chapter 8. dynamics ii: motion in a plane - northern … · copyright © 2008 pearson education,...
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Chapter 8. Dynamics II: Motion in a PlaneChapter 8. Dynamics II: Motion in a PlaneA roller coaster doing a loop-the-loop is a dramatic example of circular motion. But why doesn’t the car fall off the track when it’s upside down at the top of the loop? To answer this question, we must study how objects move in circles.Chapter Goal: To learnhow to solve problems about motion in a plane.
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Topics:Topics:• Dynamics in Two Dimensions • Velocity and Acceleration in Uniform
Circular Motion • Dynamics of Uniform Circular Motion • Circular Orbits • Fictitious Forces• Why Does the Water Stay in the Bucket?• Nonuniform Circular Motion
Chapter 8. Dynamics II: Motion in a PlaneChapter 8. Dynamics II: Motion in a Plane
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Chapter 8. Reading QuizzesChapter 8. Reading Quizzes
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Circular motion is best analyzed in a coordinate system with
A. x- and y-axes. B. x-, y-, and z-axes.C. x- and z-axes. D. r-, t-, and z-axes.
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Circular motion is best analyzed in a coordinate system with
A. x- and y-axes. B. x-, y-, and z-axes.C. x- and z-axes. D. r-, t-, and z-axes.
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This chapter studies
A. uniform circular motion.B. nonuniform circular motion.C. orbital motion.D. Both a and b. E. All of a, b, and c.
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This chapter studies
A. uniform circular motion.B. nonuniform circular motion.C. orbital motion.D. Both a and b. E. All of a, b, and c.
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For uniform circular motion, the net force
A. points toward the center of the circle.B. points toward the outside of the circle.C. is tangent to the circle.D. is zero.
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For uniform circular motion, the net force
A. points toward the center of the circle.B. points toward the outside of the circle.C. is tangent to the circle.D. is zero.
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Chapter 8. Basic Content and ExamplesChapter 8. Basic Content and Examples
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Dynamics in Two DimensionsDynamics in Two Dimensions
Suppose the x- and y-components of acceleration are independent of each other. That is, ax does not depend on y or vy, and ay does not depend on x or vx. Your problem-solving strategy is to1.draw a pictorial representation — a motion diagram (if needed) and a free-body diagram, and2.use Newton’s second law in component form.
The force components (including proper signs) are found from the free-body diagram.
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Dynamics in Two DimensionsDynamics 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.
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Projectile MotionProjectile MotionIn the absence of air resistance, a projectile has only one force acting on it: the gravitational force, FG = mg, in the downward direction. If we choose a coordinate system with a vertical y-axis, then
Consequently, from Newton’s second law, the acceleration is
The vertical motion is free fall, while the horizontal motion is one of constant velocity.
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Uniform Circular MotionUniform Circular MotionThe acceleration of uniform circular motion points to the center of the circle. Thus the acceleration vector has only a radial component ar. This acceleration is conveniently written in the rtz-coordinate system as
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EXAMPLE 8.2 The acceleration of an atomic EXAMPLE 8.2 The acceleration of an atomic electronelectron
QUESTION:
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EXAMPLE 8.2 The acceleration of an atomic EXAMPLE 8.2 The acceleration of an atomic electronelectron
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EXAMPLE 8.2 The acceleration of an atomic EXAMPLE 8.2 The acceleration of an atomic electronelectron
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Dynamics of Uniform Circular MotionDynamics of Uniform Circular MotionThe usefulness of the rtz-coordinate system becomes apparent when we write Newton’s second law in terms of the r-, t-, and z-components, as follows:
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
QUESTION:
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.4 Turning the corner IEXAMPLE 8.4 Turning the corner I
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
QUESTION:
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
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EXAMPLE 8.6 A rock in a slingEXAMPLE 8.6 A rock in a sling
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Circular OrbitsCircular Orbits
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An object moving in a circular orbit of radius r at speed vorbit will have centripetal acceleration of
That is, if an object moves parallel to the surface with the speed
Circular OrbitsCircular Orbits
then the free-fall acceleration provides exactly the centripetal acceleration needed for a circular orbit of radius r. An object with any other speed will not follow a circular orbit.
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Fictitious ForcesFictitious Forces• If you are riding in a car that makes a sudden stop, you may feel as if a force “throws” you forward toward the windshield. • There really is no such force. • Nonetheless, the fact that you seem to be hurled forward relative to the car is a very real experience! • You can describe your experience in terms of what are called fictitious forces. • These are not real forces because no agent is exerting them, but they describe your motion relative to a noninertial reference frame.
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Centrifugal ForceCentrifugal Force
If the car you are in turns a corner quickly, you feel “thrown” against the door. The “force” that seems to push an object to the outside of a circle is called the centrifugal force. Itdescribes your experience relative to a noninertial reference frame, but there really is no such force.
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Why Does the Water Stay in the Bucket?Why Does the Water Stay in the Bucket?
• Imagine swinging a bucket of water over your head. If you swing the bucket quickly, the water stays in. But you’ll get a shower if you swing too slowly.• The critical angular velocity ωc is that at which gravity alone is sufficient to cause circular motion at the top.
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Why Does the Water Stay in the Bucket?Why Does the Water Stay in the Bucket?
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Why Does the Water Stay in the Bucket?Why Does the Water Stay in the Bucket?
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Why Does the Water Stay in the Bucket?Why Does the Water Stay in the Bucket?
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Nonuniform Circular MotionNonuniform Circular Motion
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Nonuniform Circular MotionNonuniform Circular Motion• The force component (Fnet)r creates a centripetal acceleration and causes the particle to change directions. • The component (Fnet)t creates a tangential acceleration and causes the particle to change speed. • Force and acceleration are related to each other through Newton’s second law.
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Problem-Solving Strategy: Circular-Motion Problem-Solving Strategy: Circular-Motion ProblemsProblems
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Problem-Solving Strategy: Circular-Motion Problem-Solving Strategy: Circular-Motion ProblemsProblems
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Problem-Solving Strategy: Circular-Motion Problem-Solving Strategy: Circular-Motion ProblemsProblems
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Problem-Solving Strategy: Circular-Motion Problem-Solving Strategy: Circular-Motion ProblemsProblems
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Chapter 8. Summary SlidesChapter 8. Summary Slides
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General PrinciplesGeneral Principles
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General PrinciplesGeneral Principles
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General PrinciplesGeneral Principles
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Important ConceptsImportant Concepts
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Important ConceptsImportant Concepts
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ApplicationsApplications
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ApplicationsApplications
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Chapter 8. Clicker QuestionsChapter 8. Clicker Questions
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A. slow down and curve downward.B. slow down and curve upward.C. speed up and curve downward.D. speed up and curve upward.E. move to the right and down.
This acceleration will cause the particle to
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This acceleration will cause the particle to
A. slow down and curve downward.B. slow down and curve upward.C. speed up and curve downward.D. speed up and curve upward.E. move to the right and down.
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Rank in order, from largest to smallest, the centripetal accelerations (ar)ato (ar)e of particles a to e.
A. (ar)b > (ar)e > (ar)a > (ar)d > (ar)c B. (ar)b > (ar)e > (ar)a = (ar)c > (ar)d
C. (ar)b = (ar)e > (ar)a = (ar)c > (ar)d
D. (ar)b > (ar)a = (ar)c = (ar)e > (ar)d
E. (ar)b > (ar)a = (ar)a > (ar)e > (ar)d
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Rank in order, from largest to smallest, the centripetal accelerations (ar)ato (ar)e of particles a to e.
A. (ar)b > (ar)e > (ar)a > (ar)d > (ar)c B. (ar)b > (ar)e > (ar)a = (ar)c > (ar)d
C. (ar)b = (ar)e > (ar)a = (ar)c > (ar)d
D. (ar)b > (ar)a = (ar)c = (ar)e > (ar)d
E. (ar)b > (ar)a = (ar)a > (ar)e > (ar)d
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A block on a string spins in a horizontal circle on a frictionless table. Rank order, from largest to smallest, the tensions Ta to Te acting on blocks a to e.
A. Tb > Ta > Td > Tc > Te
B. Te > Tc = Td > Ta = Tb
C. Te > Td > Tc > Tb > Ta
D. Td > Tb = Te > Tc > Ta
E. Td > Tb > Te > Tc > Ta
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A block on a string spins in a horizontal circle on a frictionless table. Rank order, from largest to smallest, the tensions Ta to Te acting on blocks a to e.
A. Tb > Ta > Td > Tc > Te
B. Te > Tc = Td > Ta = Tb
C. Te > Td > Tc > Tb > Ta
D. Td > Tb = Te > Tc > Ta
E. Td > Tb > Te > Tc > Ta
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A. n > w.B. n < w.C. n = w.D.We can’t tell about n without knowing v.
A car is rolling over the top of a hill at speed v. At this instant,
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A car is rolling over the top of a hill at speed v. At this instant,
A. n > w.B. n < w.C. n = w.D.We can’t tell about n without knowing v.
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A ball on a string is swung in a vertical circle. The string happens to break when it is parallel to the ground and the ball is moving up. Which trajectory does the ball follow?
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A ball on a string is swung in a vertical circle. The string happens to break when it is parallel to the ground and the ball is moving up. Which trajectory does the ball follow?