688 2.209 -3.375 4.419 -3.250 4.169 -0.500 0.975

7.00 -10.25 15.58 18.65 -1.813 2.472 -3.625 4.944 -3.500 4.675 -0.500 1.050

7.50 -12.06 18.05 21.71 -1.938 2.753 -3.875 5.506 -3.750 5.219 -0.500 1.125

8.00 -14.00 20.80 25.07 -2.063 3.053 -4.125 6.106 -4.000 5.800 -0.500 1.200

8.50 -16.06 23.85 28.76 -2.188 3.372 -4.375 6.744 -4.250 6.419 -0.500 1.275

9.00 -18.25 27.23 32.78 -2.313 3.709 -4.625 7.419 -4.500 7.075 -0.500 1.350

9.50 -20.56 30.93 37.15 -2.438 4.066 -4.875 8.131 -4.750 7.769 -0.500 1.425

10.00 -23.00 35.00 41.88 -2.563 4.441 -5.125 8.881 -5.000 8.500 -0.500 1.500 17

Kinematics (2D) Direction of velocity Direction of acceleration 18

Kinematics (2D)

Projectile Motions: I, II & III

I II

III

21

Kinematics (2D)

Projectile Motion

2.00 m/s

a = (0) i + (g) j ￫ ￫ ￫

Horizontal and vertical motions

analyzed separately. 22

Kinematics (2D)

Question: How fast must the motorcycle leave the cliff-top?

[Quick Quiz 1]

Is this a motion

with a constant

acce lera t ion or

with a vary ing

acceleration?

24

Kinematics (2D)

This figure tells you a lot!

27

Kinematics (2D)

Further Look at Projectile Motion

(2) ax = 0

ay = g = 9.80 m/s2

(1) Choose an origin &

an x-y coordinate system

vx = constant

(3) vy = 0

(4) y = 0

28

Kinematics (2D)

Projectile Motion

[Quick Quiz 2]

Is this a motion

with a constant

acceleration or

with a varying

acceleration?

?f ina l v

29

Kinematics (2D)

30

x = v0x t (1)

vx = v0x (2)

vx2 = v0x

2 (3)

y = v0y t + ½ (g) t 2 (4)

vy = v0y + (g) t (5)

vy2 v0y

2 = 2 (g) y (6)

Kinematics (2D)

Example 1: A Hunter

A hunter aims directly at a target (on the same

level) 65.0 m away. Note that the magnitude of

gravitational acceleration on the Earth is g = 9.80 m/s2.

(a) (10 pts) If the bullet leaves the gun at a speed

o f 145 m/s , by how much wil l i t miss the

target?

(b) (15 pts) At what angle should the gun be aimed

so the target will be hit?

33

Kinematics (2D)

Example 1(a)

d

R = 65.0 m y

x t = 0, v0 = 145 m/s

Step 3: Write down kinematic equations.

Step 4: Solve them.

Step 2: Think about which principle(s) of

physics apply in this problem. (

kinematic eqs.)

t = T

ISEE

ISEE

ISEE

Step 1: Draw a diagram (or picture) of the

situation, with coordinate axes.

35

Kinematics (2D)

Example 1: A Hunter

A hunter aims directly at a target (on the same

level) 65.0 m away. Note that the magnitude of

gravitational acceleration on the Earth is g = 9.80 m/s2.

(a) (10 pts) If the bullet leaves the gun at a speed

o f 145 m/s , by how much wil l i t miss the

target?

(b) (15 pts) At what angle should the gun be aimed

so the target will be hit?

37

Kinematics (2D)

Example 1(b)

Step 1: Draw a diagram (or picture) of the

situation, with coordinate axes.

Step 3: Write down kinematic equations.

Step 4: Solve them.

Step 2: Think about which principle(s) of physics

apply in this problem. ( kinematic eqs.)

R = 65.0 m y

x t = 0, v0 = 145 m/s

q

t = T

39

Kinematics (2D)

Accidentally, you get the same answer! ( q = 0.868o)

Your claim: “I should get a full credit because the final answer

is correct.”

My response: “No”

What is wrong with this?

COMMON MISTAKE #1

R

d

tan q = d / R ,

where d is from

part (a).

Example 1(b) – Cont’d

40

Kinematics (2D)

Wrong True (W-T)/(T)

R v0 q q

[m] [m/s] [deg] [deg] [%]

65.0 145 0.867886 0.868086 0.023%

65.0 80.0 2.84901 2.85609 0.25%

65.0 40.0 11.2583 11.7305 4.0%

65.0 26.0 25.2277 35.2214 28%

65.0 25.3 26.4543 42.1839 37%

Example 1(b) – Cont’d

COMMON MISTAKE #1

You will know the correct answer

in a formula form later. 41

Kinematics (2D)

R = 65.0 m y

x t = 0, v0 = 145 m/s

COMMON MISTAKE #2

Example 1(b) – Cont’d

t = T(a) = R/v0 = 0.448 s

Question: What’s wrong with T(a) = R/v0 ?

Hint #1: Remember vector kinematics

Hint #2: Remember the velocity is a vector quantity

How to find T?

Hint #3: v0x = ? [145 m/s is a wrong choice.]

44

Kinematics (2D)

(a) d = 0.985 m

(b) q = 0.5 x sin1(9.80 x 65.0/1452) = 0.868°

Note that the following is a wrong approach for part (b):

q = tan1(0.985/65.0) = 0.868°

even though the answer numerically agrees with the

correct one. Below is an exercise of two approaches by

changing the magnitude of the velocity (v0), but using

the same distance (R) of 65.0 m, where you see a larger

discrepancy as v0 decreases.

Example 1 - Summary

45

Kinematics (2D)

y = 0

Vy = 0

y = 1.00m

(1) Choose an origin &

an x-y coordinate system

(2) ax = 0

ay = g = 9.80 m/s2

(3) Use kinematic eqs.

in x and y separately.

Vx = constant

47

Kinematics (2D)

A boy on a small hill aims his water-balloon slingshot horizontally,

straight at second boy hanging from a tree branch a distance d

away. At the instant the water balloon is released, the second boy

lets go and falls from the tree, hoping to avoid being hit. Show

that he made the wrong move.

48

Kinematics (2D)

A boy on a small hill aims his water-balloon slingshot upward,

directly at second boy hanging from a tree branch. At the instant

the water balloon is released, the second boy lets go and falls from

the tree, hoping to avoid being hit. Show that he made the wrong

move.

49

Kinematics (2D)

Same

Concept

200 m, given x?

d, given H?

H?

?

50

Kinematics (2D)

In-Class Test #1

6

51

Answer: ______

Kinematics (2D)

Example 1+1

Kinematics (2D)

A boy on a small hill aims his water-balloon slingshot horizontally,

straight at second boy hanging from a tree branch a distance d

away. At the instant the water balloon is released, the second boy

lets go and falls from the tree, hoping to avoid being hit. Show that

he made the wrong move.

Kinematics (2D)

A boy on a small hill aims his water-balloon slingshot upward,

directly at second boy hanging from a tree branch. At the instant

the water balloon is released, the second boy lets go and falls from

the tree, hoping to avoid being hit. Show that he made the wrong

move.

•Motion of the water-balloon is …

exactly the same problem as in Example 1

•Plus: the 2nd object (2nd boy hanging from a tree branch)

•Principle: Kinematic eqs. for each boy See the textbook

carefully. 53

Kinematics (2D)

t = 7.6 s

Example 2: A projectile is launched from ground level to the top of a cliff

which is R = 195 m away and H = 155 m high. The projectile lands on top of

the cliff T = 7.60 s after it is fired. Use 2sinq cosq = sin2q if necessary. The

acceleration due to gravity is g = 9.80 m/s2 pointing down. Ignore air friction.

a. Find the initial velocity of the projectile (magnitude v0 and direction q).

b. Find a formula of tanq in terms of g, R, H and T.

54

Kinematics (2D)

Chap. 3: Kinematics

(2D) - Part II

55

Recap: Kinematics (1D)

1. Vector Kinematics

2. Projectile Motion

3. Uniform Circular Motion

4. Relative Velocity

Kinematics (2D)

How to study Chap. 3 HWs

Key Categories Examples Exercises, Problems

I 1 body, with q = 0 3.6 10, 71

II 1 body, with q 3.7, 3.8, 3.9 19, 21, 47, 52, 54, 60, 62,

63, 64, 65, 67, 71

III 2 bodies with 2 a’s 3.10 11, 51, 56

C Circular Motion 3.11, 3.12 25, 28, 30

R Relative Velocity 3.12, 3.13, 3.14 32, 34

56

Version 2

Kinematics (2D)

Uniform Circular Motion

1. Kinematics

2. Coordinates: r-f

3. Newton’s 2nd Law (Future)

57

Kinematics (2D)

Kinematics: Acceleration Rate Change in Velocity

t2

t1

Center-seeking acceleration as Dt0

t1

t2

Dt

58

Kinematics (2D)

Center-seeking Acceleration

vvlvl

vv

lvv

l

v

v

rtrtr

t

r

r

t D

D

D

D

D

D

DD

D

D

D 0

rr

a2

radˆ

v

r̂

59

Kinematics (2D)

Example Problem 1 The radius of the semicircular drive in front of the

Administration Building is 200 m. How fast could a sports car

negotiate the turn, assuming that it could achieve the magnitude

of a radial acceleration arad = 0.8g? Ignore the stop sign at the

intersection with East Gate Drive.

60

Kinematics (2D)

Example Problem 2 People advocating space colonies say that it will be simple to

simulate gravity. One need only build a space module in the

form of a doughnut (torus) and let it revolve at an appropriate

rate. Consider a torus in circumference C = 10,000 m; it could

easily accommodate 25 dwellings plus landscaping and

gardens. How many times would the torus have to revolve

each hour in order that the magnitude of the radial

acceleration at the rim equal g?

62

rr

arad ˆv2

Kinematics (2D)

Problem: Relative Velocity in 1D

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. A motor scooter is being ridden on the flatcar. Find the velocity (magnitude and direction) of the scooter relative to the flatcar if its velocity relative to the observer on the ground is: a) 20.0 m/s to the right; b) 4.00 m/s to the left; c) zero.

64

VP/A = VP/B + VB/A ￫ ￫ ￫

Kinematics (2D)

2.5 m/s

q

VP/A=?

Bank

Raft

Person

VP/B

Shore

Boat

VW/S=?

VB/W = 2.40 m/s

VB/S

VB/A

67