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2-D KINEMATICS

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Jonathan White

on 23 October 2013

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Transcript of 2-D KINEMATICS

VECTORS: MOTION AND FORCES IN TWO - DIMENSIONS
Use of Scaled Vector Diagrams to Determine a Resultant
The magnitude and direction of the sum of
two or more
vectors can also be determined by use of an accurately drawn scaled vector diagram

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.

1. Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
2. Pick a starting location and draw the first vector to scale in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
3. Starting from where the head of the first vector ends, draw the second vector to scale in the indicated direction. Label the magnitude and direction of this vector on the diagram.
4. Repeat steps 2 and 3 for all vectors that are to be added
5. Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as Resultant or simply R.
6. Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
7. Measure the direction of the resultant using the counterclockwise convention discussed earlier in this lesson..

CONTINUED
R2 = (6.756… m)2 + (4.6 m)2
R2 = 45.655… m2 + 21.16 m2
R2 = 66.815… m2
R = SQRT(66.815… m2 )
R = 8.174 … m
R = ~8.2 m
Θ = tan-1 (1.46889…) = 55.7536… °
Θ = ~56°

Non-Horizontally Launched Projectile Problems
Example
A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.

Horizontal Component
vix = vi•cos(Theta)
vix = 25 m/s•cos(45 deg)
vix = 17.7 m/s

Vertical Component
viy = vi•sin(Theta)
viy = 25 m/s•sin(45 deg)
viy = 17.7 m/s

Horizontal Information
x = ???
vix = 17.7 m/s
vfx = 17.7 m/s
ax = 0 m/s/s

Vertical Information
y = ???
viy = 17.7 m/s
vfy = -17.7 m/s
ay = -9.8 m/s/s

Topic 7
LESSON 1: VECTORS - FUNDAMENTALS AND OPERATIONS
VECTORS AND DIRECTION
Vector quantities
are often represented by scaled vector diagrams. Vector diagrams depict a vector by use of
an arrow drawn to scale in a specific direction.
Proper Drawing of VECTORS
Conventions for Describing Directions of Vectors


There are a variety of conventions for describing the direction of any vector. The two conventions that will be discussed and used in this unit are described as follows:
1. The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from east, west, north, or south.

FOR EXAMPLE:

N64W - reads as 64 degrees W of N....
2. The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "tail" from due East.
Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east.
Representing the Magnitude of a Vector
The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale.
In the previous two units, we focused on adding vectors in either the horizontal or vertical direction...and then applied this to free-body diagrams.
In this unit, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions (2-D motion!).
2-D Vector Addition
The Pythagorean Theorem
The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors that make a right angle to each other.
To see how the method works, consider the following problem:

Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.
YOUR TURN!
Using Trigonometry to Determine a Vector's Direction
The direction of a resultant vector can often be determined by use of trigonometric functions.
These three trigonometric functions can be applied to the hiker problem in order to determine the direction of the hiker's overall displacement.
The measure of an angle as determined through use of SOH CAH TOA is not always the direction of the vector. The following vector addition diagram is an example of such a situation.
Test your understanding of the use of SOH CAH TOA to determine the vector direction by trying the following two practice problems. In each case, find the value of the resultant vector, as well as it's direction.
Vector Components
Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a component.


Vector Resolution

The process of determining the magnitude of a vector is known as vector resolution. The two methods of vector resolution that we will examine are:

1. The parallelogram method
2. The trigonometric method
Parallelogram Method of Vector Resolution
The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector.
Trigonometric Method of Vector Resolution
The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector.
As our first example, consider the following vector addition problem:

Example 1:
A student drives his car 6.0 km, North before making a right hand turn and driving 6.0 km to the East. Finally, the student makes a left hand turn and travels another 2.0 km to the north. What is the magnitude of the overall displacement of the student?

Component Method of Vector Addition: Addition of Three of More Right Angle Measures
When these three vectors are added together in head-to-tail fashion, the resultant is a vector that extends from the tail of the first vector (6.0 km, North, shown in red) to the arrowhead of the third vector (2.0 km, North, shown in green). The head-to-tail vector addition diagram is shown below.
As can be seen in the diagram, the resultant vector (drawn in black) is not the hypotenuse of any right triangle - at least not of any immediately obvious right triangle. But would it be possible to force this resultant vector to be the hypotenuse of a right triangle? The answer is Yes! To do so, the order in which the three vectors are added must be changed.
After rearranging the order in which the three vectors are added, the resultant vector is now the hypotenuse of a right triangle. The lengths of the perpendicular sides of the right triangle are 8.0 m, North (6.0 km + 2.0 km) and 6.0 km, East. The magnitude of the resultant vector (R) can be determined using the Pythagorean theorem.
R2 = (8.0 km)2 + (6.0 km)2
R2 = 64.0 km2+ 36.0 km2
R2 = 100.0 km2
R = SQRT (100.0 km2)
R = 10.0 km
Example 2:
Mac and Tosh are doing the Vector Walk Lab. Starting at the door of their physics classroom, they walk 2.0 meters, south. They make a right hand turn and walk 16.0 meters, west. They turn right again and walk 24.0 meters, north. They then turn left and walk 36.0 meters, west. What is the magnitude of their overall displacement?

When these four vectors are added together in head-to-tail fashion, the resultant is a vector that extends from the tail of the first vector (2.0 m, South, shown in red) to the arrowhead of the fourth vector (36.0 m, West, shown in green). The head-to-tail vector addition diagram is shown below.
R2 = (22.0 m)2 + (52.0 m)2
R2 = 484.0 m2 + 2704.0 m2
R2 = 3188.0 m2
R = SQRT (3188.0 m22)
R = 56.5 m
Tangent(Θ) = Opposite/Adjacent
Tangent(Θ) = 52.0/22.0
Tangent(Θ) = 2.3636 …
Θ = tan-1 (2.3636 …)
Θ = 67.067 …°
Θ =67.1°
67.1 + 90 = 157.71
Addition of Non-Perpendicular Vectors
Now suppose that your task involves adding two non-perpendicular vectors together. We will call the vectors A and B. Vector A is a nasty angled vector that is neither horizontal nor vertical. And vector B is a nice, polite vector directed horizontally. The situation is shown below.
And so the problem of A + B has been transformed into a problem in which all vectors are at right angles to each other. With all vectors being at right angles to one another, their addition leads to a resultant that is at the hypotenuse of a right triangle.
Max plays middle linebacker for South's football team. During one play in last Friday night's game against New Greer Academy, he made the following movements after the ball was snapped on third down. First, he back-pedaled in the southern direction for 2.6 meters. He then shuffled to his left (west) for a distance of 2.2 meters. Finally, he made a half-turn and ran downfield a distance of 4.8 meters in a direction of 240° counter-clockwise from east (30° W of S) before finally knocking the wind out of New Greer's wide receiver. Determine the magnitude and direction of Max's overall displacement.
THE SOLUTION
Cameron Per (his friends call him Cam) and Baxter Nature are on a hike. Starting from home base, they make the following movements.
A: 2.65 km, 140° CCW
B: 4.77 km, 252° CCW
C: 3.18 km, 332° CCW

Determine the magnitude and direction of their overall displacement.
4.38 km at 80.9 degrees south of west...or 260.9 degrees CCW from due east.


Relative Velocity and Riverboat Problems
On occasion objects move within a medium that is moving with respect to an observer. For example, an airplane usually encounters a wind - air that is moving with respect to an observer on the ground below. As another example, a motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. That is to say, the speedometer on the motorboat might read 20 mi/hr; yet the motorboat might be moving relative to the observer on shore at a speed of 25 mi/hr.
Motion is relative to the observer.
Analysis of a Riverboat's Motion
The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat. If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction.
Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found as follows:

(4.0 m/s)2 + (3.0 m/s)2 = R2
16 m2/s2 + 9 m2/s2 = R2

25 m2/s2 = R2

SQRT (25 m2/s2) = R

5.0 m/s = R
A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.

1. What is the resultant velocity of the motorboat?
2. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
3. What distance downstream does the boat reach the opposite shore?
A motorboat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.

1. What is the resultant velocity of the motorboat?
2. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
3. What distance downstream does the boat reach the opposite shore?
4. What is the direction of the resulting vector (the actual path the boat travels)?
C) Describing Projectiles With Numbers:
1) Horizontal and Vertical Velocity
So far , we know the following about projectiles...

A projectile is any object upon which the only force is gravity,
Projectiles travel with a parabolic trajectory due to the influence of gravity,
There are no horizontal forces acting upon projectiles and thus no horizontal acceleration,
The horizontal velocity of a projectile is constant (a never changing in value),
There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,
The vertical velocity of a projectile changes by 9.8 m/s each second,
The horizontal motion of a projectile is independent of its vertical motion.


Lesson 2: Projectile Motion
a) What is a Projectile?
b) Characteristics of a Projectile's Trajectory
c) Describing Projectiles with Numbers
1) Horizontal and Vertical Components of Velocity
2) Horizontal and Vertical Components of Displacement
d) Initial Velocity Components
e) Horizontally Launched Projectiles - Problem-Solving
f) Non-Horizontally Launched Projectiles - Problem-Solving
A) What is a Projectile?
The most common example of an object that is moving in two dimensions is a projectile.
A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles.


Projectile Motion and Inertia
Many students have difficulty with the concept that the only force acting upon an upward moving projectile is gravity. Their conception of motion prompts them to think that if an object is moving upward, then there must be an upward force. And if an object is moving upward and rightward, there must be both an upward and rightward force.
B) Characteristics of a Projectile's Trajectory
Many projectiles not only undergo a
vertical motion
, but also undergo a
horizontal motion
. That is, as they move upward or downward they are also moving horizontally.
There are the two components of the projectile's motion - horizontal and vertical motion. And since perpendicular components of motion are independent of each other, these two components of motion can (and must) be discussed separately.

Horizontally Launched Projectiles
The presence of gravity does not affect the horizontal motion of a projectile
. The force of gravity acts downward and is unable to alter the horizontal motion.
There must be a horizontal force to cause a horizontal acceleration. (And we know that there is only a vertical force acting upon projectiles.)
The vertical force acts perpendicular to the horizontal motion and will not affect it since perpendicular components of motion are independent of each other.
Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.
Non-Horizontally Launched Projectiles
Once more,
the presence of gravity does not affect the horizontal motion of the projectile
. The projectile still moves the same horizontal distance in each second of travel as it would in the absence of gravity.
The force of gravity is a vertical force and does not affect horizontal motion; perpendicular components of motion are independent of each other.
Suppose that the cannonball is launched horizontally with no upward angle whatsoever and with an initial speed of 20 m/s. The important concept depicted in the above vector diagram is that
the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by 9.8 m/s every second.
But what if the projectile is launched upward at an angle to the horizontal? How would the horizontal and vertical velocity values change with time? Again, the important concept depicted in the above diagram is
that the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by 9.8 m/s every second.

2) Horizontal and Vertical Displacement
The vertical displacement of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity. Thus, the vertical displacement (y) of a projectile can be predicted using the same equation used to find the displacement of a free-falling object undergoing one-dimensional motion.

.y = 0.5 • g • t2
Vertical Displacement of a Horizontally Launched Projectile


Where g is -9.8 m/s/s and t is the time in seconds. The above equation pertains to a projectile with no initial vertical velocity and as such predicts the vertical distance that a projectile falls if dropped from rest.
The
horizontal displacement
of a projectile is only influenced by the speed at which it moves horizontally (vix) and the amount of time (t) that it has been moving horizontally. Thus, if the horizontal displacement (x) of a projectile were represented by an equation, then that equation would be written as

x = vix • t
Now consider displacement values for a projectile launched at an angle to the horizontal (i.e., a non-horizontally launched projectile).
How will the presence of an initial vertical component of velocity affect the values for the displacement?
The diagram below depicts the position of a projectile launched at an angle to the horizontal. The projectile still falls 4.9 m, 19.6 m, 44.1 m, and 78.4 m below the straight-line, gravity-free path. These distances are indicated on the diagram below.
The projectile still falls below its gravity-free path by a vertical distance of 0.5*g*t^2. However, the gravity-free path is no longer a horizontal line since the projectile is not launched horizontally. In the absence of gravity, a projectile would rise a vertical distance equivalent to the time multiplied by the vertical component of the initial velocity (viy• t). In the presence of gravity, it will fall a distance of 0.5 • g • t2. Combining these two influences upon the vertical displacement yields the following equation.

y = viy • t + 0.5 • g • t2
Initial Velocity Components
If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and a vertical component. The horizontal velocity component (vx) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component (vy) describes the influence of the velocity in displacing the projectile vertically.
Determination of the Time of Flight
The time for a projectile to rise vertically to its peak (as well as the time to fall from the peak) is dependent upon vertical motion parameters.
The time for a projectile to rise to its peak is a matter of dividing the vertical component of the initial velocity (viy) by the acceleration of gravity.
Tup = Viy/g...so the total flight time would simply involve the following 2 X Tup...


Determination of Horizontal Displacement
The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity. As discussed in the previous part of this lesson, the horizontal displacement of a projectile can be determined using the equation x = vix • t

If a projectile has a time of flight of 8 seconds and a horizontal velocity of 20 m/s, then the horizontal displacement is 160 meters (20 m/s • 8 s). If a projectile has a time of flight of 8 seconds and a horizontal velocity of 34 m/s, then the projectile has a horizontal displacement of 272 meters (34 m/s • 8 s). The horizontal displacement is dependent upon the only horizontal parameter that exists for projectiles - the horizontal velocity (vix).



Determination of the Peak Height
A non-horizontally launched projectile with an initial vertical velocity of 39.2 m/s will reach its peak in 4 seconds. The process of rising to the peak is a vertical motion and is again dependent upon vertical motion parameters (the initial vertical velocity and the vertical acceleration). The height of the projectile at this peak position can be determined using the equation
y = viy • t + 0.5 • g • t2 where viy is the initial vertical velocity in m/s, g is the acceleration of gravity (-9.8 m/s/s) and t is the time in seconds it takes to reach the peak. This equation can be successfully used to determine the vertical displacement of the projectile through the first half of its trajectory (i.e., peak height) provided that the algebra is properly performed and the proper values are substituted for the given variables. Special attention should be given to the facts that the t in the equation is the time up to the peak and the g has a negative value of -9.8 m/s/s.


Horizontally Launched Projectile Problems
There are two types of projectile problems that we will focus on :

Problem Type 1:

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

Problem Type 2:

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.
Three common kinematic equations that will be used for both type of problems include the following:
Equations for the Horizontal Motion of a Projectile
The previous equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations.
Equations for the Vertical Motion of a Projectile
For the vertical components of motion, the three equations are:
Example
A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

Horizontal Information
x = ???
vix = 2.4 m/s
ax = 0 m/s/s

Vertical Information
y = -0.60 m
viy = 0 m/s
ay = -9.8 m/s/s

This equation, y = viy•t +0.5•ay•t2, can be used to solve for the time:
-0.60 m = (0 m/s)•t + 0.5•(-9.8 m/s/s)•t2
-0.60 m = (-4.9 m/s/s)•t2
0.122 s2 = t2
t = 0.350 s

The first horizontal equation (x = vix•t + 0.5•ax•t2) can then be used to solve for "x."

x = (2.4 m/s)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2
x = (2.4 m/s)•(0.3499 s)
x = 0.84 m




A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
Horizontal Info and Vertical Info
x = 35.0 m
y = -22.0 m
vix = ???
viy = 0 m/s
ax = 0 m/s/s
ay = -9.8 m/s/s
Use y = viy • t + 0.5 • ay • t2 to solve for time; the time of flight is 2.12 seconds.

Now use x = vix • t + 0.5 • ax • t2 to solve for vix

Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the vix can be found to be 16.5 m/s.
The unknown quantities are the
horizontal displacement
, the
time of flight
,
and the
height of the football at its peak
. From the vertical information in the table on the previous slide and this kinematic equation (
vfy = viy + ay*t
), it becomes obvious that the time of flight of the projectile can be determined. By substitution of known values, the equation takes the form of

-17.7 m/s = 17.7 m/s + (-9.8 m/s/s)•t
-35.4 m/s = (-9.8 m/s/s)•t
3.61 s = t

The total time of flight of the football is 3.61 seconds
With the time determined, information as listed in the problem and the horizontal kinematic equations can be used to determine the horizontal displacement (x) of the projectile. The first equation (
x = vix•t + 0.5•ax•t2
) listed among the horizontal kinematic equations is suitable for determining x.

x = (17.7 m/s)•(3.6077 s) + 0.5•(0 m/s/s)•(3.6077 s)2
x = (17.7 m/s)•(3.6077 s)
x = 63.8 m

The horizontal displacement of the projectile is 63.8m.
Finally, the problem statement asks for
the height of the projectile at is peak
. This is the same as asking, "
what is the vertical displacement (y) of the projectile when it is halfway through its trajectory?
" In other words,
find y when t = 1.80 seconds (one-half of the total time)
. To determine the peak height of the projectile (y with t = 1.80 sec), the first equation (
y = viy•t +0.5•ay•t2
) listed among the vertical kinematic equations can be used.

y = (17.7 m/s)•(1.80 s) + 0.5*(-10 m/s/s)•(1.80 s)2
y = 31.9 m + (-15.9 m)
y = 15.9 m


A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
As a first step, use trigonometric function to determine the initial velocity components:

Horiz: vix =(12 m/s) • cos (28 deg) =10.6 m/s
Vert: viy =(12 m/s) • sin (28 deg) = 5.6 m/s

Then set up an x-y table, listing the known information.

Horizontal Info::
x = ???
ax = 0 m/s/s
vix = 10.6 m/s

Vertical Info:
viy = 5.6 m/s
vy-peak = 0 m/s
y = 0.0 m (total); ypeak = ???
vfy = -5.6 m/s
ay = -9.8 m/s/s

Use
vfy = viy + ay • ttotal
and vertical info to solve for time; the time of flight is
1.1 seconds
(rounded from 1.1497 s).

Now use
x = vix • t + 0.5 • ax • t2
to solve for x. Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 10.6 m/s for vix and 1.1 s for t, the x can be found to be
12.2 m
(rounded from 12.1817 m)

Finally, use
tup = 0.5748 s
and the equation
ypeak = viy • tup + 0.5 • ay • tup2
with vertical info to find the y at the peak. Substituting and solving yields
1.6 m
(rounded from 1.6193 m)
Lesson 3 : Forces in Two Dimensions
a) Addition of Forces
b) Resolution of Forces
c) Equilibrium and Statics
d) Net Force Problems Revisited
e) Inclined Planes
f) Double Trouble (a.k.a. Two Body Problems)
a) Addition of Forces



A pack of five Artic wolves are exerting five different forces upon the carcass of a 500-kg dead polar bear.
B) Resolution of Forces
Another Example:
c) Equilibrium and Statics
When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium. The forces are considered to be balanced if the rightward forces are balanced by the leftward forces and the upward forces are balanced by the downward forces. This however does not necessarily mean that all the forces are equal to each other. Consider the two objects pictured in the force diagram shown on the right. Note that the two objects are at equilibrium because the forces that act upon them are balanced; however, the individual forces are not equal to each other. The 50 N force is not equal to the 30 N force.




If an object is at rest and is in a state of equilibrium, then we would say that the object is at "
static equilibrium
." "Static" means stationary or at rest.
Force A: 3.4 N at 161 deg.
Force B: 9.2 N at 70 deg.
Force C: 9.8 N at 270 deg.
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