Chapter 3: Vectors and Motion in Two
Dimensions
Tuesday, September 17, 2013
10:00 PM
Vectors are useful for describing physical quantities that have both
magnitude and direction, such as position, displacement, velocity,
acceleration, force, and many other quantities.
Make sure that you are able to operate with vectors rapidly and
accurately. Practice now will pay off in the rest of the course.
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There are 4 main parts to this chapter:
vectors
relative motion
projectile motion
uniform circular motion
Describing Vectors
There are two main ways to describe a vector: either by specifying the
magnitude and direction of a vector, or to specify its components. The
following example shows how to obtain the magnitude and direction of
a vector given its components.
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Thus, the vector in the previous diagram could be described using its
components as (5, 3), or we could describe it equivalently by saying it
is the vector with magnitude 5.8 and angle 31 degrees relative to the
positive x-axis. The following example shows how to convert from
magnitude and direction to components.
Operations with vectors
• multiplying a vector by a scalar
Note that multiplying a vector by a positive scalar that is greater
than 1 just stretches the vector while maintaining the same
direction; if the positive scalar is between 0 and 1, then the vector
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direction; if the positive scalar is between 0 and 1, then the vector
is compressed. Additionally, if the scalar is negative, then the
direction is flipped, but the final vector and the initial vector are
still parallel.
• adding vectors
triangle method for
adding vectors
parallelogram method
for adding vectors;
produces same results
as triangle method
• subtracting vectors
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By drawing graphs, you will be able to see that the rules for combining
vectors using their components (as just illustrated above) yields the same
results as combining vectors graphically, as illustrated further above
Make sure you can combine vectors (scalar multiple, addition,
subtraction) both graphically and using components. Also make
sure that you can determine the components of a vector given
its magnitude and direction, and also determine the magnitude
and direction of a vector given its components. It will help you
a lot if you are able to do such calculations rapidly and
accurately.
Definitions of velocity and acceleration for motion in more than one
dimension:
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Accelerated motion on a ramp
Example: An object slides from rest down a 5-m long ramp.
a. Determine how long it takes for the object to reach the bottom.
b. Determine the speed of the object when it reaches the bottom.
Solution:
Relative Motion
Discussion: A trip on an airplane cruising at a constant speed.
Example: You paddle across a 1-km wide river at a constant speed,
with the front of your canoe pointing directly west, which is
perpendicular to the banks of the canoe. If you were in still water,
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a.
b.
c.
d.
your constant speed would have been 5 km/h, but the current in
the river drags you north. If you were not paddling, your speed due
to the current would be a constant 2 km/h.
Determine your velocity.
Determine how long it takes you to cross the river.
Determine how far up the river you end up.
In which direction would you have to paddle in order to cross the
river to a point directly across your initial position?
Solution:
Similar problems arise in many other types of motion problems. In
airplane navigation, one speaks of "air speed" and "ground speed." In
situations where one of the motions is not at a constant velocity,
additional complications arise. For example, the Earth rotates, which
means its surface does not move at a constant velocity (though the
speed may be approximately constant). This means that air or water
currents that move north or south tend to be deflected, which leads to
the well-known cyclone patterns of air movements in the northern and
southern hemispheres, and to similar patterns of water currents. Think
back to the last time you were moving about on a merry-go-round or
carousel and you'll get a sense for the complications.
Projectile Motion
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Projectile Motion
Examples: a thrown or struck ball, a person jumping through the air,
snowboarder or skier flying off a hill side, cannonball, etc.
Examples that are not projectiles: rockets, airplanes, missiles, etc.; objects
that have their own source of propulsion are not projectiles.
Simplifying assumption: no air resistance.
With this assumption, the path of a projectile is parabolic.
The key to solving a projectile motion problem is to separate the motion
into two components, one horizontal and one vertical. The horizontal
component of velocity is constant; the vertical component of velocity
changes at a constant rate, which is the acceleration due to gravity. The
horizontal and vertical components of the motion are INDEPENDENT, as
the following picture shows:
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Key equations for solving projectile motion problems:
Example: a ball projected horizontally from the top of a cliff
A ball is thrown horizontally from a 10-m high cliff at an initial speed
of 30 m/s. Determine
a. how long it takes for the ball to hit the ground.
b. how far away from the base of the cliff the ball lands.
c. the speed of the ball when it hits the ground.
Solution:
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Example: A baseball hit at an angle.
a.
b.
c.
d.
e.
A baseball is struck at an initial height of 1 m, with an initial speed of
30 m/s at an angle of 40 degrees above the horizontal. Determine
how long it takes for the ball to reach its peak height.
the peak height of the ball.
how long it takes for the ball to hit the ground.
how far away the ball lands.
The outfield wall is 2.5 m high and 110 m away from where the ball is
hit. Does the ball pass over the wall for a home run?
Solution:
Problems for those who are interested: (some of these are discussed in
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Problems for those who are interested: (some of these are discussed in
the textbook)
• determine formulas for the range (i.e., maximum horizontal distance) and
maximum height of a projectile
• for a fixed initial speed, determine the projection angle that maximizes the
range of a projectile (a) on flat ground, and (b) if the object is projected on
slanted ground; Part (b) is considerably more challenging than Part (a);
Galileo used a symmetry argument to solve Part (a), but you calculus
lovers will be able to solve it easily using calculus
• for a fixed distance between thrower and receiver, determine the angle of
projection that minimizes the speed of the projectile when it reaches the
receiver
Circular Motion
fundamental quantities used to describe circular motion:
period T: amount of time needed to complete one revolution (unit: s,
min, h, etc.)
frequency f: number of revolutions per unit time (unit: s-1 (i.e.,
"revolutions per second"), rpm, etc.)
angular position; polar coördinates; radian measure
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the radian is a "unitless unit" of angle measure; if s and r are both
measured in the same distance unit (say metres), then the units cancel
for a full circle, s = 2 r, so the angle (in radians) for a full circle is
thus, 1 rev = 360° = 2 rad
since 2 is a little greater than 6, this means that 1 rad is a little less than 60°
• direction of acceleration for uniform circular motion
a confusing example: driving around a circular curve in a car
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The following figure is relevant (and the conclusions are
valid) for circular motion with CONSTANT speed only:
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• formula for centripetal acceleration
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Let's derive a neat formula for centripetal acceleration. In Part (a)
of the figure, let s represent the length of the circular arc (dashed
arc), which is nearly the same as the length of the vector labelled
d. Then,
From the first figure in Part (c),
We can't be sure of this formula for centripetal acceleration because
of the approximations used, and because some of the quantities
were average values. Nevertheless, the formula is correct, although it
will take calculus to prove that it is exact (see the end of the notes
below).
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below).
Example: A flywheel with radius 30 cm rotates at a constant rate of 200
rpm.
a. Determine the period of rotation.
b. Determine the speed of a point on the rim of the flywheel.
c. Determine the acceleration of a point on the rim of the flywheel.
For calculus lovers only, here is a precise derivation of the formula for
centripetal acceleration. For an object moving in a circle of radius r at a
constant speed, the position function can be written as
Now differentiate twice to obtain an expression for the acceleration,
using the chain rule (remember that theta is a function of time and the
radius of the circle is constant):
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Note that the acceleration is in the direction opposite to the position vector.
The position vector points from the centre of the circle to the location of the
moving object; thus, the acceleration vector points towards the centre of
the circle, which is why it's called a "centripetal" (centre-seeking)
acceleration. The magnitude of the centripetal acceleration is
Note that
Thus
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