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Uniform Circular Motion

Uniform Circular Motion. Constant speed, but change direction  acceleration. If acceleration  net force. If motion in circle at const speed, force towards the center. Centripetal (center seeking) acceleration a c = v 2 / r.

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Uniform Circular Motion

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  1. Uniform Circular Motion • Constant speed, but change direction  acceleration. • If acceleration net force. • If motion in circle at const speed, force towards the center. Centripetal (center seeking) acceleration ac = v2 / r

  2. A puck is traveling at a constant speed around a circle on a table. What would we have to know to find the tension in the string?

  3. Clicker! You are whirling a tennis ball on a string around in circles when the string suddenly snaps. What direction does the tennis ball fly? (the figure below is a top view). The ball will fly along a tangential direction.

  4. You are whirling a 0.5 kg ball on a string around in circles. If the radius of the circle is 1m, and the ball takes 1s to complete a circular path, what is the speedof the ball? Clicker! • 1 m/s • 2 m/s • 4p m/s • 2p m/s • 4p2 m/s 1m v = 2pr/T = (2p * 1m )/1s = 2p m/s

  5. You are whirling a 0.5 kg ball on a string around in circles. If the radius of the circle is 1m, and the ball takes 1s to complete a circular path, what is the centripetal acceleration of the ball? Clicker! • 1 m/s^2 • 2 m/s^2 • 4p m/s^2 • 2p m/s^2 • 4p2 m/s^2 1m a = v2/r = (2p2 m/s)/ 1m = 4p2 m/s^2

  6. You are whirling a 0.5 kg ball on a string around in circles. If the radius of the circle is 1m, and the ball takes 1s to complete a circular path, what is the centripetal force experience by the ball? Clicker! • 1 N • 2 N • 2p2 N • 2p N • 4p2 N 1m Fc = m ac = 0.5 kg * 4p2 m/s^2 = 2p2 N

  7. Clicker! A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the tension in the string? • 0.25 g = 2.45 N (2) 0.75 g = 7.35 N • (3) 0.5 g = 4.9 N (4) None of the above. T T = mg = 0.5 kg . 9.8 m/s^2 = 4.9 N Because the same rope is tied to the puck, the tension is the same along the rope.

  8. Clicker! A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the centripetal force experienced by the puck? • 0.25 g = 2.45 N (2) 0.75 g = 7.35 N • (3) 0.5 g = 4.9 N (4) None of the above. Fc T Because the same rope is tied to the puck, the tension is equal to the centripetal force of the puck.

  9. Clicker! A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the centripetal acceleration experienced by the puck? (Tension = 4.9 N) • 19.6 m/s^2 (2) 9.8 m/s^2 • (3) 30 m/s^2 (4) 2 m/s^2. Fc = m ac . Therefore,ac = Fc/m = 4.9 N/ 0.25 kg = 19.6 m/s^2

  10. Clicker! A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the speed of the puck? • 19.6 m/s (2) 4.4 m/s • (3) 30.0 m/s (4) 3.4 m/s. ac = v2/r v = a r = 19.6 m/s^2 * 1.0 m = 4.4 m/s

  11. Clicker! A motorcycle stunt driver performs a vertical loop the loop as shown. Which point will have the largest centripetal force ? (5)The same centripetal force at all points. W FN FN W At (3), Fc = FN +W At (1), Fc = FN -W

  12. Curved Paths Curved paths can be considered as small parts of circles. ac

  13. Satellites gravity is only force causing centripetal acceleration r to center of circle Independent of mass of satellite Fc =

  14. Example.What will be the speed of a satellite orbiting 100 km above the Earth surface?

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