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Copyright R. Janow – Spring 2012 Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed.: Ch. 4.1 – 4.4 Position, velocity, acceleration vectors Average.

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Presentation on theme: "Copyright R. Janow – Spring 2012 Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed.: Ch. 4.1 – 4.4 Position, velocity, acceleration vectors Average."— Presentation transcript:

1 Copyright R. Janow – Spring 2012 Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed.: Ch. 4.1 – 4.4 Position, velocity, acceleration vectors Average & Instantaneous Velocity Average & Instantaneous Acceleration Two Dimensional Motion with Constant Acceleration (Kinematics) Projectile Motion (Free Fall) Uniform Circular Motion Tangential and Radial Acceleration Relative Velocity and Relative Acceleration 4.1Position, Velocity and Acceleration Vectors 4.2Two Dimensional Motion with Constant Acceleration 4.3Projectile Motion 4.4A particle in Uniform Circular Motion 4.5Tangential and Radial Acceleration 4.6Relative Velocity and Relative Acceleration

2 Copyright R. Janow – Spring 2012 Motion in two and Three Dimensions Extend 1 dimensional kinematics to 2 D and 3 D Kinematic quantities become 3 dimensional Motions in the 3 perpendicular directions can be analyzed independently Vectors needed to manipulate quantities Constant acceleration Kinematic Equations hold component-wise for each dimension Same for z In vector notation each equation is 3 separate ones for x, y, z

3 Copyright R. Janow – Spring 2012 Position and Displacement A particle moves along its path as time increases Displacement Positions Trajectory: y = f(x) Parameterized by time always points from the origin to the particle’s location Path does not show time dependence. Slopes of (tangents to) x(t), y(t), z(t) graphs would show velocity components Average Velocity Same direction as Rate of change of position

4 Copyright R. Janow – Spring 2012 Instantaneous Velocity x y parabolic path Free Fall Example is tangent to the path in x,y i.e, to a plot of y vs x t x t y y(t) is parabolic Components of are tangent to graphs of corresponding components of the motion, viz: v x is tangent to x(t) v y is tangent to y(t)

5 Copyright R. Janow – Spring 2012 Displacement in 3 Dimensions

6 Copyright R. Janow – Spring 2012 Average Acceleration x y 2D Trajectory: y = f(x) Parameterized by time Velocity vectors are tangent to the trajectory at A & B Move them tail to tail is parallel to Instantaneous Acceleration If the solution is known find: v x (t), v y (t) by differentiating x(t), y(t) a x (t), a y (t) by differentiating v x (t), v y (t) is NOT tangent to the path a x IS tangent to a graph of v x (t) a y IS tangent to a graph of v y (t) a x affects v x and x, but not v y or y Similarly for a y Rate of change of velocity

7 Copyright R. Janow – Spring 2012 Summary – 3D Kinematics Formulas Definitions: Same for z Easiest to use in Cartesian coordinates – The x, y, z, dimensions move independently - Choose a x, a y, a z and initial conditions independently Cartesian Scalar Form - Constant Acceleration Vector Form - Motion with Constant Acceleration - 2D or 3D PARABOLAS

8 Copyright R. Janow – Spring 2012 Kinematic Equations in 2D: Graphical Representation vxvx vyvy

9 Copyright R. Janow – Spring 2012 Projectile Motion: Motion of a particle under constant, downward gravitation only Assume: Free fall along y direction (up/down) with horizontal motion as well a y = constant = - g, a x, a z = 0 Velocity in x-z direction is constant. Trajectory (a parabola) lies in a plane. Can choose it to be x-y. Motion is 2D, not 3D. Usually pick initial location at origin: x 0 = 0, y 0 = 0 at t=0 Initial conditions:  v 0 has x and y components v x0 = v 0 cos(  ), v y0 = v 0 sin(  )  v x is constant, no drag or non-gravitation forces usually  If v x = 0, motion is strictly vertical, range = 0 x y path v 0x v 0y  Equations: No acceleration along x Acceleration = - g along y

10 Copyright R. Janow – Spring 2012 Trajectory of a Projectile: y as a function of x Launch Range R (return to launch altitude) v y at E = -v yi Maximum Height vRvR

11 Copyright R. Janow – Spring 2012 Example: Show Projectile Trajectory is a Parabola (in x) Initial Conditions: x y path v ix v iy ii Eliminate time from Kinematics Equations: Parabolic in x Limiting Cases: As  i  0, tan  i  0, cos  i  1, As  i  90 o, tan  i  infinity, cos  i  0,

12 Copyright R. Janow – Spring 2012 Example: Find the Range of a Projectile (Level Ground) The range R is the horizontal distance traveled as a projectile returns to it’s launch height y i = 0. y x Set x = R and y = 0 in the preceding trajectory formula: Rearrange and put into standard quadratic form: R factors. Roots are: Trigonometric Identity: “Range Formula” Extremes of Range versus launch angle  i

13 Copyright R. Janow – Spring 2012 Example: Time of Flight for a Projectile Find the time of flight t R for range R – the time to return to the original launch height y i = 0. Set y = 0 in the y-displacement formula: The roots are: This is a quadratic equation in t R – factor it: and: Find maximum height y max reached at time t max Set v y,max = 0 at t max Find speed v f at time t R - passing y = 0 on the way down Same speed at the same altitude v xf = v xi = v i cos(  i ) (constant)  i =  f

14 Copyright R. Janow – Spring 2012 Projectile Range Launch with same initial speed, vary angle Maximum range at  = 45 o Complementary angles produce the same range, but different maximum heights and times of flight

15 Copyright R. Janow – Spring 2012 Problem Solving Strategy Text (page 43) defines a general 4 step method: Conceptualize Categorize Analyze Finalize Additional problem solving hints for mechanics: Choose coordinate system(s). Try to make choices that simplify representing the problem. For example, where acceleration a is constant try choosing an axis along the direction of a. Make a sketch showing axes, origin, particles. Choose names for the important quantities that will help you remember what they mean. Be sure to distinguish quantities of the same type, such as v, v x, v y, v i, v f, …. List the known quantities and the given initial conditions, like v xi, v yi, a x, a y,… Show these on your sketch. Choose equations to use for describing the motion in the problem. Time connects the x, y, z dimensions and is sometimes to be eliminated via algebra.

16 Copyright R. Janow – Spring 2012 path for g = 0 xmxm ymym  s What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps? Assume: hunter fires at the instant monkey lets go and aims directly at falling monkey

17 Copyright R. Janow – Spring 2012 What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps? When should hunter fire? When should monkey jump? Coordinates, variables: Bullet: x b (t), y b (t) Monkey: s, y m (t) Initial Conditions: Monkey at (s, h) Bullet at (0,0) x y  h s To hit monkey: y b (t) = y m (t) at same time t 1 when bullet is at s For bullet at s : For monkey: Monkey should wait until hunter fires before jumping Hunter should aim directly at monkey after he jumps Both know Physics, so everybody waits forever Equate positions:

18 Copyright R. Janow – Spring 2012 Example: Find the arrow’s speed An arrow is shot horizontally by a person whose height is not known. It strikes the ground 33 m away horizontally, making an angle of 86 o with the vertical. What was the arrow’s speed as it was released? 86 o R = 33 m v0v0 vfvf R = v 0 t t = R / v 0 t = flight time = time to hit ground Final velocity components: No acceleration along x V y0 = 0, t as above Divide equations to eliminate v f : Rearrange:

19 Copyright R. Janow – Spring 2012 Uniform Circular Motion A particle in Uniform Circular Motion moves in a circular path with a constant speed  The radius vector is rotating. The velocity vector is always tangent to the particle’s path (as usual). It is perpendicular to the (rotating) radius vector. The particle is accelerating, since the direction of the velocity is constantly changing. The centripetal acceleration vector always points toward the center of the circle. The magnitudes of all three vectors are constant, but their directions are changing All three vectors are rotating with the same constant angular velocity and period. The period of revolution T is: The frequency f is the reciprocal of the period f = 1/T

20 Copyright R. Janow – Spring 2012 Centripetal Acceleration Magnitude Formula The position vectors and velocity vectors both change due to changes in their directions: The triangles for  r and  v are both right isosceles, with the same angle  at their apexes. They are similar: tail to tail The definition of the (magnitude of) the average acceleration is: In the limit  t  0: and The magnitude of the centripetal acceleration is It always points to the center of the circular motion

21 Copyright R. Janow – Spring 2012 Example: Find the Earth’s centripetal acceleration in its orbit about the Sun Assume uniform circular motion The radius of the Earth’s orbit is r = 1.5 x m Need to find the speed v T = 1 year = 365X24x3600 s = x 10 7 s Example: Passengers on an amusement park ride move in circles whose radius is 5 m. They complete a full circle every 4 seconds. What acceleration do they feel, in “g”s”? Need to find the speed v dimensionless

22 Copyright R. Janow – Spring 2012 Uniform Circular Motion Formulas via Calculus Calculate time derivative. Constant r means dr/dt = 0 Note: r d  /dt = v = magnitude of the speed = 2  x r / T for constant v Unit vector tangent to circle Unit vector along r v is constant Note: Unit vector pointing inward along r


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