physics ch 3 notes by 43X4vjro

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									Physics        Chapter 3 notes

Vectors
•Any quantity that requires both magnitude (how much) and direction (which way) for a
complete description is called a vector.
•Examples include: displacement, velocity, acceleration, force, momentum, torque, current, etc.

Scalars
•A quantity which can be described by magnitude only (not involving direction) is called a
scalar.
•Examples include:    distance, time, speed, mass, volume, temperature, energy, money.

Mathematics
•In general, the mathematics of scalars is simply the regular addition, subtraction, multiplication,
and division, with which you are familiar.
•Vector mathematics is much more complicated because of the directions involved.
* Vector Addition
•For this course we will complete vector addition by use of scale diagrams.
•Vectors are represented by arrows. The length of the arrow represents the magnitude of the
vector. The direction of the arrow represents the direction of the vector.
•Adding vectors along parallel directions is simple. If the vectors are in the same direction they
add; if they are in opposite directions, they subtract.
•Vectors that lie at other angles are slightly more complicated.
•* The procedure we will use in this course is called the “head-to-tail” method. (Note:   The text
uses the “parallelogram rule” which is different!)

“Head-to-tail” Method
•Select a starting point for your diagram.
•Draw the first vector in the proper direction to scale.
•Use the endpoint of the first vector as the starting point of the next vector.
•Continue adding vectors in this manner until the last vector has been placed.
•The vector sum (called the resultant) is located by drawing a vector from the original starting
point to the final endpoint.
•The resultant can then be measured directly from the diagram.
•The length of the resultant represents the magnitude.
•The direction of the resultant (measured from the original starting point) is the direction.
Resultant
• The vector sum of all forces acting on an object.
• The net force.
Components
•Just as two vectors can be added to form one vector, a single vector can be separated into a pair
of vectors that are at right angles to each other.
•This pair of vectors are called components.
•Any vector may be represented by horizontal and vertical components.
•For vectors at angles, it is sometimes much easier to work with the components of the vectors
rather than the original vector.
• We will discuss components in more detail later in the course.
Projectiles
•A projectile is any object that is propelled into motion by some means and then continues in
motion under the influence of gravity (and air resistance.)
•The path of a projectile is called the trajectory.
Projectile Motion
•Most projectiles follow a curved path which appears to be very complicated. However, these
trajectories will seem much simpler if we study the components of the motion separately.
•The horizontal motion of a projectile is very similar to the simple horizontal motion previously
discussed.
•For a “freely falling” projectile there is no force acting in the horizontal direction at all.
•The horizontal component of a free fall projectile is constant.
• To calculate the horizontal displacement of the projectile at any given time, simply use the
distance formula:
              distance = speed x time
•The vertical motion for a projectile is just like the motion described previously for a vertically
falling object.
•The vertical component of the motion changes because of the gravitational acceleration.
• To calculate the vertical displacement of a projectile at any given time, use the three step
method discussed previously…
• 1.   Calculate the final vertical velocity  Vf = Vo + a t
• 2.   Calculate the average vertical velocity 
                         Va = (Vf + Vo) / 2
• 3.   Calculate the distance 
                         d = Va x t
•The curved path trajectory of a projectile is simply the combination of the constant horizontal
motion and the gravitational accelerated vertical motion.

Example #1 A ball is thrown off a 45 meter tall building with an original horizontal speed of 20
meters per second. How far from the base of the building does the ball strike the ground?
• We start the problem by considering the vertical motion.
• The ball has an original vertical velocity of zero (it is thrown sideways!)
• After is it released, it begins to fall.
• We must determine the amount of time the ball is in the air before hitting the ground.
•The ball begins at a height of 45 meters, then falls. We must use the vertical displacement
calculations to determine how much time is required for the ball to fall 45 meters.


       Time           Final velocity   Average velocity      distance

                                                              5x1
          1          0 + 10 x 1 = 10    0 + 10 / 2 = 5
                                                               =5
                                                              10 x 2
          2          0 + 10 x 2 = 20 0 + 20 / 2 = 10
                                                               = 20
                                                              15 x 3
          3          0 + 10 x 3 = 30   0 + 30 / 2 = 15
                                                               = 45


•It takes the ball three seconds to reach the ground!
• Use the horizontal distance formula to calculate the final position.
• d = s x t = 20 x 3 = 60 meters

Example #2 A cannon is fired from ground level. The original horizontal component of velocity
is 30 m/s. The original vertical component of velocity is 20 m/s. How far from where it is fired
does the cannonball strike the ground?
• Start off with the vertical motion to determine the amount of time the cannonball stays in the
air.
• The cannonball starts at ground level, moves upward, then falls back to the ground.
•The time the ball is in the air is the time at which the final height of the ball is zero.


                      Final vertical   Average vertical
      Time                                                     distance
                        velocity          velocity


                                                                15 x 1
        1            20 + -10x1 = 10    (20+10)/2 = 15
                                                                 = 15


                                                                10 x 2
        2            20 + -10x2 = 0     (20+0)/2 = 10
                                                                 = 20


                      Final vertical   Average vertical
      Time                                                     distance
                        velocity          velocity


                       20 + -10x3          (20-10)/2             5x3
        3
                          = -10               =5                 = 15


                       20 + -10x4          (20-20)/2             0x4
        4
                          = -20               =0                  =0


•The cannonball is in the air four seconds!
• Use the horizontal distance formula to calculate the final position.
• d = s x t = 30 x 4 = 120 meters
Satellites
•A satellite is simply a projectile that falls around the earth rather than into the earth.
•The speed of the satellite must be great enough to ensure the falling distance matches the
earth’s curvature.
•We learned in a previous chapter that an object falling from rest will have a vertical
displacement of 5 meters in one second.
•The earth curves 5 meters downward for each 8000 meters in horizontal distance.
•A satellite must travel 8000 m/s to stay above the ground.
•At this speed, atmospheric friction would burn most materials to a crisp.
•Satellites are placed at high elevations to put the satellite beyond the earth’s atmosphere, not to
be beyond earth’s gravity!
•  For a satellite close to the Earth, the period (the time required for a complete orbit,) is about
90 minutes!
• For higher altitudes, the period is longer.
•Communication satellites are placed at an altitude such that their rotational period is 24 hours,
so they remain at a fixed point in the sky.
•The moon is at a distance such that its rotational period is about 27.3 days.


Conceptual Physics
Projectile Motion Chapter 3 – Homework Sets


Day 1 – Pages 28-31
         1, 2, 19, 20, 21, 23, 29, 30, 32, 39


Day 2 – Pages 31-34
         8, 9, 10, 11, 41, 42, 44, 45


Day 3 – Pages 35-39
         11, 12, 15, 16, 17, 18, 28, 35, 37

								
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