Motion in Two Dimensions: Position, Velocity, and Acceleration

1. 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 Physics 111 Lecture 03Motion in Two DimensionsSJ 8th Ed.: Ch. 4.1 – 4.4 4.1 Position, Velocity and Acceleration Vectors 4.2 Two Dimensional Motion with Constant Acceleration 4.3 Projectile Motion 4.4 A particle in Uniform Circular Motion 4.5 Tangential and Radial Acceleration 4.6 Relative Velocity and Relative Acceleration

2. 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 Motion in two and Three Dimensions Extend 1 dimensional kinematics to 2 D and 3 D • Kinematic quantities become 3 dimensional • Vectors needed to manipulate quantities • Motions in the 3 perpendicular directions • can be analyzed independently

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

4. Free Fall Example y parabolic path is tangent to the path in x,y i.e, to a plot of y vs x x Components of are tangent to graphs of corresponding components of the motion, viz: x vx is tangent to x(t) t y t y(t) is parabolic vy is tangent to y(t) Instantaneous Velocity

5. Displacement in 3 Dimensions

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

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

8. vy vx Kinematic Equations in 2D: Graphical Representation

9. path y Equations: No acceleration along x Acceleration = - g along y v0y q x v0x Projectile Motion: Motion of a particle under constant, downward gravitation only Assume: • Free fall along y direction (up/down) with horizontal motion as well • ay = constant = - g, ax, az = 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: x0 = 0, y0 = 0 at t=0 • Initial conditions: • v0 has x and y components vx0 = v0cos(q), vy0 = v0sin(q) • vx is constant, no drag or non-gravitation forces usually • If vx = 0, motion is strictly vertical, range = 0

10. Maximum Height RangeR (return to launch altitude) Launch vy at E= -vyi Trajectory of a Projectile: y as a function of x vR

11. Parabolic in x path y Limiting Cases: As qi 0, tanqi  0, cosqi  1, viy qi As qi 90o, tanqi  infinity, cosqi  0, x vix Example: Show Projectile Trajectory is a Parabola (in x) Initial Conditions: Eliminate time from Kinematics Equations:

12. y x R factors. Roots are: Trigonometric Identity: Extremes of Range versus launch angle qi “Range Formula” 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 yi = 0. Set x = R and y = 0 in the preceding trajectory formula: Rearrange and put into standard quadratic form:

13. Find speed vf at time tR - passing y = 0 on the way down Same speed at the same altitude vxf = vxi = vicos(qi) (constant) qi = qf Find maximum height ymax reached at time tmax Set vy,max = 0 at tmax Example: Time of Flight for a Projectile Find the time of flight tR for range R – the time to return to the original launch height yi = 0. Set y = 0 in the y-displacement formula: This is a quadratic equation in tR – factor it: The roots are: and:

14. Projectile Range Launch with same initial speed, vary angle Maximum range at q= 45o Complementary angles produce the same range, but different maximum heights and times of flight

15. 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, vx, vy, vi, vf, …. • List the known quantities and the given initial conditions, • like vxi, vyi, ax, ay,… 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. path for g = 0 ym q xm 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. h q s For bullet at s: y Equate positions: x Hunter should aim directly at monkey after he jumps Monkey should wait until hunter fires before jumping Both know Physics, so everybody waits forever 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: xb(t), yb(t) Monkey: s, ym(t) Initial Conditions: Monkey at (s, h) Bullet at (0,0) To hit monkey: yb(t) = ym(t) at same time t1when bullet is at s For monkey:

18. R = v0t v0 t = R / v0 vf R = 33 m 86o Final velocity components: No acceleration along x Vy0 = 0, t as above Divide equations to eliminate vf: Rearrange: 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 86o with the vertical. What was the arrow’s speed as it was released? t = flight time = time to hit ground

19. 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 Uniform Circular Motion • A particle inUniform 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.

20. The definition of the (magnitude of) the average acceleration is: and In the limit Dt  0: The magnitude of the centripetal acceleration is It always points to the center of the circular motion Centripetal Acceleration Magnitude Formula tail to tail The position vectors and velocity vectors both change due to changes in their directions: The triangles for Dr and Dv are both right isosceles, with the same angle Dq at their apexes. They are similar:

21. Need to find the speed v Need to find the speed v 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”? dimensionless 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 1011 m T = 1 year = 365X24x3600 s = 3.156 x 107 s

22. Calculate time derivative. Constant r means dr/dt = 0 Note: r dq/dt = v = magnitude of the speed = 2p x r / T for constant v Unit vector tangent to circle v is constant Note: Unit vector pointing inward along r Uniform Circular Motion Formulas via Calculus Unit vector along r