1 / 12

Forces, Uniform Circular Motion, and Gravity Review

Forces, Uniform Circular Motion, and Gravity Review. October 7. m 1. 47. 60. m 1. m 2. 0. m 2. a 1. F. a 2 = 2 a 1. F. a 3. F. 0.

roscoe
Télécharger la présentation

Forces, Uniform Circular Motion, and Gravity Review

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. Forces, Uniform Circular Motion, and Gravity Review October 7

  2. m1 47 60 m1 m2 0 m2 a1 F a2 = 2a1 F a3 F 0 A force F acts on mass m1giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2= 2a1. What is the acceleration if the same force pushes on both blocks? • 4/3 a1 • 3/2 a1 • 1/2 a1 • 3/4 a1 • 2/3 a1

  3. 99 Astronauts can not weigh themselves on a conventional scale (no normal force is produced). Discuss with your neighbor ways of measuring the mass of an astronaut while in space. • . • .

  4. 47 60 0 0 When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that? • this slows your initial velocity which is already upward • you don’t go up, you’re too heavy • you’re not really pulling down -- it just seems that way • the rope actually pulls you up • you are pulling the ceiling down

  5. m1 m1 A B m2 m2 47 F 60 F 0 0 If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger? • Case A • Case B • Same in both cases

  6. a 47 60 T3 T2 T1 3m 2m m 0 0 Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? • T1 > T2 > T3 • T1 < T2 < T3 • T1 = T2 = T3 • all tensions are zero • tensions are random

  7. 47 Normal 60 0 Net Force Weight 0 A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? • The component of the gravity force parallel to the plane increased • coeff. of static friction decreased • normal force exerted by the board decreased • both #1 and #3 • all of #1, #2 and #3

  8. 60 4 5 1 3 2 47 0 0 A ping pong ball is shot into a circular tube which is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow? • 1 • 2 • 3 • 4 • 5

  9. 3 5 2 1 4 47 60 0 0 A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? • 1 • 2 • 3 • 4 • 5

  10. top v 47 60 R 0 0 You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is Fc equal to? • Fc =FT - mg • Fc =FT + FN - mg • Fc =FT + mg • Fc =FT • Fc =mg

  11. 2d 2m 5 Earth 4 d 3 2 1 m 47 60 0 0 A planet of mass m is a distance d from Earth. Another planet of mass 2m is a distance 2d from Earth. Which force vector best represents the direction of the total gravitation force on Earth? • 1 • 2 • 3 • 4 • 5

  12. Answers • 5 – 2/3 a since m1+m2=3/2m1 • Apply known force and measure acceleration (this assumes inertial mass = gravitational mass, which is now experimental fact) • 4 • 1 – Case A – Fcontact must push a larger mass for same acceleration • 1 • 4 – 1 and 3 • 2 • 1 – mg, friction, normal • 3 – if velocity is just right so mv^2/R=mg, then tension=0 and momentarily in freefall • 2 – gravity is directly proportional to m and inversely proportional to R^2, so bigger planet farther away in this case has ½ the gravitational force, so does not add an equal component to total force as the smaller, closer planet

More Related