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PHY183-Lecture06

# PHY183-Lecture06 - Physics for Scientists& Engineers 1...

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2011 Lecture 6 Motion in Two and Three Dimensions January 26, 2011 Physics for Scientists&Engineers 1 1 1 January 26, 2011 Physics for Scientists&Engineers 1 2 2 Three Dimensional Coordinate Systems Now we move from one dimensional motion to motion in two and three dimensions We will work in Cartesian coordinates January 26, 2011 Physics for Scientists&Engineers 1 3 3 Right Handed Coordinate System Conventional assignment for right-handed coordinate system (more on 3D coordinate systems later in this semester) January 26, 2011 Physics for Scientists&Engineers 1 4 4 Clicker Quiz In three dimensions, velocity can change without changing its magnitude A. True B. False C. Can’t tell January 26, 2011 Physics for Scientists&Engineers 1 5 5 Unit Vectors Vector representation in terms of unit vectors: ˆ ˆ ˆ A = ax x + a y y + a z z 2D case ˆ x = (1,0,0) ˆ y = (0,1,0) z = (0,0,1) ˆ y Projection of A on the y axis provides its component ay A ˆ ˆ A = ax x + a y y ˆ ax x ˆ x x ˆ ay a y y ˆ y ax January 26, 2011 Physics for Scientists&Engineers 1 6 6 Position, Velocity, Acceleration in 3D We can write the position vector as ˆ ˆˆ r = ( x, y, z ) = xx + yy + zz We can write the velocity vector as ˆ ˆ ˆ v = (v x , v y , v z ) = v x x + v y y + v z z vx = dx dy dz vy = vz = dt dt dt We can write the acceleration vector as ˆ ˆ ˆ a = (a x , a y , az ) = a x x + a y y + az z dv y dvx dvz ax = ay = az = dt dt dt January 26, 2011 Physics for Scientists&Engineers 1 7 7 Velocity and Acceleration in a Plane Any change in velocity produces acceleration, even when the magnitude of the velocity does not change aave = ∆v v2 − v1 = ∆t t2 − t1 v2 > v1 v2 and v1 same direction January 26, 2011 v2 < v1 v2 and v1 same direction Physics for Scientists&Engineers 1 v2 = v1 v2 and v1 different directions 8 8 Ideal Projectile Motion In some special cases of 3D motion, the horizontal projection of the flight path is a straight line • Accelerations in x and y directions are zero • vx and vy are constant In this special case, the motion in three dimensions can be described as motion in two dimensions Ideal projectile motion • Any object released with some initial velocity that then moves solely under the influence of gravity January 26, 2011 Physics for Scientists&Engineers 1 9 9 Bouncing Ball Strobe light photography shows trajectory of ball in flight January 26, 2011 Physics for Scientists&Engineers 1 10 10 Ideal Projectile Motion (2) Ideal projectile motion neglects • • • • Air resistance Wind speed Spin Etc. Realistic situations such as the flight of a golf ball or a tennis ball are not well described by ideal projectile motion and require more sophisticated techniques We will discuss these effects but we will not go into the gory details January 26, 2011 Physics for Scientists&Engineers 1 11 11 Projectile Motion in Two Dimensions Position vector Velocity vector ˆ ˆ r = ( x, y) = xx + yy ˆ ˆ v = (v x , v y ) = v x x + v y y ˆ Acceleration vector a = (0, − g ) = − gy 2 different types of motion along the two directions: • Free fall for the vertical direction (y-axis) • Motion with constant velocity (zero acceleration) in the horizontal direction (x-axis) January 26, 2011 Physics for Scientists&Engineers 1 12 12 Independence of x- and y-motion In ideal projectile motion, the motion in x- and ydirections are independent of each other Not a trivial statement • Needs to be shown experimentally • Otherwise our mathematical description will have to be modified modified Only true for ideal projectile motion, in the case that we can neglect wind resistance • With wind resistance, the drag force is proportional to v2 and thus make the motions in x- and y-directions dependent on each other. January 26, 2011 Physics for Scientists&Engineers 1 13 13 Two Objects In this demo, we will release two cubes from the same height h above the floor Cube 1 will be released at rest Cube 2 will be given a horizontal velocity Cube 1 Cube 2 v h Floor January 26, 2011 Physics for Scientists&Engineers 1 14 14 Clicker Quiz What will happen in the cube drop? • A) Cube 1 will hit the floor first • B) Cube 2 will hit the floor first • C) Cube 1 and Cube 2 will hit the floor at the same time January 26, 2011 Physics for Scientists&Engineers 1 15 15 Clicker Quiz Explanation Both cubes will hit the floor at the same time The horizontal component of the initial velocity of Cube 2 will not influence its vertical motion Both Cube 1 and Cube 2 will experience the same acceleration in the vertical direction The cubes will hit the floor at the same time • The cubes will not land at the same spot • The cubes will not have the same speed at the time they hit the floor January 26, 2011 Physics for Scientists&Engineers 1 16 16 Free Fall a = ay = − g , g = 9.81 m/s2 12 y = y0 + vy 0t + at 2 y = y0 + vy t vy = vy 0 + at 1 vy = ( vy + vy 0 ) 2 2 2 vy = vy 0 + 2a ( y − y0 ) January 26, 2011 Physics for Scientists&Engineers 1 17 17 Clicker Quiz A heavy ball is launched horizontally with a speed of 14.5 m/s from a height of 4.91 m. How long after being launched will the ball hit the floor? • • • • • A) B) C) D) E) 1.00 s 1.41 s 1.72 a 2.95 s 3.14 s 12 y − y0 = −h = − gt 2 t= 2y = g 2 ( 4.91 m ) = 1.00 s 2 9.81 m/s January 26, 2011 Physics for Scientists&Engineers 1 18 18 Equations for Ideal Projectile Motion Acceleration along each axis: ax ( t ) = 0 a y (t ) = − g v x (t ) = v x 0 x(t ) = x0 + vx 0t Horizontal motion: constant velocity Vertical motion: free fall v y (t ) = v y 0 − gt y (t ) = y0 + v y 0t − 1 gt 2 2 Use notation convention: vx 0 ≡ vx (t = 0); January 26, 2011 Physics for Scientists&Engineers 1 v y 0 ≡ v y (t = 0) 19 19 Shoot the Monkey Scenario: • A zookeeper needs to tranquilize a monkey that has escaped. The monkey is hanging from a branch in a tree. The zookeeper aims directly at the monkey and pulls the trigger on his dart gun. The sound of the dart gun startles the monkey, who lets go of the branch when he hears hears the gun go off. Clicker Question: • Will the dart hit the monkey? A. Yes B. No C. Maybe January 26, 2011 Physics for Scientists&Engineers 1 20 20 Shoot the Monkey Illustration January 26, 2011 Physics for Scientists&Engineers 1 21 21 Slow Motion Replay Because everything happened so fast, here is a super slo’ mo’ replay January 26, 2011 Physics for Scientists&Engineers 1 22 22 Shape of a Projectile’s Trajectory We have equations for the x and y coordinates of a projectile as a function of time Let’s get an expression for the trajectory of a projectile in the x-y plane, take x0 = 0 x = vx 0 t t x vx 0 y y0 x vy 0 vx 0 1 x g 2 vx 0 2 1 y = y0 + vy 0 t − gt2 2 gather powers of x y = y0 + vy 0 g x − 2 x2 2vx 0 vx 0 January 26, 2011 Physics for Scientists&Engineers 1 23 23 Parabolic Trajectory The trajectory follows an equation of the form y = c + bx + ax2 Which is the equation for a parabola • Projectile motion takes the form of a parabola y = y0 + vy 0 g x − 2 x2 vx 0 2 vx 0 We can completely specify the motion by stating ( y0 , vx 0 , vy 0 ) or (y0 , v0 ) or (y0 , v0 , θ0 ) January 26, 2011 Physics for Scientists&Engineers 1 24 24 Parabolic Trajectories IRL January 26, 2011 Physics for Scientists&Engineers 1 25 25 Parabolic Trajectories Because the projectile motion is a parabola and a parabola is symmetric we observe • It takes the same amount of time and travels the same distance from its launch point to the top of its trajectory as from the top of its trajectory trajectory back to launch level • The speed of the projectile at a given height on its way up to the top of its trajectory is the same as its speed at that same height going back down January 26, 2011 Physics for Scientists&Engineers 1 26 26 Time Dependence of the Velocity Vector The x-component of the velocity is constant vx = vx 0 The y-component of the velocity is vy = vy 0 − gt The dependence of v on y is 2 2 v = vx + vy 2 2 v = vx 0 + vy 0 − 2 g (y − y0 ) 2 v = v0 − 2 g (y − y0 ) January 26, 2011 Physics for Scientists&Engineers 1 27 27 Clicker Question Consider projectile motion. At the top of the trajectory A. vx = 0, a = 0 B. vy = 0, a = 0 C. vx = 0, a = g D. vy = 0, a = -g E. vx = 0, a = -g January 26, 2011 Physics for Scientists&Engineers 1 28 28 ...
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