Presentation is loading. Please wait.

Presentation is loading. Please wait.

Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed. : Ch. 4

Similar presentations

Presentation on theme: "Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed. : Ch. 4"— Presentation transcript:

1 Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed. : Ch. 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 4.1 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 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 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 Position and Displacement
always points from the origin to the particle’s location A particle moves along its path as time increases Trajectory: y = f(x) Parameterized by time Displacement Positions Average Velocity Same direction as Rate of change of position Path does not show time dependence. Slopes of (tangents to) x(t), y(t), z(t) graphs would show velocity components

4 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 y y(t) is parabolic Components of are tangent to graphs of corresponding components of the motion, viz: vx is tangent to x(t) vy is tangent to y(t)

5 Displacement in 3 Dimensions

6 ax affects vx and x, but not vy or y
Average Acceleration Rate of change of velocity 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: vx(t), vy(t) by differentiating x(t), y(t) 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) ax affects vx and x, but not vy or y Similarly for ay

7 Summary – 3D Kinematics Formulas
Definitions: Vector Form - Motion with Constant Acceleration - 2D or 3D PARABOLAS Same for z 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

8 Kinematic Equations in 2D: Graphical Representation
vx vy

9 No acceleration along x
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 x y path v0x v0y q Equations: No acceleration along x Acceleration = - g along y

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

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

12 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: y x Rearrange and put into standard quadratic form: R factors. Roots are: Trigonometric Identity: “Range Formula” Extremes of Range versus launch angle qi

13 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: 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

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 What should monkey (a Physics 111 student) do
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 path for g = 0 xm ym q s

17 What should monkey (a Physics 111 student) do
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: x y q h s 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 t1 when bullet is at s For bullet at s: For monkey: Equate positions: Monkey should wait until hunter fires before jumping Hunter should aim directly at monkey after he jumps Both know Physics, so everybody waits forever

18 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? 86o R = 33 m v0 vf R = v0t t = R / v0 t = flight time = time to hit ground Final velocity components: No acceleration along x Vy0 = 0, t as above Divide equations to eliminate vf: Rearrange:

19 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 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: The definition of the (magnitude of) the average acceleration is: In the limit Dt  0: and The magnitude of the centripetal acceleration is It always points to the center of the circular motion

21 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 Need to find the speed v T = 1 year = 365X24x3600 s = x 107 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 Uniform Circular Motion Formulas via Calculus
Unit vector along r 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

Download ppt "Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed. : Ch. 4"

Similar presentations

Ads by Google