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Motion in 2-D

Motion in 2-D. Projectiles Circular Motion Updated 10.12.2014. Projectile Motion - what do you notice?. Two kinds of motion blended together. Vertical - acceleration Horizontal - constant velocity. Vertical d = v i t + ½ gt 2 vf = v i + gt vf = v i 2 +2gd. Horizontal V = d/t.

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Motion in 2-D

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  1. Motion in 2-D • Projectiles • Circular Motion • Updated 10.12.2014

  2. Projectile Motion - what do you notice?

  3. Two kinds of motion blended together • Vertical - acceleration • Horizontal - constant velocity

  4. Vertical d = vit + ½ gt 2 vf= vi + gt vf = vi2 +2gd Horizontal V = d/t The Return of the Big 5 (in part)

  5. How long will it take a ball to fall when thrown at 20. m/s horizontally from a height of 15 m? How far horizontally will it travel? • Draw the situation and label with given information • Decide upon formulae to use • Solve using formulae and showing UNITS!

  6. Time to fall d = vit + ½ gt 2 Solve for t Initial v in the vertical is 0 so t = sr (2d/g) T = sr (2(15m)/9.8m/s2) T = 1.7 s

  7. How far horizontally will it travel? d = vt d = 20m/s (1.7s) d = 34 m

  8. A ball is launched at 4.5 m/s at 66° above the horizontal.What are the maximum height and flight time of the ball? Establish a coordinate system with the initial position of the ball at the origin.

  9. Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight.

  10. Known Unknown dyinit = 0 h max = dy = ? q = 66 o total flight time, t = ? vinit= 4.5 m/s a y = - g

  11. Known Unknown dyinit = 0 h max = dy = ? q = 66 o total flight time, t = ? vinit= 4.5 m/s a y = - g Find Vy using sin: Sin q = opp/hyp = Vy/4.5m/s Vy = 4.5m/s sin 66o Vy = 4.1m/s Vyf = Vi + gt Vyf = 0 (it stops) t = -Vi/-g = 4.1m/s/9.8m/s2 = t = 0.42 s Double this for up and down motion to get so 2 x 0.42 s total flight time = 0.84 s

  12. Find the y-component of vi. v yi = v i (sin q ) = 4.5 m/s (sin 66) = 4.1 m/s dy = ? dy = ½ g t2 = ½ (9.8 m/s2)(0.42s)2 dy = 0.86 m = 8.6 x 10-1 m

  13. Ch 6: 2D Motion Assignment • Projectiles: 1, 2, 3, 4, 51-53 Projectile Simulation http://www3.interscience.wiley.com:8100/legacy/college/halliday/0471320005/simulations6e/index.htm

  14. Circular Motion V An object moving in a circle at a constant speed is accelerated Centripetal acceleration depends upon the object’s speed and the radius of the circle Centripetal force causes centripetal acceleration. Fc ac Ch 6 Circular Motion Problems: 12-20, 49,61-63

  15. Definitions • Centripetal Force - a center seeking force that appears to pull an object toward the center of the circle along which it is moving • Centripetal Acceleration - acceleration toward the center ac = v2/r

  16. Newton’s Law enters in… • Since f = ma • The centripetal force is determined by substituting • ac = v2/r a = f/m • fc = mv2/r

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