One dimensional motion and gravity When we drop an object it’s velocity continues to increase. That means there is an acceleration. Near the earths surface the value of this acceleration is g = 9.8m/s2. This is due to the attractive force of gravity. g varies slightly over the earth because the earth is not a perfect sphere and because of the rotation of the earth. On the moon g is much smaller (~1/6) For CONSTANT acceleration, The equations of motion are: 2 v = v0 + at , d = v0t + ½(at ) 2 2 + v = v0 + 2ad , d = ½(v +v0)t (starting at d =0) For this case all quantities are + at is the change in v for time interval t And g is rounded up to 10.0 m/s which is 2% off and might fail CHIP Week 2 Physics 214 1 Spring 2009 One dimensional motion and gravity d, vo, v, a, t For CONSTANT acceleration, For constant acceleration, if you The equations of motion are: are told the values of three of these quantities, you can solve for a v = v + at does not involve d 0 fourth quantity --- and not have to 2 solve for the fifth quantity if you d = v0t + ½(at ) does not involve v don’t need to, by picking the proper 2 2 v = v0 + 2ad does not involve t equation. d = ½(v0 + v)t does not involve a Displacement = avg. velocity * time (starting at d =0) Combine 1st and 4th eqns. Æ 2d eq. Combine eqs 1 & 4 differently, get 3d eqn., ‘cuz (v-v0)/a = t, plug into eq.4 2 2 d = ½(v0+v)(v-v0)/a = (v -v0 )/2a is same as eqn. 3 Week 2 Physics 214 2 Spring 2009 Throwing vertically In the example shown a ball is thrown vertically.The acceleration is minus + 9.8m/s2 and the motion is symmetric. v = v0 + at at t=0 v = v0 At the top v = 0 and t = v0/9.8 At the bottom t = 2v /9.8 and v = - v 0 0 g = -9.8m/s2 time up and down = 2 x time to top 2 d = v0t + 1/2at so d=0 when t = 0 or t = -2v0/a or 2v0/9.8 since a = -9.8 2 WHY? d = v0t+1/2at = t(v0+at/2) So d = 0 if t=0 or if ( ) = 0 Week 2 Physics 214 3 Spring 2009 Throwing vertically In the example shown a ball is thrown vertically.The acceleration is minus + 9.8m/s2 and the motion is symmetric. v = v0 + at at t=0 v = v0 At the top v = 0 and t = v0/9.8 At the bottom t = 2v /9.8 and v = - v 0 0 g = -9.8m/s2 time up and down = 2 x time to top 2 d = v0t + 1/2at so d=0 when t = 0 or t = -2v0/a or 2v0/9.8 since a = -9.8 2 2 v = v0 + 2ad At d = 0 v = +v0 (t = 0) Or v = -v0 ( t = 2v0/9.8) Week 2 Physics 214 4 Spring 2009 QUIZ Ch 3 E6 Ball is dropped What is the change in velocity between 1 and 4 seconds? A. 9.8 m/s B. 19.6 m/s v g C. 29.4 m/s D. 39.2 m/s + After 1 sec v = 9.8 t = 9.8 m/s After 4 sec v = 9.8 x 4 = 39.2 m/s Change in velocity is 29.4 m/s . Week 2 Physics 214 5 Spring 2009 Projectile Motion If we throw an object so that it’s initial velocity is horizontal then ignoring friction it will continue to move with this velocity in the horizontal direction but it will also start to fall so that it’s trajectory will be curved. Week 2 Physics 214 6 Spring 2009 Motion in two dimensions If we take two axes at right angles we can analyze the motion along each axis separately and determine properties of the whole motion. We will only deal with cases where there is a constant velocity along one axis and a constant acceleration along the other axis. This means that for up Axis 2 usually y Axis 1 usually x down axis 1 v1 = constant and d1 = v1t note: a1=0 2 2 2 axis 2 v2 = v02 + at d = v02t + ½ at v2 = v02 + 2ad Week 2 Physics 214 7 Spring 2009 Projectile Motion We will focus on projectiles with a = 9.8m/s2 9.8 v0v θ v1 R = range At the highest point the vertical velocity is zero vv = v0v + at so t = v0v/9.8 since a = -g 2 h = v0vt + ½ at = vertical height above launch At impact t = 2v0v/9.8 and R = v1 x 2v0v/9.8 And the vertical velocity is minus v0v o Maximum range R for 45 degree upward launch, easiest to see with trig. Week 2 Physics 214 8 Spring 2009 Projectile Motion We will focus on projectiles with a = 9.8m/s2 9.8 v2=vv v0v θ v1 R = range At impact t = 2v0v/9.8 and R = v1 x 2v0v/9.8 2 2 v1=v0cos(θ) v0v=v0sin(θ) R=2c.s.v0 /9.8m/s 2sin.cos = sin(2θ) , max when 2θ = 90o o Maximum range R for 45 degree upward launch, easiest to see with trig. Week 2 Physics 214 9 Spring 2009 Trajectories With no friction there are always two angles which give the same range for the same starting velocity 450 + X and 450 –X R varies as sin(2θ) http://www.physics.purdue.edu/academic_programs/courses/phys214/movies.php (anim0002.mov) http://faculty.tcc.fl.edu/scma/carrj/Java/baseball4.html http://www.mhhe.com/physsci/physical/giambattista/proj/projectile.html http://www.physics.purdue.edu/class/applets/phe/projectile.htm Week 2 Physics 214 10 Spring 2009 Throwing a ball horizontally from a building Use + down so v1 g is positive and h is positive 2 g h = v02t + ½ at h v = v02 + at 2 2 v = v02 + 2ah v2 2 R v02 = 0, t = 2h/a R = v1t = v1sqrt(2h/a) NOTE h is measured down Week 2 Physics 214 11 Spring 2009 Summary Chapter 3 Any motion in a plane can analyzed using two axes at right angles. The motion along each axis is independent of the motion along the other axis so the two dimensional motion can be analyzed as two one dimensional motions linked by time. Special case axis 1 v1 = constant and d1 = v1t 2 2 2 axis 2 v2 = v02 + at , h = v02t + 1/2at ,v = v02 + 2ah 2 v0v 9.8m/s v1 v1 v1 v For projectile motion a = 9.8m/s2 0v Week 2 Physics 214 12 Spring 2009 1D-20 Independence of Vertical & Horizontal Motions (Drop-Kick) One ball drops from rest. The other ball is simultaneously projected horizontally Which ball will hit the ground first ? Listen to the SOUND when they hit the ground and when they bounce. THE VERTICAL & HORIZONTAL MOTIONS ARE INDEPENDENT. THE HORIZONTAL VELOCITY DOES NOT AFFECT THE VERTICAL MOTION. THE VERTICAL FALL TIME IS THE SAME AS LONG AS THE BALLS DROP SIMULTANEOUSLY FROM THE SAME HEIGHT. Week 2 Physics 214 13 Spring 2009 1D-21 Independence of Vertical and Horizontal Motions A ball is projected vertically from a cart traveling horizontally The trajectory in the cart frame The trajectory in the room frame THE HORIZONTAL MOTION OF THE BALL IS UNAFFECTED BY ITS VERTICAL MOTION. Week 2 Physics 214 14 Spring 2009 1G-03 Measurement of g Measuring g by dropping an object d = ½ gt2 t = sqrt ( 2d/g ) g = 2d/t2 Let d = 1 m, What difficulties might be t = sqrt(2/9.8) = 0.45 s encountered in measuring h & t of fall ? Hand timing would not be accurate because of the short fall time Week 2 Physics 214 15 Spring 2009 1K-11 Coin and Feather DROPPING A COIN AND A FEATHER ? DO ALL OBJECTS HAVE THE SAME ACCELERATION WHEN DROPPED ? IN AIR, WEIGHT AND SURFACE AREA MAKE OBJECTS FALL AT DIFFERENT SPEEDS BECAUSE OF AIR FRICTION. Week 2 Physics 214 16 Spring 2009 1D-22 Water Jets & Projectile Motion PROJECTILE MOTION OF A WATER JET What angle gives the maximum range? g NEGLECTING FRICTION THE RANGE IS A MAXIMUM AT 450. TWO DIFFERENT ANGLES CAN GIVE THE SAME RANGE (ANGLES SYMMETRIC ABOUT 45°). A LARGER ANGLE MEANS A LONGER TIME OF FLIGHT, BUT LESS HORIZONTAL VELOCITY. A SMALLER ANGLE MEANS A LARGER HORIZONTAL VELOCITY, AND LESS FLIGHT TIME. THE TRAJECTORY IS SYMMETRIC. Week 2 Physics 214 17 Spring 2009 1D-23 Shoot the Monkey The monkey falls out of the tree at the instant the gun is fired v0y/v0x = h/d WHERE SHOULD ONE AIM, ABOVE, BELOW OR AT? y=0 . |--------------d-----------------| Turn off g, bullet hits monkey at height h. Turn on g, both bullet and monkey fall same amount 2 Ignoring friction y = v0yt – 1/2gt , t = d/v0x , v0yt = v0yd/v0x = h So at x = d y = h – ½ gt2 In the same time the monkey falls ½ gt2 So the bullet always hits the monkey no matter what the value of v0, Unless range R is less than d! THE VERTICAL MOTION IS INDEPENDENT OF THE HORIZONTAL MOTION THE EFFECT OF FRICTION IS MINIMIZED BY USING A HEAVY TARGET Week 2 Physics 214 18 Spring 2009 Questions Chapter 3 Q1 A small piece of paper is dropped and flutters to the floor.