Presentation is loading. Please wait.

Presentation is loading. Please wait.

Kinematics of 3- or 2-dimensional motion z x y Position vector: Average velocity: Instantaneous velocity: Average acceleration: Instantaneous acceleration:

Similar presentations


Presentation on theme: "Kinematics of 3- or 2-dimensional motion z x y Position vector: Average velocity: Instantaneous velocity: Average acceleration: Instantaneous acceleration:"— Presentation transcript:

1 Kinematics of 3- or 2-dimensional motion z x y Position vector: Average velocity: Instantaneous velocity: Average acceleration: Instantaneous acceleration: a || → magnitude of velocity a ┴ → direction of velocity

2 Equations of 3-D Kinematics for Constant Acceleration Result: 3-D motion with constant acceleration is a superposition of three independent motions along x, y, and z axes.

3 Projectile Motion a x =0 → v x =v 0x =const a y = -g → v y = v oy - gt x = x 0 + v ox t y = y o + v oy t – gt 2 /2 v 0x = v 0 cos α 0 v 0y = v 0 sin α 0 tan α = v y / v x Exam Example 6: Baseball Projectile Data: v 0 =22m/s, α 0 =40 o x0x0 y0y0 v 0x v 0y axax a y xyvxvx vyvy t 00 ? ?0-9.8m/s 2 ????? Find: (a) Maximum height h; (b) Time of flight T; (c) Horizontal range R; (d) Velocity when ball hits the ground Solution: v 0x =22m/s·cos 40 o =+17m/s; v 0y =22m/s·sin40 o =+14m/s (a)v y =0 → h = (v y 2 -v 0y 2 ) / (2a y )= - (14m/s) 2 / (- 2 · 9.8m/s 2 ) = +10 m (b)y = (v 0y +v y )t / 2 → t = 2y / v 0y = 2 · 10m / 14m/s = 1.45 s; T = 2t =2.9 s (c)R = x = v 0x T = 17 m/s · 2.9 s = + 49 m (d)v x = v 0x, v y = - v 0y (examples 3.7-3.8, problems 3.12)

4 Motion in a Circle (a)Uniform circular motion: v = const (b) Non-uniform circular motion: v ≠ const Centripetal acceleration: Magnitude: a c = v 2 / r Direction to center:

5 Exam Example 7: Ferris Wheel (problems 3.29) Data: R=14 m, v 0 =3 m/s, a || =0.5 m/s 2 Find: (a) Centripetal acceleration (b) Total acceleration vector (c) Time of one revolution T Solution: (a) Magnitude: a c =a ┴ = v 2 / r Direction to center: (b) θ (c)

6 Relative Velocity c Flying in a crosswind Correcting for a crosswind

7 Exam Example 8: Relative motion of a projectile and a target (problem 3.56) Data: h=8.75 m, α=60 o, v p0 =15 m/s, v tx =-0.45 m/s 0 y x Find: (a) distance D to the target at the moment of shot, (b) time of flight t, (c) relative velocity at contact. Solution: relative velocity (c) Final relative velocity: (b) Time of flight (a) Initial distance

8 Principles of Special Theory of Relativity ( Einstein 1905 ): 1.Laws of Nature are invariant for all inertial frames of reference. (Mikelson-Morly’s experiment (1887): There is no “ether wind” ! ) 2. Velocity of light c is the same for all inertial frames and sources. Relativistic laws for coordinates transformation and addition of velocities are not Galileo’s ones: Contraction of length: Slowing down of time:Twin paradox Slowing and stopping light in gases (predicted at Texas A&M) yy’ x x’x’ V Proved by Fizeau experiment (1851) of light dragging by water Lorentz transformation


Download ppt "Kinematics of 3- or 2-dimensional motion z x y Position vector: Average velocity: Instantaneous velocity: Average acceleration: Instantaneous acceleration:"

Similar presentations


Ads by Google