Presentation is loading. Please wait.

Presentation is loading. Please wait.

CHAPTER 7 NOTES KONICHEK. I. Projectile motion- A Projectile- this is the object being launched A Projectile- this is the object being launched B. trajectory-The.

Similar presentations


Presentation on theme: "CHAPTER 7 NOTES KONICHEK. I. Projectile motion- A Projectile- this is the object being launched A Projectile- this is the object being launched B. trajectory-The."— Presentation transcript:

1 CHAPTER 7 NOTES KONICHEK

2 I. Projectile motion- A Projectile- this is the object being launched A Projectile- this is the object being launched B. trajectory-The path the projectile follows. B. trajectory-The path the projectile follows.

3  II. Independence of motion in 2 dimensions  A. The horizontal motion of a object is independent of the vertical motion.  1. A bullet shot and one dropped will strike the ground at the same time if they are from the same height.  a. The only force acting on the bullet once outside the barrel of the gun is gravity.

4 B. Equations to represent the motion- neglect air resistance B. Equations to represent the motion- neglect air resistance 1. X=VxT- this is horizontal displacement 1. X=VxT- this is horizontal displacement 2. T 2 = 2Y/g -time for the shot to hit the ground 2. T 2 = 2Y/g -time for the shot to hit the ground

5 EXAMPLE PROBLEM(P136)  A stone thrown horizontally at 15m/s at from a cliff 44m high.   SINCE THE GROUND ID ZERO THE OBJECT IS REPRESENTED BY A -44M  A)How long does it take to reach the bottom?  B) how far from the base of the cliff does the stone strike?  C) sketch the trajectory

6 SOLUTION  A A A A . T 2 = 2(height of cliff)/g  T 2 =3(-44m)/9.8m/s 2 = 3s  B.  X=Vt-  15m/(3s)=45m 

7 DON’T WORRY THERES AN EASIER WAY C. Objects launched at an angle 1. The horizontal component is small since no other forces are acting on it(Vcosθ) 2. Vertical components are larger at the beginning a. Object slows as it’s going up, at the top it stops(Vsinθ) b. the vertical component then increases again but in the opposite sign when falling. c. When the object reaches launch height it is traveling at the same velocity as the launch velocity

8 THESE WILL MAKE YOUR LIFE EASIER  Magical projectile motion equations  R x = V 2 sin2θ/g  H= -(vsinθ) 2 /2g  T= -2Vsinθ/g

9 3. Sample problem - a ball in flight has an initial velocity of 4.47m/s at an angle of 66° above the horizontal. Find A) How long it took the ball to land. B) How high the ball flew. C) Find the range

10 III. Periodic motion-The motion of the object repeats itself. A pendulum, a Yo Yo III. Periodic motion-The motion of the object repeats itself. A pendulum, a Yo Yo A. circular motion-The product of 2 forces acting on an object. A. circular motion-The product of 2 forces acting on an object. 1. F 1 - the outward force of inertia- tangent to the motion of the object. 1. F 1 - the outward force of inertia- tangent to the motion of the object. 2. F 2 - the inward force called centripetal force. 2. F 2 - the inward force called centripetal force.

11 F 1 IS THAT IMAGINARY FORCE CALLED CENTRIFUGAL FORCE F1F1 F2F2

12  B. Centripetal acceleration- This is the acceleration towards the center of the circle  1. A c = V 2 /r  2. V= 2πr/T velocity for one time period  3. Substituting into the equation a= (2πr/t) 2 / r  a=4π 2 r/T 2 

13  C. Centripetal forces - the inward force causing uniform circular motion  1 F= m( 4π 2 r/T 2 )

14 IV. Simple Harmonic Motion. _ when an object is moved from an equilibrium position, and is released, moves through an equilibrium position. A. Period- this is the time necessary for one complete cycle. B Pendulum- an object suspended on a string which undergoes simple harmonic motion. 1. Period of pendulum T= 2π√lg, so g= 4πl/T 2


Download ppt "CHAPTER 7 NOTES KONICHEK. I. Projectile motion- A Projectile- this is the object being launched A Projectile- this is the object being launched B. trajectory-The."

Similar presentations


Ads by Google