kinematics
TRANSCRIPT
UCVTS AIT Physics
Kinematics
• Kinematics– the branch of mechanics that deals with the study of the
motion of objects without regard to the forces that cause the motion
• Displacement– A vector that points from an object’s initial position (Xo) to its
final position (X) and has a magnitude equal to the shortest distance between the 2 positions
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Vector Example
• A particle travels from A to B along the path shown by the dotted red line– This is the distance
traveled and is a scalar
• The displacement is the solid line from A to B– The displacement is
independent of the path taken between the two points
– Displacement is a vector
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Kinematics• Kinematics in 1 Dimension
– We live in a 3-dimensional world, so why bother analyzing 1-dimensional situations? Well, because any translational (straight-line, as opposed to rotational) motion problem can be separated into one or more 1-dimensional problems. Problems are often analyzed this way in physics (and remember throughout your future career; a complex problem can often be reduced to a series of simpler problems).
– The first step in solving a kinematics problem is to set up a coordinate system. This defines an origin (a starting point) as well as positive and negative directions. We'll also need to distinguish between scalars and vectors (which we have done already last week….remember? I know you do…). A scalar is something that has only a magnitude, like area or temperature, while a vector has both a magnitude and a direction, like displacement or velocity.
– In analyzing the motion of objects, there are four basic parameters to keep track of. These are • time (t)• displacement (x or y)• velocity (v)• acceleration (a).
• Time is a scalar, while the other three are vectors. In 1 dimension, however, it's difficult to see the difference between a scalar and a vector. The difference will be more obvious in 2 dimensions.
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Kinematics• Kinematics in 1 Dimension
– (CJ6 Section 2.1) Displacement– The displacement represents the distance traveled, but it is a vector, so it also gives the direction. If you start in a particular
spot and then move north 5 meters from where you started, your displacement is 5 m north. If you then turn around and go back, with a displacement of 5 m south, you would have traveled a total distance of 10 m, but your net displacement is zero, because you're back where you started. Displacement is the difference between your final position (x) and your starting point (xi) : It is a vector that points from an object’s initial position to it’s final position.
– (CJ6 Section 2.2) Speed and Velocity– Imagine that on your way to class one morning, you leave home on time, and you walk at 3 m/s east towards campus. After
exactly one minute you realize that you've left your physics assignment at home, so you turn around and run, at 6 m/s, back to get it. You're running twice as fast as you walked, so it takes half as long (30 seconds) to get home again.
– There are several ways to analyze those 90 seconds between the time you left home and the time you arrived back again. One number to calculate is your average speed, which is defined as the total distance covered divided by the time. If you walked for 60 seconds at 3 m/s, you covered 180 m. You covered the same distance on the way back, so you went 360 m in 90 seconds.
– Average speed = distance / elapsed time = 360 / 90 = 4 m/s. – The average velocity, on the other hand, is given by:
• In this case, your average velocity for the round trip is zero, because you're back where you started so the displacement is zero
distance
total time
: ,
instantaneous velocity
:
avg speed
xaveragevelocity v x is thedisplacement vector v has samedirectionas x
tv
v t is smalltv
averageacceleration a v is a vectort
:x
averagevelocity vt
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Kinematics
• Acceleration– An object accelerates whenever its velocity changes. Going back to the example we used above, let's say instead of instantly
breaking into a run the moment you turned around:• you steadily increased your velocity from 3m/s west to 6 m/s west in a 10 second period.
– If your velocity increased at a constant rate, you experienced a constant acceleration of 0.3 m/s per second (or, 0.3 m/s2).
– We can figure out the average velocity during this time. If the acceleration is constant, which it is in this case, then the average velocity is simply the average of the initial and final velocities. The average of 3 m/s west and 6 m/s west is 4.5 m/s west. This average velocity can then be used to calculate the distance you traveled during your acceleration period, which was 10 seconds long. The distance is simply the average velocity multiplied by the time interval, so 45 m.
– Similar to the way the average velocity is related to the displacement, the average acceleration is related to the change in velocity: the average acceleration is the change in velocity over the time interval (in this case a change in velocity of 3 m/s in a time interval of 10 seconds). The instantaneous acceleration is given by:
• As with the instantaneous velocity, the time interval is very small (unless the acceleration is constant, and then the time interval can be as big as we feel like making it).
– On the way out, you traveled at a constant velocity, so your acceleration was zero. On the trip back your instantaneous acceleration was 0.3 m/s2 for the first 10 seconds, and then zero after that as you maintained your top speed. Just as you arrived back at your front door, your instantaneous acceleration would be negative, because your velocity drops from 6 m/s west to zero in a small time interval. If you took 2 seconds to come to a stop, your acceleration is -6 / 2 = -3 m/s2.
va
t
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Kinematics
• Equations of Kinematics when Acceleration is Constant– When the acceleration of an object is constant, calculations of the distance
traveled by an object, the velocity it's traveling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined.
– Note that the equations apply under these conditions: – the acceleration is constant – the motion is measured from t = 0 – the equations are vector equations, but the variables are not normally written in
bold letters. The fact that they are vectors comes in, however, with positive and negative signs.
– The equations are:
0
0 0
20 0
2 20 0
1( )
21
2
2 ( )
v v at
x x v v t
x x v t at
v v a x x
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Kinematics• Applications of the Equations of Kinetics
– 1. Make a drawing to represent the situation being studied– 2. Decide which directions are positive and negative– 3. Make a chart and write down all known values and what the question is asking for.– 4. Verify that that the given information contains at least 3 of the 5 kinetic variables.– 5. If the motion of the object is divided into segments, remember that the final velocity of one segment is the initial velocity of the next segment.
– Do Example 4 problem in CJ6 page 31
– Do Example 6 Problem in CJ6 on page 35
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Kinematics
• Applications of the Equations of Kinetics
– Do Example 6 Problem in CJ6 on page 35• Find the displacement of the jet when v=62m/s
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Kinematics• Freely Falling Bodies (Free Fall)
– Objects falling straight down under the influence of gravity are excellent examples of objects traveling at constant acceleration in one dimension. This also applies to anything you throw straight up in the air which, because of the constant acceleration downwards, will rise until the velocity drops to zero and then will fall back down again.
– The acceleration experienced by a dropped or thrown object while it is in flight comes from the gravitational force exerted on the object by the Earth. If we're dealing with objects at the Earth's surface, which we usually are, we call this acceleration g, which has a value of 9.8 m/s2. This value is determined by three things: the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant.
– A typical one-dimensional free fall question (free fall meaning that the only acceleration we have to worry about is g) might go like this. – You throw a ball straight up. It leaves your hand at 12.0 m/s.
• How high does it go? • If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight? • How fast is it traveling when you catch it?
– Origin = height at which it leaves your hand Positive direction = up
– (a) At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation: – This gives: 0 = 144 + 2 (-9.8) x – Solving for x gives x = 7.35 m, so the ball goes 7.35 m high.
• (b) To analyze the rest of the problem, it's helpful to remember that the down half of the trip is a mirror image of the up half. In other words, if, while going up, the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its maximum height, and then double that to get the total time.
• Another way to do it is simply to plug x = 0 into the equation: • This gives 0 = 0 + 12 t - 4.9 t2 • A factor of t can be canceled out of both terms, leaving: • 0 = 12 - 4.9 t, which gives a time of t = 12 / 4.9 = 2.45 s.
– (c) The answer for part (c) has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation:
– v = vo + a t which gives: • v = 12 - 9.8 (2.45) = -12 m/s.
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Kinematics• Freely Falling Bodies (Free Fall)
– Do Example 12 (CJ6 page 41)
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Kinematics
• 2D Kinematics– Motion can be described in terms of time t and the x and y components of the
displacement, acceleration and initial and final velocity vectors.– Treat the x and y motion separately (each occurs as if the other was not
happening!)• Combine x and y motions at the end of the problem using Pythagorean
Theorem and trigonometry
0 0
0 0
0 0
2 20 0
2 2 20 0
1 1( ) ( )
2 21 1
2 2
2
x y
f x f y
x y
f x x x f y y y
x f x y f y
x x y y
f x x x f y
x component y component
x displacement y
a acceleration a
v final velocity v
v initial velocity v
t elapsed time t
v v a t v v a t
x v v t y v v t
x v t a t y v t a t
v v a x v v
2 2y ya y
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Kinematics• Projectile Motion
– Idealized kind of motion that occurs when a moving object experiences only the acceleration due to gravity, acting vertically downward. We will use this type of motion as the basis for our study of 2D Kinematics.
– For the trajectory of projectile motion, ay has a magnitude of 9.8m/s2 in the x direction there is no acceleration (ax=0)
– The x motion takes place as if the y motion isn't happening, and the y motion takes place independent of whatever is happening in the x direction.
– One good example of this is the case of two objects (e.g., baseballs) which are released at the same time. One is dropped so it falls straight down; the other is thrown horizontally. As long as they start at the same height, both objects will hit the ground at the same time, no matter how fast the second one is thrown.
• Do example 2 (page 60) and example 3 (page 62)
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Physics• Problem Solving Steps
– 1. Don’t Panic! Every problem has a solution.• Well, at least the problems this year in Physics
– 2. READ the problem, READ the problem!– 3. Construct an informative diagram of the physical situation (sketch).– 4. Identify and list the given information in variable form (make a table of values)– 5. Identify and list the unknown information in variable form. – 6. Identify and list the equation which will be used to determine the unknown information from
the known variables. – 7. Substitute known values into the equation and use appropriate algebraic steps to solve for
the unknown • or solve for the unknown variable first and then substitute known values (this is generally preferable and
easier)– 8. Check your answer to ensure that it is reasonable and mathematically correct (include UNITS
in you solution and final answer)