One Dimensional Kinematics: Problem Solving Kinematics in Two-Dimensions: Law of Addition of...

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One Dimensional Kinematics: Problem Solving Kinematics in Two- Dimensions: Law of Addition of Velocities Projectile Motion 8.01 W02D1

Transcript of One Dimensional Kinematics: Problem Solving Kinematics in Two-Dimensions: Law of Addition of...

Page 1: One Dimensional Kinematics: Problem Solving Kinematics in Two-Dimensions: Law of Addition of Velocities Projectile Motion 8.01 W02D1.

One Dimensional Kinematics:Problem Solving

Kinematics in Two-Dimensions:Law of Addition of Velocities

Projectile Motion

8.01W02D1

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Today’s Reading Assignment:

Young and Freedman:

3.1-3.3, 3.5

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One-Dimensional Kinematics:Review

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Summary: Kinematics

Acceleration is a non-constant function of time

Change in velocity

Change in position

,0

0

( ) ( )t t

x x x

t

v t v a t dt′=

′=

′ ′− = ∫

( )xa t

0

0

( ) ( )t t

x

t

x t x v t dt′=

′=

′ ′− = ∫

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Special Case: Constant Acceleration

Acceleration

Velocity

Position

v

x(t) =vx,0 + axt

20 ,0

1( )

2x xx t x v t a t= + +

ax=constant

x(t) −x0 = (vx,0 + ax ′t )d ′t

′t =0

′t =t

∫ =vx,0t+12

axt2

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Problem Solving Strategies: One-Dimensional Kinematics

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Strategy: One Dimensional Kinematics

1. Define an appropriate coordinate system

2. Integrate or differentiate

A coordinate system consists of:

1. An origin at a particular point in space2. A set of coordinate axes with scales and labels3. Choice of positive direction for each axis with a set of unit vectors 4. Choice of type: Cartesian or Polar or Spherical

Example: Cartesian One-Dimensional Coordinate System

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Gravitational Free Fall: Constant Acceleration

Special Case:

Choose coordinate system with x-axis vertical, origin at ground, and positive unit vector pointing upward.

Constant Acceleration:

Velocity

Position:

ax =−g=−(9.8 m⋅s-2 )

v

x(t) =vx,0 −gt

x(t) =x0 + vx,0t−

12

gt2

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Concept Question: One Dimensional Kinematics

A person standing at the edge of a cliff throws one ball straight up and another ball straight down, each at the same initial speed. Neglecting air resistance, which ball hits the ground below the cliff with the greater speed:

1. ball initially thrown upward;

2. ball initially thrown downward;

3. neither; they both hit at the same speed.

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Table Problem: Car and Bicycle

A car is driving through a green light at t = 0 located at

x = 0 with an initial known speed vc,o. At a known time t1 the car slows down and comes to a rest at an unknown time t2.

The acceleration of the car as a function of time is given by

where b is a known positive constant. A bicycle rider is riding at an unknown constant speed vb,o and at t = 0 is a distance d behind the car. The bicyclist reaches the car when the car just comes to rest. Find the speed of the bicycle. Discuss with your group what quantities you need in order to determine the speed of the bicycle. Then make a plan for solving this problem.

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Law of Addition of Velocities

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Reference Systems

Use coordinate system as a ‘reference frame’ to describe the position, velocity, and acceleration

of objects.

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Inertial Reference Frames

Two reference frames.

Origins need not coincide.

One moving object has different

position vectors in different frames

Relative velocity between the two reference frames

is constant since the relative acceleration is zero

1 2= +r R rrr r

d dt=V Rr r

d dt= =A V 0r rr

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Law of Addition of Velocities Suppose the object is moving; then, observers in

different reference frames will measure different velocities

Velocity of the object in Frame 1:

Velocity of the object in Frame 2:

Velocity of an object in two different reference frames

1 1d dt=v rrr

2 2d dt=v rrr

drr

1

dt=

drRdt

+drr2

dt

1 2= +v V vrr r

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Table Problem: Law of Addition of Velocities

Suppose two cars, Car 1, and Car 2, are traveling along roads that are perpendicular to each other. Reference Frame A is at rest with respect to the ground. Reference Frame B is at rest with respect to Car 1. Choose unit vectors such that Car 1 is moving in the positive y-direction, and Car 2 is moving in the positive x-direction in reference Frame A.

a) What is the vector description of the velocity of Car 2 in Reference Frame B?

b) What is the magnitude of the velocity of Car 2 as observed in Reference Frame B?

c) What angle does the velocity of Car 2 make with respect to the positive x-direction as observed in Reference Frame B?

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Two-Dimensional Motion

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Vector Description of Motion

Position

Displacement

Velocity

Acceleration

ˆ ˆ( ) ( ) ( )t x t y t= +r i jr

( ) ( )ˆ ˆ ˆ ˆ( ) ( ) ( )x y

dx t dy tt v t v t

dt dt= + ≡ +v i j i j

r

( )( ) ˆ ˆ ˆ ˆ( ) ( ) ( )yxx y

dv tdv tt a t a t

dt dt= + ≡ +a i j i j

r

ˆ ˆ( ) ( ) ( )t x t y tΔ =Δ +Δr i jr

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Summary Constant Acceleration

0, 0, ,x x x y y yv v a t v v a t= + = +

x =x0 + vx0t+

12

axt2 , y=y0 + vy0t+

12

ayt2

2a

x(x −x0 ) =vx

2 −vx,02

Components of Velocity:

Components of Position:

Eliminating t:

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Demo:

Relative Motion Gun A7Video Link:

http://tsgphysics.mit.edu/front/index.php?page=demo.php?letnum=A%207&show=0

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Kinematics in Two DimensionsProjectile Motion

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Projectile Motion: Constant Acceleration

A projectile is fired from a height y0 with an initial speed v0 at an angle θ above the horizontal. Ignore air resistance.

Choice of coordinate system: Cartesian

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Components of Acceleration: Constant and Zero

y-component:

x-component:

a

y=−g

ax=0

29.8 m/sg =

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Kinematic Equations –Zero Acceleration: x-components Acceleration :

Velocity :

Position :

0xa =

0 ,0( ) xx t x v t= +

,0( )x xv t v=

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Kinematic Equations: Constant Acceleration: y-components :Acceleration:

Velocity:

Position:

ya g=−

20 ,0

1( )

2yy t y v t gt= + −

,0( )y yv t v gt= −

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Concept Question: Top of Flight

Consider the path of a ball moving along a path through the air under the action of the gravitational force. You may neglect the effects of air friction. As it reaches the highest point in its arc, which of the following statement is true?

1) The magnitudes of the velocity and acceleration are zero.

2) The magnitude of the velocity is at a minimum but not equal to zero.

3) The magnitude of the velocity is equal to zero, and the magnitude of the acceleration is constant and not equal to zero.

4) The magnitude of the velocity is at a minimum but not equal to zero and the magnitude of the acceleration is zero.

5) Neither the magnitudes of acceleration or velocity has yet attained its minimum value.

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Concept Q.: Which Hits First?A person simultaneously throws two objects in the air. The objects leave the person’s hands at different angles and travel along the parabolic trajectories indicated by A and B in the figure below. Which of the following statements best describes the motion of the two objects? Neglect air resistance.

1. The object moving along the trajectory A hits the ground before the object moving along the trajectory B.

2. The object moving along the higher trajectory A hits the ground after the object moving along the lower trajectory B.

3. Both objects hit the ground at the same time.

4. There is not enough information specified in order to determine which object hits the ground first.

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Table Problem: Hitting a Baseball

A batter hits a baseball into the air with an initial speed, v0 = 50 m/s , and makes an angle 0 = 30o with respect to the horizontal. How far does the ball travel if it is caught at exactly the same height that it is hit from? When the ball is in flight, ignore all forces acting on the ball except for gravitation. Let g = 9.8 m/s2.

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Initial Conditions Initial position:

Initial velocity:

Velocity components:

Initial speed:

Direction:

0 0 0ˆ ˆx y= +r i j

r

0 ,0 ,0ˆ ˆ( ) x yt v v= +v i j

r

,0 0 0 ,0 0 0cos , sinx yv v v v = =

( )1/ 22 20 0 ,0 ,0| | x yv v v= = +v

r

,010

,0

tan y

x

v

v − ⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

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Orbit Equation

,02 2,0 2

,0 ,0

1 1( ) ( ) ( )

2 2y

yx x

v gy t v t gt x t x t

v v= − = −

0 0( , ) (0,0)x y =

1tandy

dx − ⎛ ⎞= ⎜ ⎟

⎝ ⎠

The slope of the curve y(t) vs. x(t) at any point determines the direction of the velocity

,0,0

( )( ) x

x

x tx t v t t

v= ⇒ =

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Concept Q.: 2-Dim Kinematics An object moves along a parabolic orbit under the influence of gravity. At each point along the orbit,

1. the magnitude of the velocity can be determined from the slope of the tangent line to the graph of y vs. x but not the direction.

• the magnitude and direction of the velocity can be determined from the slope of the tangent line to the graph of y vs. x.

• neither the magnitude nor the direction of the velocity can be determined from the slope of the tangent line to the graph of y vs. x.

• the direction of the velocity can be determined from the slope of the tangent line to the graph of y vs. x but not the magnitude.

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Worked Example: Stuffed Animal and the Gun

A stuffed animal is suspended at a height h above the ground. A physics demo instructor has set up a projectile gun a horizontal distance d away from the stuffed animal. The projectile is initially a height s above the ground. The demo instructor fires the projectile with an initial velocity of magnitude v0

just as the stuffed animal is released. Find the angle the projectile gun must be aimed in order for the projectile to strike the stuffed animal. Ignore air resistance.

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Demo:

Stuffed Animal and Gun A6 Video Link

http://tsgphysics.mit.edu/front/index.php?page=demo.php?letnum=A%207&show=0

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Table Problem: Catching a Softball

A softball is hit over a third baseman’s head. The third baseman, as soon as the ball is hit, turns around and runs straight backwards a constant speed of 7 m/s for a time interval 2 s and catches the ball at the same height it left the bat. The third baseman was initially a distance 18 m from home plate. What was the initial speed and angle of the softball when it left the bat?

 

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Next Reading Assignment: W02D2

Young and Freedman:

4.1-4.6, 5.1-5.3