Lecture 4 H-ITT Recap
PHY2053, Fall 2013, Lecture 5
Vector problem strategy
● Read problem carefully, the devil is in the details
● Draw a vector diagram of the problem, following the
details
● Translate the vector diagram into vector equation
● Carefully extract quantity of interest from equation
[carefully ≣ OCD on conventions – easiest place to slip]
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PHY 2053, Lecture 5:
Motion in a Plane
With Constant
Acceleration
PHY2053, Fall 2013, Lecture 5
The Superposition Principle
● Applies to regions in space in which the acceleration
does not depend on the position of the object
● Motion along one axis is independent of
[does not affect] the motion along any other axis
● Consequence:
● The object is moving in the x-y plane over Δt
● The final position of the object will be the same as if
● the object first moved only along the x-axis for Δt
● the object then moved only along the y-axis for Δt
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Experiment:
Relative motion cart
PHY2053, Fall 2013, Lecture 5
2-d Motion: Constant Acceleration
• Kinematic Equations of Motion (Vector Form)
Acceleration Vector (constant)
Velocity Vector (function of t)
Position Vector (function of t)
The velocity vector and position
vector are a function of the time t.
r
Warning! These
a = a x xˆ + a y yˆ
equations are only
valid if the
r
r r
acceleration is
v (t ) = v0 + at
constant.
r
r r
r2
1
r (t ) = r0 + v0t + 2 at
Velocity Vector at
time t = 0.
Position Vector at
time t = 0.
r
v0 = v x 0 xˆ + v y 0 yˆ
r
r0 = x0 xˆ + y0 yˆ
The components of the acceleration vector, ax and ay, are constants.
The components of the velocity vector at t = 0, vx0 and vy0, are constants.
The components of the position vector at t = 0, x0 and y0, are constants.
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PHY2053,
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Lecture
5
PHY 2053
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2-d Motion: Constant Acceleration
• Kinematic Equations of Motion (Component Form)
ax =
ay =
constant
v x (t ) = v x 0 + a x t
2
1
x(t ) = x0 + v x 0t + 2 a x t
constant
Warning! These
equations are only
valid if the
acceleration is
constant.
v y (t ) = v y 0 + a y t
2
1
y (t ) = y0 + v y 0t + 2 a y t
The components of the acceleration vector, ax and ay, are constants.
The components of the velocity vector at t = 0, vx0 and vy0, are constants.
The components of the position vector at t = 0, x0 and y0, are constants.
• Ancillary Equations
2
x
2
xo
2
y
2
yo
Valid at any time t
v (t ) − v = 2a x ( x(t ) − x0 )
v (t ) − v = 2a y ( y (t ) − y0 )
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University
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PHY2053,
Fall
2013,
PHY2053,
Fall
2013, Lecture
Lecture 5!
5
PHY 2053
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99!
Data Analysis:
Using Accelerometers To
Determine Displacement
PHY2053, Fall 2013, Lecture 5
The Principle
● An accelerometer sensor reports the acceleration
along an axis – three accelerometers for 3D reporting
● Superposition principle for motion
● Take accelerometer readings very often [50 / second]
● Acceleration is ~ constant during such a short interval
● Along each axis [x,y,z] and for each interval, apply
a 2
v
=
v
+
a
t
xf
xi
xf = xi + vxi t +
t
2
● xf , vxf for one given interval become the xi , vxi of the
next interval, repeat from first to last interval
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The challenge
● Will post both raw data and CSV formatted data on
eLearning page
● The accelerometer will always report Earth’s
gravitational acceleration, , which needs to be
subtracted
● The accelerometer will always be slightly tilted, so the
gravitational acceleration will not always point in the
exact z direction
● The subtraction of will have to be done in each of the
three directions – “calibration” periods in data
● When you compute x(t), y(t), z(t), plot x vs y and see if
you can reconstruct the trajectory of the iPod
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Practical Applications
● This general calculation principle is used:
● In accelerometer-based pedometers
[FitBit, tri-axis pedometers etc.]
● Vehicle collision reconstruction:
● Airbag control systems deploy airbags based on
accelerometer readings
● Accelerometer readings preceding a crash
“candidate” are stored and can be read out later
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Projectile Motion
PHY2053, Fall 2013, Lecture 5
Projectile Motion
y-axis
• Near the Surface of the Earth
v0
θ
In this case, ax= 0 and ay= -g, vx0 = v0cosθ
θ, vy0 = v0sinθ
θ, x0 = 0, y0 =h.
ay = −g
ax = 0
v x (t ) = v0 cos θ
(constant)
h
x-axis
v y (t ) = v0 sin θ − gt
x(t ) = (v0 cos θ )t
1
2
y (t ) = h + (v0 sin θ )t − gt
2
• Maximum Height H
The time, tmax, that the projective reaches its maximum height occurs when
vy(tmax) = 0. Hence,
t max
For a fixed v0 the largest H
occurs when θ = 90o!
v0 sin θ
=
g
(v0 sin θ )
H = y (t max ) = h +
2g
R. Field 9/6/2012
PHY2053,
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Projectile Motion
y-axis
• Near the Surface of the Earth (h = 0)
In this case, ax= 0 and ay= -g, vx0 = v0cosθ
θ, vy0 = v0sinθ
θ, x0 = 0, y0 = 0.
v x (t ) = v0 cos θ
x(t ) = (v0 cos θ )t
v y (t ) = v0 sin θ − gt
v0
θ
x-axis
1
2
y (t ) = (v0 sin θ )t − gt
2
• Maximum Height H
The time, tmax, that the projective reaches its maximum
height occurs when vy(tmax) = 0. Hence,
v0 sin θ
(v0 sin θ ) 2
tmax =
g
H = y (t max ) =
2g
• Range R (maximum horizontal distance traveled)
The time, tf, that it takes the projective reach the
ground occurs when y(tf) = 0. Hence,
1
2
0 = y (t f ) = (v0 sin θ )t f − gt
2
0
2
f
2v0 sin θ
tf =
g
2
0
2v sin θ cos θ v sin 2θ
R = x(t f ) = (v0 cos θ )t f =
=
g
g
R. Field 9/6/2012
PHY2053,
Fall
2013,
University
of Florida
PHY2053,
Fall
2013, Lecture
Lecture 55!
PHY 2053
Page 5
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16!
Example: BMG .50 Trajectory
http://www.hornady.com/assets/files/ballistics/2013-Metric-Ballistics.pdf
Assuming simple projectile
motion, from the trajectory table
positions at 100, 150 and 200 m,
motion, compute:
● The bullet velocity at 150 m
● The muzzle height relative
to the bullet height at 100 m
PHY2053, Fall 2013, Lecture 5
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Example: BMG .50 Trajectory
http://www.hornady.com/assets/files/ballistics/2013-Metric-Ballistics.pdf
Assuming simple projectile
motion, from the trajectory table
positions at 100, 150 and 200 m,
motion, compute:
● The bullet velocity at 150 m
● The muzzle height relative
to the bullet height at 100 m
PHY2053, Fall 2013, Lecture 5
17
Example: BMG .50 Trajectory
http://www.hornady.com/assets/files/ballistics/2013-Metric-Ballistics.pdf
Assuming simple projectile
motion, from the trajectory table
positions at 100, 150 and 200 m,
motion, compute:
● The bullet velocity at 150 m
● The muzzle height relative
to the bullet height at 100 m
PHY2053, Fall 2013, Lecture 5
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Experiment:
Monkey Falling Out of Tree
PHY2053, Fall 2013, Lecture 5