# Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 4. Kinematics in Two Dimensions Chapter Goal: To learn to solve.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 4. Kinematics in Two Dimensions Chapter Goal: To learn to solve problems about motion in a plane.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Ch. 4 - Student Learning Objectives To identify the acceleration vector for curvilinear motion. To compute two-dimensional trajectories. To read and interpret graphs of projectile motion. To solve problems of projectile motion using kinematics equations. To understand the kinematics of uniform and non-uniform circular motion.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration in two dimensions This acceleration vector will cause the particle to: A: slow down in the direction it’s going. B.slow down and curve to the left C.reverse direction D.turn left at a constant speed

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens when acceleration is constant along one axis and zero along the other?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another A toy rocket ship is launched with a constant vertical acceleration of 8 m/s 2 from a rolling cart with an initial horizontal velocity of 2 m/s. Draw a graph of this particle’s trajectory for t = 0,1, 2,3,4 and 5 s. Label the x,y coordinates of each point. What is the shape of the trajectory?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another A toy rocket ship is launched with a constant vertical acceleration of 8 m/s 2 from a rolling cart with an initial velocity of 2 m/s. What does the slope at a given point indicate?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Slope of the trajectory graph The value of the slope equals (dy/dx) which is NOT equal to the magnitude of the velocity. If the slope is horizontal, the object has not stopped, its going in a horizontal direction The angle of the slope indicates the direction of motion; therefore: tan θ = v y /v x

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Compare these two graphs The position vs time graph (left) looks similar to the trajectory graph. The value of the slope for the p vs t graph = ds/dt. This IS the value of the velocity (when that slope is horizontal, v=0). The value of the slope of the p vs t graph does not indicate direction.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A rocket-powered hockey puck moves on a frictionless horizontal table. It starts at the origin. At t=5s, v x = 40 cm/s. a.In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis. b.How far from the origin is the puck at t = 5s?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A rocket-powered hockey puck moves on a frictionless horizontal table. It starts at the origin. At t=5s, v x = 40 cm/s. a.In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis. 62 degrees b.How far from the origin is the puck at t = 5s? 180 cm

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Projectile motion on another planet A student on planet Exidor throws a ball that follows a parabolic trajectory as shown. The ball’s position is shown at 1- second intervals until t = 3s. At t=1s, the ball’s velocity is the value shown. a.Determine the ball’s velocity at t = 0,2, and 3 seconds. b.Determine the value of g on Exidor. c.Determine the angle at which the ball was thrown.

Projectile motion on another planet |g| = 2 m/s 2 θ = tan -1 (v 0y /v 0x ) = 63 0

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Throwing a ball A ball is thrown up toward a cliff of height h with a speed of 30 m/s and an ang/e of 60 degrees above horizontal. It lands on the edge of the cliff 4.0 seconds later. A.How high is the cliff? B.What was the maximum height of the ball above the ground C.What is the impact speed?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Throwing a ball A ball is thrown up toward a cliff of height h with a speed of 30 m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 4.0 seconds later. A.How high is the cliff? – 25.5 m B.What was the maximum height of the ball above the ground? 34.4 m C.What is the impact speed? – 20 m/s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The water hose Water from a garden hose pointed 25° above horizontal, lands directly on a sunbather lying on the ground 4.4 m away in the horizontal direction. The nozzle is held 1.4 m above the ground. At what speed does the water leave the nozzle?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The water hose Water from a garden hose pointed 25° above horizontal, lands directly on a sunbather lying on the ground 4.4 m away in the horizontal direction. The nozzle is held 1.4 m above the ground. At what speed does the water leave the nozzle? 5.78 m/s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion Period (T) – the time it takes to go around the circle once. The SI units of T are seconds (per cycle, or per revolution) The reciprocal of T are rps, revolutions per second

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion – motion around a circle at constant speed For a particle traveling in circular motion, we can describe it’s position by distance from origin, r, and angle θ, in radians: θ = s/r where s is the arc length (meters) and r is the radius of the circular motion (also in meters). The unit of radians is dimensionless. There are 6.28 (2π) radians in a circle. By convention, counterclockwise direction is positive.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion - Velocity Translational Angular pos: s metersθ = s/r radians vel: ds/dt = v (m/s)dθ/dt = v/r = ω rads/s (s -1 )

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moves cw around a circle at constant speed for 2.0 s. It then reverses direction and moves ccw at half the original speed until it has traveled through the same angle. Which is the particle’s angle-versus-time graph?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A convenient coordinate system for circular motion – the rtz system

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Circular Motion – Velocity Vector Components Consider the velocity vector The components are v r = 0, therefore |v t | = where r denotes radial component and t denotes tangential component. For tangential components, ccw = positive, cw = negative. For radial components, positive is into the center of the circle.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion – direction of acceleration vector

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration of Uniform Circular motion Consider the acceleration vector, graphical vector subtraction shows that for uniform circular motion: a t = 0 (v t constant) a r = The magnitude of a r is: a r = v 2 /r or a r = ω 2 r a r is centripetal acceleration The magnitude of a r is constant, but direction is not Cannot use a r in kinematic equations

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rank in order, from largest to smallest, the centripetal accelerations (a r ) a to (a r ) e of particles a to e.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #29, more or less Your roommate spins the 60-cm diameter bicycle wheel while working on his bicycle. You notice that a stone is stuck in the tread. It goes by 3 times every second. Find T, ω (in SI units), v t and a r for the stone:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #29 - answer Your roommate spins the 60-cm diameter bicycle wheel while working on his bicycle. You notice that a stone is stuck in the tread. It goes by 3 times every second. Find T, ω, v t and a r for the stone: T = 0.33 s/rev ω = 19 rads/s v t = 5.7 m/s a r = 108 m/s 2

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Nonuniform Circular Motion If the object changes speed: a t = dv/dt a r = v 2 /r where v is the instantaneous speed. if a t is constant, arc length s and velocity v t can be found with kinematic equations.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Nonuniform Circular Motion For non-uniform circular motion: the acceleration vector no longer points towards the center of the circle. a t can be positive (ccw) or negative (cw) but a r is always positive.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #34 A car speeds up as it makes a circular turn from heading due south to heading due east. When exactly halfway around the curve (heading southeast) his total acceleration is 3.0 m/s 2, 20 degrees north of east. What are the magnitudes of the radial and tangential components of his acceleration? What is the cardinal direction of each component? Southeast is 45 degrees from due east.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #34 A car speeds up as it makes a circular turn from heading due south to heading due east. When exactly halfway around the curve (heading southeast) his total acceleration is 3.0 m/s 2, 20 degrees north of east. What are the magnitudes of the radial and tangential components of his acceleration? What is the cardinal direction of each component? Southeast is 45 degrees from due east. Ans: radial: 2.7 m/s 2, northeast tangential 1.27 m/s 2, southeast

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Acceleration a t = dv/dt where v = v t v t = r ω a t = r dω/dt The quantity dω/dt is defined as the angular acceleration: α = dω/dt, in rad/s 2 a t = r α

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Acceleration the graphical relationships for angular velocity and acceleration are the same as those for linear velocity and acceleration

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Comparisons for accelerated circular motion translational motionangular motion pos:s metersθ = s/r radians vel:ds/dt = v m/sdθ/dt = v/r = ω rads/s (s -1 ) acc:dv/dt = a t m/s 2 dω/dt = a t /r = α rads/s 2

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Acceleration if α is constant, θ and ω can be found with the angular equivalents of the kinematic equations.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Crankshaft slowing down A 3.0 cm diameter crankshaft that is rotating at 2500 rpm slows to a stop at a constant rate in 1.5 s. a.What is the magnitude of the tangential acceleration (a t ) of a point on the surface? b.What is the centripetal acceleration after 1.0 s? c.How many revolutions does the crankshaft make as it stops?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Crankshaft slowing down a.What is the magnitude of the tangential acceleration (a t ) of a point on the surface? 2.62 m/s 2 b.What is the centripetal acceleration after 1.0 s? 114.3 m/s 2 c.How many revolutions does the crankshaft make as it stops? 31.3 revolutions

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The fan blade is slowing down. What are the signs of ω and α, respecitively? a.Pos, pos b.Pos, neg c.Neg, pos d.Neg, neg

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Relationship between angular and tangential quantities Point 1 and 2 on the rotating wheel have different radii. θ 1 = θ 2 but s 1 ≠ s 2 ω 1 =ω 2 but v 1 ≠ v 2 α 1 = α 2 but a t1 ≠ a t2