1 / 14

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.

javan
Télécharger la présentation

Uniform Circular Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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?

More Related