Motion in Two Dimensions Physics Chapter 6 Motion in Two Dimensions

Physics Turn in Chapter 5 Homework & Worksheet & Lab Lecture Q&A

1-D Review Position: d (unit: m) (unit: m) Displacement: d = df – di Velocity: (unit: m/s) Instantaneous velocity = slope of Position-Time graph Acceleration: (unit: m/s2) Instantaneous acceleration = slope of Velocity-Time graph

1D Review (2) The 3 Great Equations of constant acceleration motion:

Projectile Motion: 2-D Choose coordinate so that the motion is confined in one plane  2-D motion x y d x y v vx vy vi vix or vxi viy or vyi ay a ax

Package Dropped From Airplane

Snapshots Horizontal Direction: velocity is _______ constant Vertical Direction: velocity is __________ increasing

Ball projected straight up from truck

Snapshots

Projectile Motion: Horizontal and Vertical Displacements

Velocity Components of Projectile

Independence of Motion From observation: The horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. Connection: Both horizontal and vertical motions are functions of time. Time connects the two independent motions. Though these two motions are independent, they are connected to each other by time.

Initial Velocity Initial velocity: vi at angle i with the horizontal Set up coordinate: Horizontal: x direction Vertical: y direction y Decompose vi into horizontal and vertical direction: viy vi i x vix

Projectile Motion Breakdown Projectile: Object launched into the air Horizontal: Constant velocity

Projectile Motion Breakdown (2) Similarly, Vertical: Constant acceleration (ay = g, downward) ay = g if downward is defined as +y direction ay = -g if upward is defined as +y direction

Symmetry of Trajectory Trajectory: Path of projectile Upward motion and downward motion are symmetric at same height. Upward total time is equal to downward total time if landing point is at same height as initial point. At the same height, speed is the same. vx stays unchanged. vy remains at the same magnitude but changes in direction. vy is upward when the projectile is going up and vy is downward when coming down

Symmetry of Trajectory (2) Velocity at any moment is tangent to the actual path. Velocity is horizontal at the top of the trajectory. v vy vx vx vy v

Highest Point of Trajectory x y Highest Point of Trajectory Minimum speed at top, but vy = 0 vx = vix = vi cos i Maximum Height is

x y Horizontal Range Two angles (complementary) with the same initial speed give the same range. Horizontal range is maximum when the launch angle is 45o. Valid only when the landing point and initial point are at the same height.

Equation of Path No time involved. x y Equation of Path No time involved. Can be used to find x or y when the other is given. Parabolic equation  Trajectory is parabolic. Valid only when upward is defined as the +y direction and origin is set at the initial point.

Example: 150-1 A stone is thrown horizontally at a speed of + 5 Example: 150-1 A stone is thrown horizontally at a speed of + 5.0 m/s from the top of a cliff 78.4 m high. a) How long does it take the stone to reach the bottom of the cliff? b) How far from the base of the cliff does the stone hit the ground? c) What are the horizontal and vertical components of the velocity of the stone just before it hits the ground? Set up coordinates as to the right. Then ay = y vi = 5m/s x g, yi = 0, xi = 0, vix = 5m/s, viy = y = 78.4m a) t = ? In y-direction: b) In x-direction:

Continues … x y vx c) Horizontal: vy v Vertical: Speed: Direction: vi = 5m/s x y Continues … vx c) Horizontal: vy  v Vertical: Speed: Direction:

Practice: A softball is tossed into the air at an angel of 50 Practice: A softball is tossed into the air at an angel of 50.0o with the vertical at an initial velocity of 11.0 m/s. What is its maximum height? y x Or Max. Height Eqn.

Practice: 152-10 A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0o below the horizontal. How far does the ball move horizontally before it hits the ground? y x

Uniform Circular Motion Circular path or circular arc Uniform = Constant speed (Constant velocity? why?) v Magnitude of velocity: constant Direction of velocity: v changing Velocity: changing

Centripetal Acceleration v: speed of particle r: radius of circle or circular arc , where v Direction of acceleration is always toward the center of circle (or circular arc)  Centripetal a a v a v for uniform circular motion at any time.

Direction of Acceleration a in the same direction as v:  Speed: increases a opposite to v:  Speed: decreases a  v:  Speed: does not change  Direction of velocity: changing

Uniform Circular Motion Period: Time for one complete cycle Frequency: Number of cycles per unit time Unit:

Centripetal Force Centripetal force is in general not a single physical force; rather, it is in general the net force. Do not draw centripetal force on force diagram (Free Body Diagram)

Example: 156-12 A runner moving at a speed of 8 Example: 156-12 A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. a) Find the centripetal acceleration of the runner. b) What agent exerts force on the runner (to round the bend)? b) The friction the ground giving to the shoes provides the centripetal force.

Practice: 156-13 A car racing on a flat track travels at 22 m/s around a curve with a 56-m radius. Find the car’s centripetal acceleration. What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping? b) The friction provides the centripetal force, so N f W

What if? If centripetal force is not provided or not large enough, the object will not be able to move in the circle it intends to move. • • Centripetal force disappears. Insufficient Centripetal force.

Relative Velocity VA,B + Velocity of A relative to B Velocity of A measured by B Velocity of A at reference frame B 1D or 2D 1D: Define positive direction 2D: vector addition (head-tail or parallelogram)

Example:159-22 You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to the bus. What is your speed relative to the street? Let direction bus moving = ”+” direction, also let you = y, bus = b, street = s, then vb,s = 2.0m/s, Vy,b = 4.0 m/s, Vy,s =?

Example: 167-71 A weather station releases a balloon to measure cloud conditions that rises at a constant 15 m/s relative to the air, but there is also a wind blowing at 6.5 m/s toward the west. What are the magnitude and direction of the velocity of the balloon? x y Let west = +x, up = +y. Balloon = b, air/wind = a, ground = g. vb,a = 15 m/s, va,g = 6.5 m/s, vb,g = ?, = ? va,g vb,a vb,g

Practice:159-24 A boat is rowed directly upriver at a speed of 2 Practice:159-24 A boat is rowed directly upriver at a speed of 2.5 m/s relative to the water. Viewers on the shore see that the boat is moving at only 0.5 m/s relative to the shore. What is the speed of the river? Is it moving with or against the boat? Let upriver = ”+” direction, also let boat = b, river/water = w, shore = s, then vb,w = 2.5 m/s, Vb,s = 0.5 m/s, Vw,s =? against