Make sure that you are able to operate with vectors rapidly and accurately. Practice now will pay off in the rest of the course.

Transcription

1 Ch3 Page 1 Chapter 3: Vectors and Motion in Two Dimensions Tuesday, September 17, :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. 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. Thus, the vector in the previous diagram could be described using its

2 Ch3 Page 2 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 is compressed. Additionally, if the scalar is negative, then the

3 Ch3 Page 3 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

4 Ch3 Page 4 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:

5 Ch3 Page 5 Accelerated motion on a ramp a. b. Example: An object slides from rest down a 5-m long ramp. Determine how long it takes for the object to reach the bottom. Determine the speed of the object when it reaches the bottom. Solution:

6 Ch3 Page 6 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 river. If you were in still water, 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. a. Determine your velocity. b. Determine how long it takes you to cross the river. c. Determine how far up the river you end up. d. In which direction would you have to paddle in order to cross the river to a point directly across your initial position?

7 Ch3 Page 7 Solution: If you wish to specify the velocity vector using its magnitude (i.e., the speed) and direction, then use Pythagoras's theorem to determine the magnitude and use trigonometry to determine the angle. The result is: Thus, the speed is 5.4 km/h, and the direction of the velocity is 22 degrees North of West. (b) Now that we know the speed, which is constant, we could determine the time needed to cross the river if only we knew the distance travelled by the canoe along the slanting path. This can be done by noting that the triangle in the figure below, which shows the displacement of the canoe, is similar (that is, has the same angles) as the triangle above that shows the velocity and its components.

8 Ch3 Page 8 However, an alternative solution for this problem, which foreshadows a very useful strategy for solving projectile-motion problems (and other problems involving two-dimensional motion), is to observe that the canoe's motion can be separated into two independent components. The paddling effort (and its associated westward component of velocity) pushes the canoe across the river; the river current (which is directed northward) pushes the canoe up the river. To determine how long it takes for the canoe to cross the river, we can focus our attention on just the westward components of displacement and velocity: (c) Because the displacement triangle is similar to the velocity triangle,

9 Ch3 Page 9 (d) Experienced paddlers will know that you have to point your canoe up river when you wish to go directly across. The velocities have to look something like this: Knowing the magnitudes of two of the velocities in the triangle enables us to determine the angle, which gives us the direction in which we have to point the canoe. Thus, to go directly across the river, we must point the canoe at an angle of 24 degrees south of west. 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

10 Ch3 Page 10 carousel and you'll get a sense for the complications. 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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12 Ch3 Page 12 Key equations for solving projectile motion problems: Note that the equations above embody the convention that up is the positive direction, and that g represents the magnitude of the acceleration due to gravity. Therefore, if you wish to adopt the convention that down is the positive direction, you should change the two negative signs in the formulas to positive signs. a. b. c. 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 how long it takes for the ball to hit the ground. how far away from the base of the cliff the ball lands. the speed of the ball when it hits the ground. Solution:

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14 Ch3 Page 14 Example: A baseball hit at an angle. 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 a. how long it takes for the ball to reach its peak height. b. the peak height of the ball. c. how long it takes for the ball to hit the ground. d. how far away the ball lands. e. 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:

15 Ch3 Page 15 (e) No, the hit ball does not pass over the outfield wall, because it only travels a horizontal distance of 91.6 m. However, if the ball had travelled 115 m, how could you determine whether the ball passes over the wall, or hits the wall before it hits the ground?

16 Ch3 Page 16 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.) basic formula relating T and f: Remember that the basic formula relating period and frequency is only valid when T and f have compatible units. For example, if you would like T to be measured in minutes (which really means minutes per revolution), then f must be measured in revolutions per minute (rpm).

17 Ch3 Page 17 angular position; polar coördinates; radian measure: 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; this makes it especially convenient for certain formulas relating angular and linear kinematical quantities (as we'll see in Chapter 7). For a full circle, s = 2 r, so the angle (in radians) for a full circle is which is consistent with what we learned in high school. Thus, 1 rev = 360 = 2 rad. Since is a little greater than 6, this means that 1 rad is a little less than direction of acceleration for uniform circular motion

18 Ch3 Page 18 A confusing example: driving around a circular curve in a car. In lecture we discussed the fact that we viscerally feel that we are thrown outwards when we go around a curve in a car; however, the acceleration is actually inward. It's understandable why this is confusing; our bodies are telling us that something is pushing us outwards, yet the following arguments indicate that the acceleration is actually inwards. It's difficult for the brain to understand something when the body feels the opposite. (You can understand why it took a few millennia for the greatest minds in humanity to figure out what is going on here.) We'll discuss circular motion at greater length in Chapters 4 and 6; for now, just focus on the vector diagrams below, and make it your goal to understand that the acceleration in uniform (that is, constant speed) circular motion is really towards the centre of the circle.

19 The following figure is relevant (and the conclusions are valid) for circular motion with CONSTANT speed only: Ch3 Page 19

20 Ch3 Page 20 formula for centripetal acceleration

21 Ch3 Page 21 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 (calculus lovers are welcome to read the passage at the end of the notes below).

22 Ch3 Page 22 a. b. c. Example: A flywheel with radius 30 cm rotates at a constant rate of 200 rpm. Determine the period of rotation. Determine the speed of a point on the rim of the flywheel. Determine the acceleration of a point on the rim of the flywheel. We can also simply say that the period is T = 0.3 s, because the "per revolution" is understood as part of the definition of the period. (b) The distance d covered by a point on the rim of the wheel in one revolution is Because the speed is constant, the speed of a point on the rim of the wheel is

23 Ch3 Page 23 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. You'll also note that we've used a fact from vector calculus, that to differentiate a vector function you just differentiate each component separately. You might like to think about why this is true. 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

24 Ch3 Page 24 Note that Thus