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Two Dimensional Dynamics Today’s lecture will cover   Chapter 4 Physics 101:  Lecture 06 Exam I Starting Wed Sept 15, W-F office hours will be in 32 Loomis. M office hours will continue in 236 Loomis
Brief Review Thus Far Newton’s Laws of motion:   F=ma Kinematics:  x = x 0  + v 0 t + ½ a t 2 Dynamics  Today we work in 2 Dimensions!
2-Dimensions X and Y are INDEPENDENT! Break 2-D problem into two  1-D problems y x
Position, Velocity and Acceleration Position, Velocity and Acceleration are Vectors! x and y directions are INDEPENDENT! x direction y direction
Velocity in Two Dimensions  5  m/s 3  m/s A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s.   What is the magnitude of the change in velocity? A) 0 m/s  B) 2 m/s  C) 2.6 m/s  D) 3 m/s  E) 5 m/s x-direction v ix  = 5  m/s v fx  = 3  m/s  cos(25)  v x  = 3cos(25)–5 =-2.28 m/s   y-direction v iy  = 0  m/s v fy  = 3  m/s  sin(25)  v y  = 3sin(25)=+1.27  m/s   y x
Acceleration in Two Dimensions 5  m/s 3  m/s A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s.   What is the average acceleration? x-direction y-direction y x
Kinematics in Two Dimensions x = x 0  + v 0x t  + 1/2 a x t 2 v x  = v 0x  +   a x t v x 2  = v 0x 2  + 2a x    x y = y 0  + v 0y t  + 1/2 a y t 2 v y  = v 0y  +   a y t v y 2  = v 0y 2  + 2a y    y x  and   y   motions are   independent ! They share a common time t Must be able to identify variables in these equations!
Train Act/Demo A flatbed railroad car is moving along a track at  constant velocity.  A passenger at the center of the car throws a ball straight up.  Neglecting air resistance, where will the ball land ? A. Forward of the center of the car B. At the center of the car C. Backward of the center of the car Demo - train x- direction ball and car start with same position and  velocity, a=0, so always have same position correct
ACT  A flatbed railroad car is accelerating down a track due to gravity. The ball is shot perpendicular to the track. Where will it land? A. Forward of the center of the car B. At the center of the car C. Backward of the center of the car x direction Ball mg sin(  ) = ma a = g sin(  ) x direction Cart mg sin(  ) = ma a = g sin(  ) Same acceleration gives same position correct y x
Projectile Motion ACT  One marble is given an initial horizontal velocity, the other simply dropped. Which marble hits the ground first? A) dropped B) pushed  C) They both hit the ground at the same time When ball hits depends on y only! y(t) = y 0  + v yo  + ½ a y  t Same for both balls!
You are a vet trying to shoot a tranquilizer dart into a monkey hanging from a branch in a distant tree. You know that the monkey is very nervous, and will let go of the branch and start to fall as soon as your gun goes off.  In order to hit the monkey with the dart, where should you point the gun before shooting? 1 Right at the monkey 2 Below the monkey 3 Above the monkey Monkey Pre-Flight  Demo - monkey 72% 14% 14% No animals were harmed in the making of this demo “ Poor monkey...I hope there's something soft for the little guy to land on.” correct
Most intriguing answer “ The jungle night is hot, but I have waited too long to miss an opportunity like this one. The rare Bare-eared Squirrel Monkey, indigenous only to Brazil, was in my sights. This monkey could lead to mankind's survival and it was all up to me to capture it. This simple creature, hanging from a branch, was the key, the key to unlock the cure. Being the well-trained vet I am, I quickly realized how skittish these monkeys are...meaning I'd only have one shot...one shot to save the world. I lined up my dart gun and, thanks to my p.h.D in physics, did a quick calculation to judge how quickly the dart would fall and how quickly the monkey will fall. Based on the weight and reaction time of the monkey and the power of my gun, I knew that the dart and monkey will drop at the same rate. I thus aimed my weapon right at the animal and fired! At the sound of the shot the little guy dropped from his branch. I watched as the dart met its target dead on. I had done it. With this creature in hand, I had the tools to save the world. The only question that remained was if I was already to late...”
x  =  x 0 y   = - 1 / 2  g  t 2  See text:  4-3 x   =  v 0  t y   = - 1 / 2  g  t 2  Shooting the Monkey...
Shooting the Monkey... y   =  v 0y   t -  1 / 2  g  t 2 Still works even if you shoot upwards! y   =  y 0  -  1 / 2  g  t 2 Dart hits the   monkey!
Projectile Motion a x  = 0  a y  = -g   x = x 0  + v 0x  t v x  = v 0x y = y 0  + v 0y  t  - ½ gt 2 v y  = v 0y   –   g t v y 2  = v 0y 2  – 2 g   y Choose direction where you “know” information Solve kinematics in that direction.  Use  t  from that direction as  t  in other direction
Throw ball to  Monkey  You throw a ball to a monkey who is on a platform 12 meters above and 5 meters to the right of you. Determine the speed and angle you should throw it such that it “just reaches” the monkey. 5 m 12 m Y-direction v f 2  –v i 2  = 2 a   y v i  = sqrt(2 9.8 12) = 15.3 m/s y = ½ (v f  + v i ) t  t = 1.56 s. X-direction v = d/t = 5 m / 1.56 s = 3.2 m/s
Summary of Concepts X and Y directions are Independent! Position, velocity and acceleration are vectors “ Share” t  F = m a  applies in both x and y direction Projectile Motion a x  = 0 in horizontal direction a y  = g in vertical direction

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Lecture06

  • 1. Two Dimensional Dynamics Today’s lecture will cover Chapter 4 Physics 101: Lecture 06 Exam I Starting Wed Sept 15, W-F office hours will be in 32 Loomis. M office hours will continue in 236 Loomis
  • 2. Brief Review Thus Far Newton’s Laws of motion:  F=ma Kinematics: x = x 0 + v 0 t + ½ a t 2 Dynamics Today we work in 2 Dimensions!
  • 3. 2-Dimensions X and Y are INDEPENDENT! Break 2-D problem into two 1-D problems y x
  • 4. Position, Velocity and Acceleration Position, Velocity and Acceleration are Vectors! x and y directions are INDEPENDENT! x direction y direction
  • 5. Velocity in Two Dimensions 5 m/s 3 m/s A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the magnitude of the change in velocity? A) 0 m/s B) 2 m/s C) 2.6 m/s D) 3 m/s E) 5 m/s x-direction v ix = 5 m/s v fx = 3 m/s cos(25)  v x = 3cos(25)–5 =-2.28 m/s y-direction v iy = 0 m/s v fy = 3 m/s sin(25)  v y = 3sin(25)=+1.27 m/s y x
  • 6. Acceleration in Two Dimensions 5 m/s 3 m/s A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the average acceleration? x-direction y-direction y x
  • 7. Kinematics in Two Dimensions x = x 0 + v 0x t + 1/2 a x t 2 v x = v 0x + a x t v x 2 = v 0x 2 + 2a x  x y = y 0 + v 0y t + 1/2 a y t 2 v y = v 0y + a y t v y 2 = v 0y 2 + 2a y  y x and y motions are independent ! They share a common time t Must be able to identify variables in these equations!
  • 8. Train Act/Demo A flatbed railroad car is moving along a track at constant velocity. A passenger at the center of the car throws a ball straight up. Neglecting air resistance, where will the ball land ? A. Forward of the center of the car B. At the center of the car C. Backward of the center of the car Demo - train x- direction ball and car start with same position and velocity, a=0, so always have same position correct
  • 9. ACT A flatbed railroad car is accelerating down a track due to gravity. The ball is shot perpendicular to the track. Where will it land? A. Forward of the center of the car B. At the center of the car C. Backward of the center of the car x direction Ball mg sin(  ) = ma a = g sin(  ) x direction Cart mg sin(  ) = ma a = g sin(  ) Same acceleration gives same position correct y x
  • 10. Projectile Motion ACT One marble is given an initial horizontal velocity, the other simply dropped. Which marble hits the ground first? A) dropped B) pushed C) They both hit the ground at the same time When ball hits depends on y only! y(t) = y 0 + v yo + ½ a y t Same for both balls!
  • 11. You are a vet trying to shoot a tranquilizer dart into a monkey hanging from a branch in a distant tree. You know that the monkey is very nervous, and will let go of the branch and start to fall as soon as your gun goes off. In order to hit the monkey with the dart, where should you point the gun before shooting? 1 Right at the monkey 2 Below the monkey 3 Above the monkey Monkey Pre-Flight Demo - monkey 72% 14% 14% No animals were harmed in the making of this demo “ Poor monkey...I hope there's something soft for the little guy to land on.” correct
  • 12. Most intriguing answer “ The jungle night is hot, but I have waited too long to miss an opportunity like this one. The rare Bare-eared Squirrel Monkey, indigenous only to Brazil, was in my sights. This monkey could lead to mankind's survival and it was all up to me to capture it. This simple creature, hanging from a branch, was the key, the key to unlock the cure. Being the well-trained vet I am, I quickly realized how skittish these monkeys are...meaning I'd only have one shot...one shot to save the world. I lined up my dart gun and, thanks to my p.h.D in physics, did a quick calculation to judge how quickly the dart would fall and how quickly the monkey will fall. Based on the weight and reaction time of the monkey and the power of my gun, I knew that the dart and monkey will drop at the same rate. I thus aimed my weapon right at the animal and fired! At the sound of the shot the little guy dropped from his branch. I watched as the dart met its target dead on. I had done it. With this creature in hand, I had the tools to save the world. The only question that remained was if I was already to late...”
  • 13. x = x 0 y = - 1 / 2 g t 2 See text: 4-3 x = v 0 t y = - 1 / 2 g t 2 Shooting the Monkey...
  • 14. Shooting the Monkey... y = v 0y t - 1 / 2 g t 2 Still works even if you shoot upwards! y = y 0 - 1 / 2 g t 2 Dart hits the monkey!
  • 15. Projectile Motion a x = 0 a y = -g x = x 0 + v 0x t v x = v 0x y = y 0 + v 0y t - ½ gt 2 v y = v 0y – g t v y 2 = v 0y 2 – 2 g  y Choose direction where you “know” information Solve kinematics in that direction. Use t from that direction as t in other direction
  • 16. Throw ball to Monkey You throw a ball to a monkey who is on a platform 12 meters above and 5 meters to the right of you. Determine the speed and angle you should throw it such that it “just reaches” the monkey. 5 m 12 m Y-direction v f 2 –v i 2 = 2 a  y v i = sqrt(2 9.8 12) = 15.3 m/s y = ½ (v f + v i ) t t = 1.56 s. X-direction v = d/t = 5 m / 1.56 s = 3.2 m/s
  • 17. Summary of Concepts X and Y directions are Independent! Position, velocity and acceleration are vectors “ Share” t F = m a applies in both x and y direction Projectile Motion a x = 0 in horizontal direction a y = g in vertical direction

Editor's Notes

  • #6: ACT A) 0 B) 2 C) 2.6 D)3 E) 5
  • #9: Do x(t) for cart and ball. Show they are the same.
  • #11: Could add y(t) equation here change to marble equation
  • #17: Vf2-vi2 = 2 a y v = d/t Vi = sqrt(2 9.8 12) = 3.2 m/s Vi = 15.3 m/s Vf = vi + a t T = 1.56 s