Chapter 3: Kinematics in Two Dimensions p
Chapter p 3: Kinematics in Two Dimensions Vectors and Scalars A scalar is a number with units. units It can be positive, positive negative, or zero. • Time: 100 s • Distance and speed are scalars, although they cannot be negative because of the way they are defined. defined A vector is a quantity which specifies magnitude and direction. • Velocity: 4 m/s East • Acceleration: 10 m/s2 to the right
Notation Vector qquantities are represented p usingg bold type or an arrow above the symbol . The g of a vector ((a scalar)) is magnitude represented using the absolute value symbol yp or without bold type: G velocity: v = v = 5 m/s East | v | = v = 5 m/s
Arrow Representation Arrows are also used to represent vectors themselves. • The length of the arrow is proportional to the magnitude. • The direction of the arrow indicates the direction of the vector. 4 m/s East
2 m/s North
Adding Vectors Only add vectors with the same units units. For example: v3 = v1 + v2 i OK, is OK bbut v3 = x 1 + v2 does not make sense Remember that a vector -A A points opposite to A. A –A
Adding g Vectors Graphically p y To add two vectors A + B, put the tail of B at the h tip i off A and d connect the h tail il off A to the h tip of B.
Use a ruler and protractor to determine the magnitude g and direction of C.
Subtracting Vectors, and Multiplication by a Scalar The negative of a vector is a vector having the same length, but pointing in the opposite direction
So we can define the subtraction of one vector from another as
A vector can also be multiplied p byy a scalar c. We define this product so that cV has the same direction as V and a magnitude c|V|.
Components of a Vector A vector can be specified either by its magnitude and direction, or by its components in a coordinate system.
rθ r,θ
rx , ry
Be Careful The components of a vector are not always positive.
The angle is not always defined relative to the x axis.
Trigonometric Functions SOH CAH TOA opposite side Ay sin θ = = hypotnuse A adjacent side Ax cos θ = = hypotnuse A opposite side Ay tan θ = = adjacent side Ax A = Ax2 + Ay2 Be careful with your calculator! A you using Are i degrees d or radians? di ?
A Ay θ Ax
Example Find the x and y components p of a vector of magnitude 5, if its angle relative to the x axis is 200°.
Ax = A cos θ = 5cos ( 200 ) = −4.70 D
Ay = A sin θ = 5sin ( 200 ) = −1.71 D
Adding g Vectors Usingg Components • Break vectors up into components. • Add components. t • Find magnitude and direction of the sum from its components.
Example 12. The vectors are shown in the figure below. Their magnitudes are given are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with x axis.
Unit Vectors A unit vector is a vector with magnitude equal to 1 and pointing in a certain defined direction. direction Example: the unit vector in the x direction is usually written: xˆ o r iˆ Now I can write a vector as its components multiplied p by y the unit vector: G A = Ax xˆ + Ay yˆ
or G A = Ax iˆ + Ay ˆj where Ax and Ay are scalars.
Example Write the vector in the figure in unit vector notation. y
θ
G v
G v = 10 x
θ = 50°
Position and Displacement Vectors The position vector r points from the origin to a particular location.
The displacement vector Δr indicates the g in position: p change Δr = rf - ri
Writing of displacement in terms of components and unit vectors,
G Δr = ( x2 − x1 )iˆ + ( y2 − y1 ) ˆj + ( z2 − z1 )kˆ We defined the instantaneous velocity as
G G Δr dr G = v = lim Δt →0 Δt dt So we can write the instantaneous velocity as,
G dx ˆ dy ˆ dz ˆ j + k = vx iˆ + v y ˆj + vz kˆ v= i+ dt dt dt
Likewise for acceleration,,
G G G Δv v2 − v1 G aav = = Δt Δt The instantaneous acceleration is given by
G G Δv dv G a = lim = Δt →0 Δt dt so
G dvx ˆ dv y ˆ dvz ˆ G ˆ G ˆ G ˆ a= i+ j+ k = ax i + a y j + az k dt dt dt
Acceleration occurs whenever there is a g in velocity. y change
If a car travels in a circle at a constant speed of g 50 mi/hr, is it accelerating? YES! Because the direction of the velocity vector is changing. changing
Kinematic Equations for Constant Acceleration in 2 Dimensions x-component
x = xo + vxot + ax t 1 2
y-component 2
vx = vxo + ax t v = v + 2ax ( x − xo ) 2 x
2 xo
y = yo + v yot + 12 a y t 2 v y = v yo + a y t 2 v y2 = v yo + 2a y ( y − yo )
Projectile Motion Motion in the x direction is independent from motion in the y direction. We use the same equations from Chapter 2, 2 but for each dimension separately.
There are not really any new equations in this p chapter.
The equations you need. 1 2 x = x0 + v0 x t + ax t 2 1 2 y = y0 + v0 y t + a y t 2 vx = v0 x + ax t v y = v0 y + a y t v = v + 2ax ( x f − xi ) 2 x
2 0x
v = v + 2a y ( y f − yi ) 2 y
2 0y
NOTE For projectile motion problems, ax (the horizontal componant of the acceleration) will usually be zero since i usually ll there th will ill be b no acceleration in the x direction
Projectile Motion • Assume that acceleration of g gravityy is constant, downward and has a magnitude of g = 9.81 m/s2 • Air resistance is ignored • The Earth’s rotation is ignored g ⇒Horizontal velocity is constant: because ax = 0 ⇒Vertical motion governed by the constant acceleration of gravity
Motion of a Projectile Launched H i t ll Horizontally The dots Th d represents the h position of the object everyy 0.05 s.
t=1s
y0 = 8 m; x0 = 0 v0y = 0; v0x = 6 m/s ay = -9.81 m/s2; ax = 0 1. Verify the position of the object at t = 1s. 2. What would be the position of the object at t = 1 s if it were dropped (v0x = 0)?
1 2 x = 0 + v0 x t y = y0 + v0 y t + a y t 2 1 2 x = 6 ×1 = 6 m y = 8 − 9.8 × (1) = 3.1 m 2 __________________________________________________________
1 2 y = 8 − 9.8 × (1) = 3.1 m 2 x=0
Example p Pitcher’s mounds are raised to compensate for the vertical drop of the h ball b ll as it i travels l 18 m to the h catcher. h (a) If a pitch is thrown horizontally with an initial speed of 32 m/s (71 mi/hr), mi/hr) how far does it drop by the time it reaches the catcher? (b) If the speed of the pitch is increased, increased does the drop distance increase, decrease or stay the same? Explain. (c) If this baseball game were to be played on the moon, would the drop distance increase, decrease, or stay the same? Explain.
(a ) x = v0 x t 18 = 32treach −catcher
treach −catcher = 0.56 s
1 y = y0 + v0 y t − 9.8t 2 2 y0 − y = 1.54 1 54 m
1 2 y − y0 = − 9.8 × ( 0.56 ) = −1.54 m 2
(b) If the speed of the pitch increases, the time gets smaller, so the drop distance is less (c) On the moon “g” is less than 9.8 m/s2, so the drop distance is less.
General Launch Angle Consider an object launched from the origin at an g θ with respect p to the horizontal. angle y
vo
θ
x
What are the x and y components of the initial velocity vector?
Example: On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2 25 m/s 2.25 / at an angle l off 35.0° 35 0° above b the h horizontal. h i l If she h is i in i flight fli h for 1.60 s, how high above the water was she when she let go of the rope? p
What is the girl girl’ss minimum speed during her flight? What is her acceleration at the top of her trajectory?
1 y = y0 + v0 sin θ t − 9.81t 2 2 1 0 = y0 + 2.25sin 35D t − 9.81t 2 2 1 2 D 0 = y0 + 2.25sin 35 (1.6 ) − 9.81(1.6 ) 2 y0 = 10.49 10 49 m
Girl’s minimum speed is at the maximum in her trajectory! C you see why? Can h ? Girl s acceleration is 9.81 Girl’s 9 81 m/s2 at the top of her trajectory. trajectory What is her acceleration at other points on her trajectory?
Range The range g R of a pprojectile j is the horizontal distance it travels before landing.
⎛ v02 ⎞ R = ⎜⎜ ⎟⎟ sin 2θ ⎝g⎠
(BE CAREFUL!, This equation only works when the initial and final elevation are the same)
You should try to derive this equation! It’s not difficult. Just remember that x=R when the projectile hits the ground. What angle θ results in the maximum range? Remember 0°≤ θ ≤90°. What if we do not ignore air resistance?
Example A pprojectile j is fired at 200 m/s at an angle g of 45 degrees g above the horizontal. How far does it travel?
⎛ v02 ⎞ R = ⎜⎜ ⎟⎟ sin 2θ ⎝g⎠
⎛ (200m/s) 2 ⎞ o sin(2(45 )) = 4081m R=⎜ 2 ⎟ 9 8m/s ⎠ ⎝ 9.8m/s
Maximum Height The maximum height (and therefore the “hang hang time time”)) of a projectile depends only on the vertical component of its initial velocity. y At ymax, the vertical velocity vy is zero. v 2y = v02 y + 2aΔy 0 = v02 sin 2 θ + 2(− g ) ymax ymax
v02 sin 2 θ = 2g
Example From the previous example, example how high does the projectile go?
ymax
v sin θ = 2g 2 o
2
( 200m / s ) =
2
ymax
2
o
sin (45 ) = 1020m 2(9.8)
Uniform Circular Motion v1
C id a particle Consider ti l moving i in i a circle. i l Is the particle accelerating?
G G Δv dv G = a = lim Δt →0 Δt ddt - The speed of the particle is constant - The direction of the ball is changing
v2
What is the direction of the acceleration?
Magnitude of Radial Acceleration This type acceleration is called centripetal or radial acceleration. The magnitude of the acceleration is given by, by v2 aR = r
Period and Frequency The amount of time that it takes the particle to make one full revolution is called the period , T. The number of times that the particle goes around the circle per q y, ff. unit time is called the frequency,
1 T= f The speed of a particle moving around in a circle is given by,
distance 2π r = = 2π rf v= time T
Example 33. A shotputter 33 h t tt throws th throws th the th shot h t (mass ( = 7.3 7 3 kg) k ) with ith an initial speed of 14.0 m/s at a 40° angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a height of 2.2 m above the ground.
Example p 18. The position of a particular particle as a function of time is given by,
G 2ˆ ˆ ˆ r = (7.60 (7 60ti + 88.85 85 j − t k )m (a) Determine the particle’s velocity and acceleration as a function of time. (b) Describe the motion in each direction.
Example p Chapter 2 #40. A space vehicle accelerates uniformly from 65 m/s at t=0 to 162 m/s at 10.0 s. How far did it move b t between t 2 0 andd t=6.0 t=2.0s t 6 0 s??
Example A baseball is hit at an angle of 50 50° relative to the horizontal with an initial speed of 120 miles/hour. ((a)) H How hi highh does d it go?? (b) How long does it spend in the air? (c) How far does it travel?
Notation Vector qquantities are represented p usingg bold type or an arrow above the symbol . The g of a vector ((a scalar)) is magnitude represented using the absolute value symbol yp or without bold type: G velocity: v = v = 5 m/s East | v | = v = 5 m/s
Arrow Representation Arrows are also used to represent vectors themselves. • The length of the arrow is proportional to the magnitude. • The direction of the arrow indicates the direction of the vector. 4 m/s East
2 m/s North
Adding Vectors Only add vectors with the same units units. For example: v3 = v1 + v2 i OK, is OK bbut v3 = x 1 + v2 does not make sense Remember that a vector -A A points opposite to A. A –A
Adding g Vectors Graphically p y To add two vectors A + B, put the tail of B at the h tip i off A and d connect the h tail il off A to the h tip of B.
Use a ruler and protractor to determine the magnitude g and direction of C.
Subtracting Vectors, and Multiplication by a Scalar The negative of a vector is a vector having the same length, but pointing in the opposite direction
So we can define the subtraction of one vector from another as
A vector can also be multiplied p byy a scalar c. We define this product so that cV has the same direction as V and a magnitude c|V|.
Components of a Vector A vector can be specified either by its magnitude and direction, or by its components in a coordinate system.
rθ r,θ
rx , ry
Be Careful The components of a vector are not always positive.
The angle is not always defined relative to the x axis.
Trigonometric Functions SOH CAH TOA opposite side Ay sin θ = = hypotnuse A adjacent side Ax cos θ = = hypotnuse A opposite side Ay tan θ = = adjacent side Ax A = Ax2 + Ay2 Be careful with your calculator! A you using Are i degrees d or radians? di ?
A Ay θ Ax
Example Find the x and y components p of a vector of magnitude 5, if its angle relative to the x axis is 200°.
Ax = A cos θ = 5cos ( 200 ) = −4.70 D
Ay = A sin θ = 5sin ( 200 ) = −1.71 D
Adding g Vectors Usingg Components • Break vectors up into components. • Add components. t • Find magnitude and direction of the sum from its components.
Example 12. The vectors are shown in the figure below. Their magnitudes are given are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with x axis.
Unit Vectors A unit vector is a vector with magnitude equal to 1 and pointing in a certain defined direction. direction Example: the unit vector in the x direction is usually written: xˆ o r iˆ Now I can write a vector as its components multiplied p by y the unit vector: G A = Ax xˆ + Ay yˆ
or G A = Ax iˆ + Ay ˆj where Ax and Ay are scalars.
Example Write the vector in the figure in unit vector notation. y
θ
G v
G v = 10 x
θ = 50°
Position and Displacement Vectors The position vector r points from the origin to a particular location.
The displacement vector Δr indicates the g in position: p change Δr = rf - ri
Writing of displacement in terms of components and unit vectors,
G Δr = ( x2 − x1 )iˆ + ( y2 − y1 ) ˆj + ( z2 − z1 )kˆ We defined the instantaneous velocity as
G G Δr dr G = v = lim Δt →0 Δt dt So we can write the instantaneous velocity as,
G dx ˆ dy ˆ dz ˆ j + k = vx iˆ + v y ˆj + vz kˆ v= i+ dt dt dt
Likewise for acceleration,,
G G G Δv v2 − v1 G aav = = Δt Δt The instantaneous acceleration is given by
G G Δv dv G a = lim = Δt →0 Δt dt so
G dvx ˆ dv y ˆ dvz ˆ G ˆ G ˆ G ˆ a= i+ j+ k = ax i + a y j + az k dt dt dt
Acceleration occurs whenever there is a g in velocity. y change
If a car travels in a circle at a constant speed of g 50 mi/hr, is it accelerating? YES! Because the direction of the velocity vector is changing. changing
Kinematic Equations for Constant Acceleration in 2 Dimensions x-component
x = xo + vxot + ax t 1 2
y-component 2
vx = vxo + ax t v = v + 2ax ( x − xo ) 2 x
2 xo
y = yo + v yot + 12 a y t 2 v y = v yo + a y t 2 v y2 = v yo + 2a y ( y − yo )
Projectile Motion Motion in the x direction is independent from motion in the y direction. We use the same equations from Chapter 2, 2 but for each dimension separately.
There are not really any new equations in this p chapter.
The equations you need. 1 2 x = x0 + v0 x t + ax t 2 1 2 y = y0 + v0 y t + a y t 2 vx = v0 x + ax t v y = v0 y + a y t v = v + 2ax ( x f − xi ) 2 x
2 0x
v = v + 2a y ( y f − yi ) 2 y
2 0y
NOTE For projectile motion problems, ax (the horizontal componant of the acceleration) will usually be zero since i usually ll there th will ill be b no acceleration in the x direction
Projectile Motion • Assume that acceleration of g gravityy is constant, downward and has a magnitude of g = 9.81 m/s2 • Air resistance is ignored • The Earth’s rotation is ignored g ⇒Horizontal velocity is constant: because ax = 0 ⇒Vertical motion governed by the constant acceleration of gravity
Motion of a Projectile Launched H i t ll Horizontally The dots Th d represents the h position of the object everyy 0.05 s.
t=1s
y0 = 8 m; x0 = 0 v0y = 0; v0x = 6 m/s ay = -9.81 m/s2; ax = 0 1. Verify the position of the object at t = 1s. 2. What would be the position of the object at t = 1 s if it were dropped (v0x = 0)?
1 2 x = 0 + v0 x t y = y0 + v0 y t + a y t 2 1 2 x = 6 ×1 = 6 m y = 8 − 9.8 × (1) = 3.1 m 2 __________________________________________________________
1 2 y = 8 − 9.8 × (1) = 3.1 m 2 x=0
Example p Pitcher’s mounds are raised to compensate for the vertical drop of the h ball b ll as it i travels l 18 m to the h catcher. h (a) If a pitch is thrown horizontally with an initial speed of 32 m/s (71 mi/hr), mi/hr) how far does it drop by the time it reaches the catcher? (b) If the speed of the pitch is increased, increased does the drop distance increase, decrease or stay the same? Explain. (c) If this baseball game were to be played on the moon, would the drop distance increase, decrease, or stay the same? Explain.
(a ) x = v0 x t 18 = 32treach −catcher
treach −catcher = 0.56 s
1 y = y0 + v0 y t − 9.8t 2 2 y0 − y = 1.54 1 54 m
1 2 y − y0 = − 9.8 × ( 0.56 ) = −1.54 m 2
(b) If the speed of the pitch increases, the time gets smaller, so the drop distance is less (c) On the moon “g” is less than 9.8 m/s2, so the drop distance is less.
General Launch Angle Consider an object launched from the origin at an g θ with respect p to the horizontal. angle y
vo
θ
x
What are the x and y components of the initial velocity vector?
Example: On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2 25 m/s 2.25 / at an angle l off 35.0° 35 0° above b the h horizontal. h i l If she h is i in i flight fli h for 1.60 s, how high above the water was she when she let go of the rope? p
What is the girl girl’ss minimum speed during her flight? What is her acceleration at the top of her trajectory?
1 y = y0 + v0 sin θ t − 9.81t 2 2 1 0 = y0 + 2.25sin 35D t − 9.81t 2 2 1 2 D 0 = y0 + 2.25sin 35 (1.6 ) − 9.81(1.6 ) 2 y0 = 10.49 10 49 m
Girl’s minimum speed is at the maximum in her trajectory! C you see why? Can h ? Girl s acceleration is 9.81 Girl’s 9 81 m/s2 at the top of her trajectory. trajectory What is her acceleration at other points on her trajectory?
Range The range g R of a pprojectile j is the horizontal distance it travels before landing.
⎛ v02 ⎞ R = ⎜⎜ ⎟⎟ sin 2θ ⎝g⎠
(BE CAREFUL!, This equation only works when the initial and final elevation are the same)
You should try to derive this equation! It’s not difficult. Just remember that x=R when the projectile hits the ground. What angle θ results in the maximum range? Remember 0°≤ θ ≤90°. What if we do not ignore air resistance?
Example A pprojectile j is fired at 200 m/s at an angle g of 45 degrees g above the horizontal. How far does it travel?
⎛ v02 ⎞ R = ⎜⎜ ⎟⎟ sin 2θ ⎝g⎠
⎛ (200m/s) 2 ⎞ o sin(2(45 )) = 4081m R=⎜ 2 ⎟ 9 8m/s ⎠ ⎝ 9.8m/s
Maximum Height The maximum height (and therefore the “hang hang time time”)) of a projectile depends only on the vertical component of its initial velocity. y At ymax, the vertical velocity vy is zero. v 2y = v02 y + 2aΔy 0 = v02 sin 2 θ + 2(− g ) ymax ymax
v02 sin 2 θ = 2g
Example From the previous example, example how high does the projectile go?
ymax
v sin θ = 2g 2 o
2
( 200m / s ) =
2
ymax
2
o
sin (45 ) = 1020m 2(9.8)
Uniform Circular Motion v1
C id a particle Consider ti l moving i in i a circle. i l Is the particle accelerating?
G G Δv dv G = a = lim Δt →0 Δt ddt - The speed of the particle is constant - The direction of the ball is changing
v2
What is the direction of the acceleration?
Magnitude of Radial Acceleration This type acceleration is called centripetal or radial acceleration. The magnitude of the acceleration is given by, by v2 aR = r
Period and Frequency The amount of time that it takes the particle to make one full revolution is called the period , T. The number of times that the particle goes around the circle per q y, ff. unit time is called the frequency,
1 T= f The speed of a particle moving around in a circle is given by,
distance 2π r = = 2π rf v= time T
Example 33. A shotputter 33 h t tt throws th throws th the th shot h t (mass ( = 7.3 7 3 kg) k ) with ith an initial speed of 14.0 m/s at a 40° angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a height of 2.2 m above the ground.
Example p 18. The position of a particular particle as a function of time is given by,
G 2ˆ ˆ ˆ r = (7.60 (7 60ti + 88.85 85 j − t k )m (a) Determine the particle’s velocity and acceleration as a function of time. (b) Describe the motion in each direction.
Example p Chapter 2 #40. A space vehicle accelerates uniformly from 65 m/s at t=0 to 162 m/s at 10.0 s. How far did it move b t between t 2 0 andd t=6.0 t=2.0s t 6 0 s??
Example A baseball is hit at an angle of 50 50° relative to the horizontal with an initial speed of 120 miles/hour. ((a)) H How hi highh does d it go?? (b) How long does it spend in the air? (c) How far does it travel?
