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Uniform Circular Motion

Uniform Circular Motion. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB. U niform = Constant Speed C ircular = The Path is a Circle (or part of a circle) M otion = Speed is not zero. Prepared by Vince Zaccone

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Uniform Circular Motion

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  1. Uniform Circular Motion Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  2. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  3. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  4. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed A rock tied to a string and whirled in a circle Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  5. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed A rock tied to a string and whirled in a circle Clothes in a dryer spinning at constant speed Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  6. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed A rock tied to a string and whirled in a circle Clothes in a dryer spinning at constant speed A passenger on a Ferris wheel Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  7. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed A rock tied to a string and whirled in a circle Clothes in a dryer spinning at constant speed A passenger on a Ferris wheel Earth orbiting the Sun (almost, but not quite true) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  8. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed A rock tied to a string and whirled in a circle Clothes in a dryer spinning at constant speed A passenger on a Ferris wheel Earth orbiting the Sun (almost, but not quite true) What do these motions have in common? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  9. Uniform = Constant Speed Circular = The Path is a Circle (or part of a circle) Motion = Speed is not zero Examples of UCM: A car driving around a circular turn at constant speed A rock tied to a string and whirled in a circle Clothes in a dryer spinning at constant speed A passenger on a Ferris wheel Earth orbiting the Sun (almost, but not quite true) What these motions have in common: Constant speed (not constant velocity) Acceleration toward the center of the circle (constant magnitude) CENTRIPETAL is the word for this Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  10. We have a formula that we will use often for circular motion. For an object moving in a circular path, the centripetal (toward the center) acceleration is given by: You might also see arad, which stands for radial acceleration Here v stands for the linear speed and R is the radius of the circular path. v v arad arad Notice that the radial acceleration is always toward the center of the circle, and the velocity is always tangent to the circle. This is Uniform Circular Motion v arad arad arad v v Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  11. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  12. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Here is a diagram of the centrifuge. The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula. 8.3 cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  13. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Here is a diagram of the centrifuge. The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula. We need to find the speed v 8.3 cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  14. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Here is a diagram of the centrifuge. The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula. We need to find the speed v Convert from rpm to m/s: 8.3 cm The circumference of the circle is the distance traveled during each revolution Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  15. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Here is a diagram of the centrifuge. The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula. We need to find the speed v Convert from rpm to m/s: 8.3 cm The circumference of the circle is the distance traveled during each revolution Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  16. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Here is a diagram of the centrifuge. The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula. We need to find the speed v Convert from rpm to m/s: 8.3 cm The circumference of the circle is the distance traveled during each revolution Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  17. Example Problem 1 A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube. Here is a diagram of the centrifuge. The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula. We need to find the speed v Convert from rpm to m/s: 8.3 cm The circumference of the circle is the distance traveled during each revolution Now we can use our formula to find acceleration: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  18. Example Problem 2 A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  19. Example Problem 2 A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom? First think about the direction of the acceleration: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  20. Example Problem 2 A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom? First think about the direction of the acceleration: At the top, the acceleration is downward (toward the center) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  21. Example Problem 2 A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom? First think about the direction of the acceleration: At the top, the acceleration is downward (toward the center), and at the bottom, the acceleration is upward (again, toward the center) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  22. Example Problem 2 A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom? First think about the direction of the acceleration: At the top, the acceleration is downward (toward the center), and at the bottom, the acceleration is upward (again, toward the center) We can find the magnitude from our formula for centripetal acceleration: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  23. Example Problem 2 A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom? First think about the direction of the acceleration: At the top, the acceleration is downward (toward the center), and at the bottom, the acceleration is upward (again, toward the center) We can find the magnitude from our formula for centripetal acceleration: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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