2-Dimensional Motion

1. 2-Dimensional Motion Part I: Projectile MotionPart II: Circular Motion

2. 2D Motion y 1D Motion x x First of All, What is 2-D Motion? • Before, we talked about motion in one dimension (just the x axis) • Now we are investigating things that can move it 2 dimensions (x and y)!

3. 2-D Motion Part I:Projectile Motion • A “projectile” is an object on which the only force acting is gravity. • These are things that fly through the air!(Examples: footballs, baseballs, cars flying off cliffs, bullets, cannon shells, and batman) • Usually they are thrown/fired/etc. but are not things like rockets, which still have something “pushing” them as they fly through the air! • As always, gravity will accelerate the object down.

4. The Dimensions are Independent • Since 2D motion is complicated, it is helpful to think of the x and y dimensions separately when we analyze motion. • This is because the x and y dimensions are completely independent. • This lets us use all of our favorite 1D equations when solving problems!

5. Notice that after this idiot jumps off the cliff his velocity in the x dimension does not change. BUT……His velocity in the y dimension is increasing due to gravity. The same is true for this cannonball:

6. Gravity • What causes projectiles to accelerate? • Gravity! • In which dimension does gravity exist? • The vertical (y)! • The acceleration is always 9.81 m/s2downward. This is called “acceleration due to gravity” or “g” • Since acceleration only exists in the y dimension this means it is always zero for the x dimension.

7. Horizontal Velocity • Since we have established that acceleration only exists in the y dimension and the acceleration in the x dimension is zero, what does this mean about the velocity of a projectile in the x dimension? • It is always constant! • This needs to be remembered (by YOU)!

8. Solving Problems • We are going to take some time to go over the problem solving process for these projectile motion problems. • It is a “modified” version of GUESS. • The best part is that you don’t have to learn any complicated equations… we will use the ones you already know from the last unit! It’s easy!

9. Givens • Of course, the first step to solving any problem is to list your givens. • Always begin by drawing a diagram representing the motion. And remember to include your arrows for initial velocity, acceleration, displacement, and the positive direction. • Remember, we are separating the variables into the two dimensions. • You will write out the same givens table as before, but this time it will be a different set for each dimension (x or y)! + a + v v d d

10. Equations • The equations you will use are the same ones. • You will just be using the variables from one dimension (x or y) at a time.

11. Time • In projectile motion, we will treat the x and y variables as completely different sets. • This means the givens we use for one set cannot be used with the other! • However, Time is a variable that is always the same for both x and y.

12. Sample Problem A cannon is placed at the edge of a cliff so the barrel is 60 meters above the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land? As always, we will start by drawing a diagram of the situation. Remember to label any givens and include a direction vector for v, d, a, and + After you have drawn your picture, fill in your Givens table. And of course our unknown variable is x. Always circle the unknown in the table. + Vx=95m/s 95m/s 0m/s vy ay ? 95m/s y =60 m 0m/s2 10.0m/s2 + ? ? ? 60 m x = ?? m

13. + Vx=95m/s 95m/s 0m/s vy ay ? 95m/s dy=60 m 0m/s2 10.0m/s2 + ? ? dx = ?? m Sample Problem A cannon is placed at the edge of a cliff so the barrel is 60 meters above the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land? Now determine the equations you will need to solve the problem. Remember, we will keep the sets of variables completely separate (except time)!!! x = vot + ½ at2 x = (95)(3.5) + ½ (0)(3.5)2 x = 332.5 m y = vot + ½ at2 t = √(2y/ a) t = √[(2)(60) / (10.0)] = 3.5 s This is the only equation we can use to solve for x. But we are missing time! We need to use the variables in the y-dimension to find time first. 3.5 s 3.5 s ? 60 m

14. Practice Problem Thelma and Louise drive their car off the edge of the grand canyon at a speed of 30 m/s and die 3.4 seconds later. (a) how tall is the canyon at that spot? Answer: 57.8 m (b) how far away from the bottom of the canyon do they land? Answer: 102 m

15. Which of the following does not effect the hang time of a projectile? • Angled fired • Vertical velocity • Horizontal velocity • Height it was fired at. [Default] [MC Any] [MC All]

16. Angled Projectile Motion • Occasionally, you will encounter problems where the projectile is initially fired at an angle. • This means that the initial velocity will have a horizontaland a vertical component. • You will need to calculate these components in order to fill in your givens table on each problem.

17. Vector Components • In order to find the components of a vector (like velocity) you will need to use those timeless Trigonometric Functions.

18. Vector Components • So we have a projectile that is fired at an angle of 30o with respect to the horizontal at a velocity of 50 m/s. • We need to think of it like the projectile is fired vertically and horizontally at the same time, giving it both components. vy V=50m/s vx Θ=30o

19. Vector Components • In order to calculate the components, we need to shift vy to make a right triangle. • Then we can use trig functions to solve for vy and vx like they are sides of a right triangle. • To solve for vx, we will use cosine because it is adjacent and we have the hypotenuse. vy V=50m/s vx Θ=30o

20. Vector Components • To solve for vx, we will use cosine because it is the adjacent side and we have the hypotenuse. • To solve for vy, use the same process but with sine. vy V=50m/s vx Θ=30o

21. Sample Problem A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of 20m/s. What is the range of the soccer ball? (how far away does it land) As always, start by drawing a diagram, including all vectors. We need to use trig functions to solve for vix and viy And as always circle your unknown variable. 6.84m/s ? +18.8m/s ? Begin to fill in what you can on your givens table 6.84m/s ? ? -18.8m/s 0 m/s2 -10.0 m/s2 ? ? ? 0 m a + vix = VcosΘ vix=(20)cos(70) vix=6.84 m/s V=20m/s vy viy = VsinΘ viy=(20)sin(70) viy=18.8 m/s Θ=70o vx dx +

22. a + V=20m/s vy Θ=70o vx dx + Sample Problem A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of 20.0m/s. What is the range of the soccer ball? (how far away does it land) dx = vixt + ½ at2 dx = (6.84)(t) + ½ (0)t2 dx = (6.84)(3.76) + 0 dx = 25.7 m vf = vi + at t = (vf – vi) / a t = (-18.8 – 18.8) / (-10.0) t = 3.76 s

23. Practice Problem A cannon shoots a cannon ball at 50° from the horizontal at a velocity of 300 m/s. What is the range (dx) of the cannon ball? Answer: ~8,900 m

24. Projectile Motion Formulas

26. 2-D Motion Part II:Circular Motion You spin me right round….

27. Circular Motion • Circular motion is exactly what is sounds like: objects moving in a circular path. • We are going to investigate how objects rotate and revolve, and why they do this…

28. Rotation vs. Revolution • Before we get started lets make a quick differentiation. • When we say “Rotation,” we will be talking about something spinning. • The axis of rotation is inside the object. • When we say “Revolution” we will be talking about something going around something else. • The axis of rotation is outside the object.

29. Angular Velocity • For an object in circular motion, it makes sense to describe its velocity in angular terms. • Meaning: we will describe the velocity of an object in motion by how many degrees (or radians) it covers in a certain amount of time. • The symbol for angular velocity is a lower-case omega: ω • Also, the units for Θ must be in radians. The equation for angular velocity:

30. Radians?!? • If you’re in Pre-Calculus (which you should be) then you have already been introduced to the idea of radians. But here is the basic idea: • One full revolution of a circle (360o) is equal to 2π Radians. • So if you have degrees and want to convert to radians the equation is: • For any problem where you need to find the angular velocity, be sure that your units are in radians!

31. Sample Problem A child on a merry-go-round completes one rotation in 60 seconds. What is the child’s angular velocity? ω = ∆Θ / t ω = (2π) / (60) ω = 0.10 rad/s ∆Θ is 2π because he completes one rotation

32. Tangential Velocity • Suppose we have 2 horses on a carosel. The black horse is 1 meter from the center and the white horse is 2 meters from the center. • Which horse has a greater angular velocity? • They have the same! They will each cover a full rotation (360o) in the same amount of time. • Which horse “feels” like they are going “faster”? • The white one! • The white horse is going faster because it has a greater Tangential Velocity. • It covers a greater distance (circumference) in the same amount of time.

33. Tangential Velocity • Hopefully you learned in math what a “Tangent” line is. • (Basically it’s a straight line that touches one point on a curve.) • When we talk about the tangential velocity of an object in motion we mean its actual velocity (distance/time). • Tangential velocity can be found using the angular velocity and radius. • It can also be found if you know the distance (circumference) traveled and time. • The effect of tangential velocity can hurt!

34. Sample Problem A student twirls a rock on a string above their head. The rock completes 2 revolutions in 4.50 seconds and the length of the string is 0.750 meters. What is the tangential velocity of the rock? As always, start by listing Givens and Unknowns: t = 4.50 s ∆Θ = 2 revs = 4π radians r = 0.750 meters. vT = ? vT = rω vT = (0.75)(?) vT = (0.75)(2.79) vT = 2.09 m/s ω = ∆Θ/t ω = (4π)/(4.5) ω = 2.79 rad/s The equation we will need to use is: vT = rω But there is a problem.. We don’t know ω! So we must use ω = ∆Θ/t

35. Question… • We know that it is possible for an object to rotate in a circle with a constant speed, but can an object rotate with a constant velocity? • The answer is NO! But why? • Because the direction is constantly changing. Remember: velocity is a vector and has a magnitude and direction! • What is it called when something has a change in velocity? • Acceleration! In this case we refer to it as Centripetal Acceleration.

36. Centripetal Acceleration • Centripetal Acceleration is what causes objects to travel in a circle. • “Centripetal” means “center pointing” • As the object is moving, the acceleration pushes toward the axis of rotation (center) and causes it to change direction and travel in a circle. • The equation for Centripetal Acceleration is:

37. Sample Problem A car is doing donuts out in the parking lot and is traveling in a perfect circle with a radius of 15 meters at a constant speed of 18 m/s. What is the magnitude of the centripetal acceleration acting on the car? Givens: vT = 18 m/s r = 15 m ac = ? ac = vT2/r ac = (18)2 / 15 ac = 21.6 m/s2

38. “g” Forces • “g” Forces are used to describe acceleration in terms of how it compares to acceleration due to gravity (g) which is 10m/s2. • When undergoing centripetal acceleration, people can undergo many g’s!

39. UCM Formulas