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Transcript of 1 Chapter 3 Motion in Two Dimensions 2 3 3.1 Position and Displacement The position of an object is...
![Page 1: 1 Chapter 3 Motion in Two Dimensions 2 3 3.1 Position and Displacement The position of an object is described by its position vector, The displacement.](https://reader035.fdocuments.net/reader035/viewer/2022062422/56649ee55503460f94bf4f07/html5/thumbnails/1.jpg)
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Chapter 3
Motion in Two Dimensions
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3.1 Position and Displacement The position of an
object is described by its position vector,
The displacement of the object is defined as the change in its position
Fig 3.1
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Average Velocity The average velocity is
the ratio of the displacement to the time interval for the displacement
The direction of the average velocity is the direction of the displacement vector, Fig 3.2
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Average Velocity, cont The average velocity between points is
independent of the path taken This is because it is dependent on the
displacement, which is also independent of the path
If a particle starts its motion at some point and returns to this point via any path, its average velocity is zero for this trip since its displacement is zero
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Instantaneous Velocity The instantaneous velocity is the limit of
the average velocity as ∆t approaches zero
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Instantaneous Velocity, cont The direction of the instantaneous
velocity vector at any point in a particle’s path is along a line tangent to the path at that point and in the direction of motion
The magnitude of the instantaneous velocity vector is the speed
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Average Acceleration The average acceleration of a particle
as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs.
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Average Acceleration, cont
As a particle moves, can be found in different ways
The average acceleration is a vector quantity directed along
v
Fig 3.3
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Instantaneous Acceleration The instantaneous acceleration is the
limit of the average acceleration as approaches zero
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3.2 Producing An Acceleration Various changes in a particle’s motion
may produce an acceleration The magnitude of the velocity The direction of the velocity
Even if the magnitude remains constant Both change simultaneously
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Kinematic Equations for Two-Dimensional Motion When the two-dimensional motion has a
constant acceleration, a series of equations can be developed that describe the motion
These equations will be similar to those of one-dimensional kinematics
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Kinematic Equations, 2 Position vector
Velocity
Since acceleration is constant, we can also find an expression for the velocity as a function of time:
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Kinematic Equations, 3 The velocity vector
can be represented by its components
is generally not along the direction of either or
Fig 3.4(a)
fi
xf xi x
yf yi y
t
v v a t
v v a t
v v a
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Kinematic Equations, 4 The position vector can also be
expressed as a function of time:
This indicates that the position vector is the sum of three other vectors:
The initial position vector The displacement resulting from The displacement resulting from
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Kinematic Equations, 5 is generally not in
the same direction as or as
and are generally not in the same direction
Fig 3.4(b)
212
212
212
fi i
fi xi x
fi yi y
t t
x x v t a t
y y v t a t
r r v a
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Active Figure3.4
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SolutionConceptualize
Establish the mental representation and thinking about what the particle is doing.
Categorize Consider that because the acceleration is only in the x direction, the moving particle can be modeled as one under constant acceleration in the x direction and one under constant velocity in the y direction.
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AnalyzeIdentify vxi = 20 m/s and ax = 4.0 m/s2. The equations of kinematics for the x direction,
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Solution At t = 5.0 s, the velocity expression from part A gives
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拋 射 3.3 Projectile Motion An object may move in both the x and y
directions simultaneously The form of two-dimensional motion we
will deal with is called projectile motion
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假 設Assumptions of Projectile Motion
The free-fall acceleration is constant over the range of motion and is directed downward
The effect of air friction is negligible
With these assumptions, an object in projectile motion will follow a parabolic path
This path is called the trajectory This is the simplification model that will be used
throughout this chapter
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Verifying the Parabolic Trajectory Reference frame chosen
y is vertical with upward positive Acceleration components
ay = -g and ax = 0
Initial velocity components vxi = vi cos and vyi = vi sin
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Projectile Motion – Velocity Equations The velocity components for the
projectile at any time t are: vxf = vxi = vi cosi = constant
vyf = vyi – gt = vi sini – gt
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Projectile Motion – Position Displacements
Combining the equations gives:
This is in the form of y = ax – bx2 which is the standard form of a parabola
2 21 12 2
( cos )
( sin )
fi xi i i
fi yi i i
x x v t v t
y y v t gt v t gt
22 2(tan )
2 cosfi ffi i
gy x x
v
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Projectile Motion Vectors
Fig 3.6
212fi it t r r v g
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Projectile Motion Diagram
Fig 3.5
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Active Figure3.5
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Projectile Motion – Implications The y-component of the velocity is zero
at the maximum height of the trajectory The accleration stays the same
throughout the trajectory
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Horizontal Range and Maximum Height of a Projectile
When analyzing projectile motion, two characteristics are of special interest
The range, R, is the horizontal distance of the projectile
The maximum height the projectile reaches is h
Fig 3.7
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Height of a Projectile, equation The maximum height of the projectile
can be found in terms of the initial velocity vector:
2 2sin2
i ivh
g
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Range of a Projectile, equation The range of a projectile can be
expressed in terms of the initial velocity vector:
This is valid only for symmetric trajectory
2 sin2i ivR
g
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More About the Range of a Projectile
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Range of a Projectile, final
The maximum range occurs at i = 45o
Complementary angles will produce the same range The maximum height will be different for
the two angles The times of the flight will be different for
the two angles
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Active Figure3.8
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Projectile Motion – Problem Solving Hints Conceptualize
Establish the mental representation of the projectile moving along its trajectory
Categorize Confirm that the problem involves the
particle in free fall (in the y-direction) and air resistance can be neglected
Select a coordinate system
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Problem Solving Hints, cont. Analyze
Resolve the initial velocity into x and y components Remember to treat the horizontal motion independently
from the vertical motion Treat the horizontal motion using constant velocity
techniques Analyze the vertical motion using constant
acceleration techniques Remember that both directions share the same time
Finalize Check your results
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Non-Symmetric Projectile Motion Follow the general rules
for projectile motion Break the y-direction
into parts up and down or symmetrical back to
initial height and then the rest of the height
May be non-symmetric in other ways
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A stone is thrown from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, as in Figure 3.9.
Fig 3.9
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To find t, we use the vertical motion, in which we model the stone as a particle under constant acceleration. We use Equation 3.13 with yf = 45.0 m and vyi = 10.0 m/s (we have chosen the top of the building as the origin):
t = 4.22 s
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Solution The y component of the velocity just before the stone strikes the ground can be obtained using Equation 3.11, with t = 4.22 s:
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An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the pictorial representation in Figure 3.10. If the plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground, where does the package strike the ground relative to the point at which it is released?
Fig 3.10
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Solution Ignore air resistanceModel this problem as a particle in two-dimensional free-fall,a particle under constant velocity in the x direction anda particle under constant acceleration in the y direction
Define the initial position xi = 0 right under the plane at the instant the package is released.
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If we know t, the time at which the package strikes the ground, we can determine xf. To find t, we turn to the equations for the vertical motion of the package. At the instant the package hits the ground, its coordinate is 100 m.
The initial component of velocity vyi of the package is zero.
From Equation 3.13,
xf = (40.0 m/s)(4.52 s) = 181 m
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An athlete throws a javelin a distance of 80.0 m at the Olympics held at the equator, where g = 9.78 m/s2. Four years later the Olympics are held at the North Pole, where g = 9.83 m/s2. Assuming that the thrower provides the javelin with exactly the same initial velocity as she did at the equator, how far does the javelin travel at the North Pole?
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50= 79.6 m
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3.4 Uniform Circular Motion Uniform circular motion occurs when an
object moves in a circular path with a constant speed
An acceleration exists since the direction of the motion is changing This change in velocity is related to an
acceleration The velocity vector is always tangent to the
path of the object
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Fig 3.11
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Centripetal Acceleration The acceleration is always
perpendicular to the path of the motion The acceleration always points toward
the center of the circle of motion This acceleration is called the
centripetal acceleration Centripetal means center-seeking
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Changing Velocity in Uniform Circular Motion The change in the
velocity vector is due to the change in direction
The vector diagram shows
Fig 3.11
fiv v v
fir r r
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avg
fi
fit t t
v v va
v rrv
avg
vt r t
rva
2
c
va
r
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Period The period, T, is the time interval
required for one complete revolution The speed of the particle would be the
circumference of the circle of motion divided by the period
Therefore, the period is
2 r
Tv
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What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?
Solution We model the Earth as a particle and approximate the Earth’s orbit as circular Although we don’t know the orbital speed of the Earth, with the help of Equation 3.18 we can recast Equation 3.17 in terms of the period of the Earth’s orbit, which we know is one year:
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3.5 Tangential Acceleration The magnitude of the velocity could also
be changing As well as the direction
In this case, there would be a tangential acceleration
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Total Acceleration
The tangential acceleration causes the change in the speed of the particle
The radial acceleration comes from a change in the direction of the velocity vector
Fig 3.12
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Total Acceleration, equations The tangential acceleration:
The radial acceleration:
The total acceleration: Magnitude The direction of the acceleration is the
same or opposite that of the velocity
2
r c
va a
r
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Active Figure3.12
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3.6 Relative Velocity Two observers moving relative to each other generally
do not agree on the outcome of an experiment For example, the observer on the side of the road
observes a different speed for the red car than does the observer in the blue car
Fig 3.13
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Relative Velocity, generalized
Reference frame S is stationary
Reference frame S’ is moving
Define time t = 0 as that time when the origins coincide
Fig 3.14
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Relative Velocity, equations The positions as seen from the two reference
frames are related through the velocity
The derivative of the position equation will give the velocity equation
This can also be expressed in terms of the observer O’
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A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The river has a current such that the water moves with uniform speed of 5.00 km/h due east relative to the ground.
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bwˆ10.0 km/hjv
wEˆ5.00 km/hiv
bE bw wE
ˆ ˆ(5.00 10.0 ) km/hi j
v v v
2 2bE bw wE
2 2 10.0 5.00 11.2 km/h
v v v
1 1wE
bw
5.00tan tan
10.0
26.6
vv
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1 1wE
bw
5.00sin sin
10.0
30.0
vv
2 2bE bw wE
2 2 10.0 5.00 8.66 km/h
v v v
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3.7 Acceleration of Autos The lateral acceleration
橫向加速度 is the maximum possible centripetal acceleration the car can exhibit without rolling over in a turn
The lateral acceleration depends on the height of the center of mass of the vehicle and the side-to-side distance between the wheels
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Beamon 於 1968 在 Mexico 創下 8.90 m 的跳遠紀錄,由照片得知,他起跳時重心高度約為 1.0 m ,在最高點處重心高度為 1.90 m ,落地時重心高度 0.15 m 。由這些數據,求 (a) 他在空中停留時間。 (b) 起跳時速度之水平分量及垂直分量。 (c) 起跳角度。
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An enemy ship is on the east side of a mountain island, as shown in Figure P3.61. The enemy ship has maneuvered to within 2500 m of the 1800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
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Problem1, 3, 10, 13, 15, 19, 20, 27, 31, 36, 38, 40, 41, 48, 51, 55, 60