Chapter 8. Dynamics II: Motion in a Plane

Chapter 8. Dynamics II: Motion in a Plane Chapter Goal: To learn how to solve problems about motion in a plane. Slide 8-2

Chapter 8 Preview Slide 8-3

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Chapter 8 Preview Slide 8-8

Dynamics in Two Dimensions Suppose the x- and y-components of acceleration are independent of each other. That is, a x does not depend on y or v y, and a y does not depend on x or v x. Your problem-solving strategy is to: 1. Draw a pictorial representation: a motion diagram (if needed) and a free-body diagram. 2. Use Newton s second law in component form: The force components (including proper signs) are found from the free-body diagram Slide 8-21

Dynamics in Two Dimensions Dynamics in Two Dimensions 3. Solve for the acceleration. If the acceleration is constant, use the two-dimensional kinematic equations of Chapter 4 to find velocities and positions: Slide 8-22

Example 8.1 Rocketing in the Wind Slide 8-23

Example 8.1 Rocketing in the Wind Slide 8-24

Example 8.1 Rocketing in the Wind Slide 8-25

Example 8.1 Rocketing in the Wind Slide 8-26

Example 8.1 Rocketing in the Wind Slide 8-27

Example 8.1 Rocketing in the Wind Slide 8-28

Example 8.1 Rocketing in the Wind Slide 8-29

Projectile Motion: Review In the absence of air resistance, a projectile moves under the influence of only gravity. If we choose a coordinate system with a vertical y-axis, then Consequently, from Newton s second law, the acceleration is Slide 8-30

Projectile Motion: Review Consider a projectile with initial speed v 0, and a launch angle of θ above the horizontal. In Chapter 4 we found that the distance it travels before it returns to the same elevation from which it was launched (the range) is: Trajectories of a projectile launched at different angles with a speed of 99 m/s. The maximum range occurs for θ = 45. All of these results neglect the effect of air resistance. Slide 8-31

Projectile Motion For low-mass projectiles on earth, the effects of air resistance, or drag, are too large to ignore. When drag is included, the angle for maximum range of a projectile depends both on its size and mass. The optimum angle is roughly 35 for baseballs. The flight of a golf ball is even more complex, because of the dimples and effects of spin. Professional golfers achieve their maximum range at launch angles of barely 15! Slide 8-32

Projectile Motion The acceleration of a typical projectile subject to drag force from the air is: The components of acceleration are not independent of each other. These equations can only be solved numerically. The figure shows the numerical solution for a 5-g plastic ball. Slide 8-33

Uniform Circular Motion When describing circular motion, it is convenient to define a moving rtz-coordinate system. The origin moves along with a certain particle moving in a circular path. The r-axis (radial) points from the particle toward the center of the circle. The t-axis (tangential) is tangent to the circle, pointing in the ccw direction. The z-axis is perpendicular to the plane of motion. Slide 8-34

Uniform Circular Motion A particle in uniform circular motion with angular velocity ω has velocity v = ωr, in the tangential direction. The acceleration of uniform circular motion points to the center of the circle. The rtz-components of and are: Slide 8-35

Example 8.2 The Ultracentrifuge Slide 8-40

Example 8.2 The Ultracentrifuge Slide 8-41

Dynamics of Uniform Circular Motion An object in uniform circular motion is not traveling at a constant velocity in a straight line. Consequently, the particle must have a net force acting on it Without such a force, the object would move off in a straight line tangent to the circle. The car would end up in the ditch! Highway and racetrack curves are banked to allow the normal force of the road to provide the centripetal acceleration of the turn. Slide 8-42

Dynamics of Uniform Circular Motion The figure shows a particle in uniform circular motion. The net force must point in the radial direction, toward the center of the circle. This centripetal force is not a new force; it must be provided by familiar forces. Slide 8-43

Example 8.4 Turning the Corner I Slide 8-48

Example 8.4 Turning the Corner I VISUALIZE The second figure below shows the top view of a tire as it turns a corner. The force that prevents the tire from sliding across a surface is static friction. Static friction pushes sideways on the tire, perpendicular to the velocity, since the car is not speeding up or slowing down. The free-body diagram, drawn from behind the car, shows the static friction pointing toward the center of the circle. Slide 8-49

Example 8.4 Turning the Corner I Slide 8-50

Example 8.4 Turning the Corner I Slide 8-51

Example 8.4 Turning the Corner I Slide 8-52

Banked Curves Real highway curves are banked by being tilted up at the outside edge of the curve. The radial component of the normal force can provide centripetal acceleration needed to turn the car. For a curve of radius r banked at an angle θ, the exact speed at which a car must take the curve without assistance from friction is. Slide 8-53

Banked Curves Consider a car going around a banked curve at a speed higher than. In this case, static friction must prevent the car from slipping up the hill. Slide 8-54

Banked Curves Consider a car going around a banked curve at a speed slower than. In this case, static friction must prevent the car from slipping down the hill. Slide 8-55

Circular Orbits The figure shows a perfectly smooth, spherical, airless planet with one tower of height h. A projectile is launched parallel to the ground with speed v 0. If v 0 is very small, as in trajectory A, it simply falls to the ground along a parabolic trajectory. This is the flat-earth approximation. Slide 8-58

Circular Orbits As the initial speed v 0 is increased, the range of the projectile increases as the ground curves away from it. Trajectories B and C are of this type. If v 0 is sufficiently large, there comes a point where the trajectory and the curve of the earth are parallel. In this case, the projectile falls but it never gets any closer to the ground! This is trajectory D, called an orbit. Slide 8-59

Circular Orbits In the flat-earth approximation, shown in figure (a), the gravitational force on an object of mass m is: Since actual planets are spherical, the real force of gravity is toward the center of the planet, as shown in figure (b). Slide 8-60

Circular Orbits An object in a low circular orbit has acceleration: If the object moves in a circle of radius r at speed v orbit the centripetal acceleration is: The required speed for a circular orbit near a planet s surface, neglecting air resistance, is: Slide 8-61

Circular Orbits The period of a low-earth-orbit satellite is: If r is approximately the radius of the earth R e = 6400 km, then T is about 90 minutes. An orbiting spacecraft is constantly in free fall, falling under the influence only of the gravitational force. This is why astronauts feel weightless in space. Slide 8-62

Gravity: Preview of Chapter 13 The circular-orbit equations of this chapter assume: The orbiting object is very small compared to the planet. The orbital radius is very close to the radius of the planet, so that a r = g. When we study orbits of astronomical objects in Chapter 13, such as the moon, we must take into account the fact that the size of g diminishes with increasing distance from the planet. It turns out that the moon, just like the space shuttle, is simply falling around the earth. Saturn s beautiful rings consist of dust particles and small rocks orbiting the planet. Slide 8-63

Fictitious Forces If you are riding in a car that makes a sudden stop, you seem to be hurled forward. You can describe your experience in terms of fictitious forces. Fictitious forces are not real because no agent is exerting them. Fictitious forces describe your motion relative to a noninertial reference frame. Slide 8-64

Centrifugal Force? The figure shows a bird seye view of you riding in a car as it makes a left turn. From the perspective of an inertial reference frame, the normal force from the door points inward, keeping you on the road with the car. Relative to the noninertial reference frame of the car, you feel pushed toward the outside of the curve. The fictitious force which seems to push an object to the outside of a circle is called the centrifugal force. There really is no such force in an inertial reference frame. Slide 8-65

Gravity on a Rotating Earth The figure shows an object being weighed by a spring scale on the earth s equator. The observer is hovering in an inertial reference frame above the north pole. If we pretend the spring-scale reading is F sp = F G = mg, this has the effect of weakening gravity. The free-fall acceleration we measure in our rotating reference frame is: Slide 8-72

Loop-the-Loop The figure shows a rollercoaster going around a vertical loop-the-loop of radius r. Note this is not uniform circular motion; the car slows down going up one side, and speeds up going down the other. At the very top and very bottom points, only the car s direction is changing, so the acceleration is purely centripetal. Because the car is moving in a circle, there must be a net force toward the center of the circle. Slide 8-73

Loop-the-Loop The figure shows the roller-coaster free-body diagram at the bottom of the loop. Since the net force is toward the center (upward at this point), n > F G. This is why you feel heavy at the bottom of the valley on a roller coaster. The normal force at the bottom is larger than mg. Slide 8-74

Loop-the-Loop The figure shows the roller-coaster free-body diagram at the top of the loop. The track can only push on the wheels of the car, it cannot pull, therefore presses downward. The car is still moving in a circle, so the net force is also downward: The normal force at the at the top can exceed mg if v top is large enough. Slide 8-75

Loop-the-Loop At the top of the roller coaster, the normal force of the track on the car is: As v top decreases, there comes a point when n reaches zero. The speed at which n = 0 is called the critical speed: This is the slowest speed at which the car can complete the circle without falling off the track near the top. Slide 8-76

Loop-the-Loop A roller-coaster car at the top of the loop. Slide 8-77

Nonuniform Circular Motion: Review The particle in the figure is speeding up as it moves around the circle. The tangential acceleration is: The centripetal acceleration is: a r = v 2 /r = ω 2 r Slide 8-84

Nonuniform Circular Motion: Review In terms of angular quantities, the equations of constant-acceleration kinematics are: Slide 8-85

Dynamics of Nonuniform Circular Motion A net force is applied to a particle moving in a circle. is likely to be a superposition of several forces, such as tension, thrust, friction, etc. The tangential component of the net force (F net ) t creates a tangential acceleration and causes the particle to change speed. The radial component (F net ) r is directed toward the center, creates a centripetal acceleration, and causes the particle to change direction. Slide 8-86

Dynamics of Nonuniform Circular Motion Force and acceleration are related through Newton s second law: Slide 8-87