1. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 1 Unit 6 Part 2 Circular Motion and Force

2. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 2 Center Seeking Acceleration  An object traveling in a circular pattern always has an acceleration acting to the center of its rotation (to the center of the circle traced out by the object).  Which way does the acceleration act on the vertically swinging ball (left) and on the man (right) on the Ferris Wheel when they are at the top of their rotation?  Which way does the acceleration act on the vertically swinging ball (left) and on the man (right) on the Ferris Wheel when they are at the bottom of their rotation?

3. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 3 FBD – Ferris Wheel  Which way is the Normal Force acting on the man (when he is sitting in the car on a seat) at the top of his rotation (N T )?  Which way is the weight acting on the man at the top?  Which way is the acceleration acting on the man at the top of the rotation?  Which way is the Normal Force acting on the man when he is at the bottom of his rotation (N B )?  Which way is the weight acting on the man at the bottom?  Which way is the acceleration acting on the man at the bottom of the rotation?

4. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 4 Apparent Weight  Daniel Boone rides in a roller coaster car without a seat belt.  If the car is moving slowly, then will he be safe when he reaches the top of the hill? Explain.  Will he remain safe as the speed of the car increases (medium)? Explain.  Will he appear to weigh more or less as his speed over the top increases? Explain.  If he is going very, very fast, then what will happen to him when he reaches the top? Why?  If he was sitting on a scale at the moment he was ejected, then how much would the scale say he weighed once he was ejected?  No matter his speed, which way is he accelerating at the top of the hill? SlowFastMedium

5. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 5 Apparent Weight  What forces are acting on Daniel Boone while he is sitting in the car?  Would you expect the upward Normal Force (the force the seat exerts onto him) to increase or decrease as his speed over the top of the hill increases? Explain.  What would the value of the Normal Force equal when the car is moving very, very fast resulting in the rider being ejected from the car? Run

6. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 6 Apparent Weight  What trend did we observe in how much the man appeared to weigh (his apparent weight) as the speed of the car over the top increased?  We observed that his apparent weight is reduced as his speed increases.  What conclusion did we draw in regards to the magnitude of the Normal Force as the speed of the car increased?  The Normal forces is reduced as his speed increases.  What similarities exist in these two trends?  Given this observed trend, what relationship do you suppose exists between the apparent weight and the Normal Force?  They are the same thing.  If the car continues to speed up as it goes over the top of the hill, then what value will the Normal Force (apparent weight) eventually reach?

7. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 7 Apparent Weight  Daniel Boone is now going through a valley on the coaster.  What forces are acting on him at the bottom of the valley?  In which direction is the acceleration acting at the bottom of the valley?  How will his Normal Force (apparent weight) change as his speed increases?  The Normal Force (apparent weight) increases as the speed increases. Slow Fast Medium

8. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 8 FBD - Vertically Swinging Ball  Which way is the Tension in the string acting on the ball when it is at the top of its rotation (T T )?  Which way is the weight acting on the ball at the top?  Which way is the acceleration acting on the ball at the top of the rotation?  Which way is the Tension in the string acting on the ball when it is at the bottom of its rotation (T B )?  Which way is the weight acting on the ball at the bottom?  Which way is the acceleration acting on the ball at the bottom of the rotation?

9. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 9 Tension and Apparent Weight  In the same manner that the Normal Force and the apparent weight are the same thing in the Ferris Wheel and Roller Coaster examples, the Tension and the apparent weight are the same thing in the vertically swinging ball example.  At which point in the rotation of the ball, top or bottom, will the speed of the ball be the greatest?  If you are spinning a ball in a vertical circle, then at which position, top or bottom, would you expect the Tension (apparent weight) to be the greatest? Explain.  Since the speed is higher at the bottom, the Tension will be higher at the bottom.

10. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 10 Circular Motion and Centripetal Acceleration  Let us take a look at a Medieval Flail.  Why did the flail continue to rotate before the chain broke?  What does the chain do to the rotating spiked ball?  The chain pulls inward on the spiked ball.  A pull is an example of a …?  A force.  What two things can a force do to a body?  Accelerate (change speed or change direction) it and/or deform it.  The force on the chain causes an inward acceleration that makes the spiked ball want to constantly change direction.  This inward acceleration is known as radial or centripetal acceleration.  Once the chain breaks, the force and its accompanying acceleration no longer exist.  As a result, the flail no longer “wants” to turn and proceeds in a tangential direction from the point where the chain broke.

11. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 11 Centripetal Force - Swinging Ball Examples  Any body rotating about a fixed point will experience a centripetal (center seeking) acceleration.  This acceleration is always directed inwards towards the center of the circle.  The velocity is always perpendicular to this acceleration.  The acceleration’s magnitude is found using the following equation where v is its tangential velocity and r is the radius of the circle.  If asked to find the centripetal force, then simply multiply the acceleration by the mass.  When viewing “Swinging Ball” problems, note that the body has a force (tension, normal,…) acting on it that is pulling the body towards the center of the circle at all times.  In such instances, the Tension (or Normal Force) is known as the “apparent weight.”  In this problem that force is the tension in the string.

12. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 12 Centripetal Force - Swinging Ball Examples #27  On this slide we will derive the equations relating the tensions in the top and bottom of a vertically swinging ball to the velocities of the ball at the top and bottom of its swing.  Notice that the tension is pulling the ball towards the center of rotation at all times.  We start with Newton’s second law and proceed as below (top in left column bottom in right column).  Think about the results.  Which tension would you expect to be the largest: top or bottom? Why?  Do our equations agree with our speculations? Explain?

13. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 13 Minimum Speed  In order to maintain its circular orbit, the ball must spin on the string fast enough to complete its circular orbit at the top of the swing.  The Tension (apparent weight) at the top would be less than that at the bottom but greater than zero.  When the ball is traveling at the minimum speed, the Tension (apparent weight) in the string will equal zero at the top of the swing.  What would happen to the ball’s orbit if it was moving at a speed less than the minimum speed?  The orbit would decay and no longer be circular. MaintainDecayMinimum

14. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 14 Centripetal Force - Swinging Ball Examples #28  Often you will be asked to calculate the minimum speed that the ball must maintain so that the ball continues moving in a circular fashion in the vertical plane.  What do you think the value of the tension would be at the top of the circle if the ball was traveling at its minimum speed?  T T = 0.  As a result, our equation would become as follows.  Notice that this equation has no final dependence on mass.  This means that any mass (a boulder, a marble, a bowling ball,…) traveling in a circle of radius r must have at least the speed given by this equation in order to continue to travel in a vertical circle.

15. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 15 Vertically Swinging Ball Problem?  Watch the animation below as the ball rolls down the hill.  What “type” of problem is one?  What is the chief defining characteristic of a swinging ball problem?  In this example, the Normal Force behaves in the same way as the Tension in a vertically swinging ball problem.  Therefore, this problem would be classified as a “Swinging Ball” problem.

16. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 16 Circular Motion and Force #29  Synchronized jet fighters perform a perfect vertical circular loop maneuver.  The jets are going 1622 km/h at the bottom of the loop.  What is the minimum diameter of the loop needed in order to prevent the pilot from experiencing no more than 5.5 g (5.5 times the normal value of Earth’s gravity)?  What is the apparent weight of the pilot at the top and bottom of the loop?  Before answering this question, identify the “type” of problem that this example is.  This problem is similar to which of the circular types discussed so far?  In which direction is the normal force acting on the pilot at the top of the loop? Bottom?  This problem is a swinging ball problem because the normal force always acts towards the center of the loop.

17. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 17 Centripetal Force – Ferris Wheel Examples  As the man riding the Ferris wheel is moving in a circle, he experiences a centripetal force (red) at all points as shown.  The seat is also exerting an upward normal force on him (blue), and his net force (green) at any point is the vectorial sum of these forces.  At the top, the centripetal force counters the normal force; therefore, the net force acts up and is small.  At the bottom, the centripetal force acts in the same direction as the normal force; therefore, the net force acts up and is large.  Where would Dr. Physics’ apparent weight be the highest: top or bottom?  When viewing “Ferris Wheel” problems, note that the force (tension, normal,…) acting on the body changes directions (towards or away from the center) as the body rotates.  Notice that the normal force acts away from the center at the top of the Ferris wheel and towards the center at the bottom of the Ferris wheel.

18. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 18 Centripetal Force – Ferris Wheel Examples #30  On this slide we will derive the equations relating the Normal Force in the top and bottom of a vertically spinning Ferris Wheel to the velocities of a body at the top and bottom of its rotation.  We start with Newton’s second law and proceed as below (top in left column bottom in right column).  Think about the results.  Which normal force would you expect to be the largest: top or bottom? Why?  Do our equations agree with our speculations? Explain?

19. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 19 Maximum Safe Speed  Previously were observed that an unsecured rider would eventually be ejected from a roller coaster (Ferris Wheel) as the roller coaster increased speed.  We also observed that the magnitude of the Normal Force (apparent weight) reduced as the speed at the top increased.  If the car continues to speed up as it goes over the top of the hill, then what value will the Normal Force (apparent weight) eventually reach?  The speed, where the Normal Force becomes zero, is the maximum safe speed of the coaster for an unsecured rider.  If the speed of the coaster increases beyond this point, then the rider will be ejected.

20. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 20 Centripetal Force – Ferris Wheel Examples #31  Often you will be asked to calculate the maximum speed that a body may have in order to remain in the Ferris Wheel without being buckled in.  What do you think the value of the normal force would be at the top of the circle if the body was traveling at this maximum speed?  N = 0.  As a result, our equation would become as follows.  Notice that this equation has no final dependence on mass.  This means that any mass (a boulder, a marble, a bowling ball,…) sitting in a Ferris wheel traveling in a circle or radius r must have a speed not greater than that given by this equation in order to keep from being thrown from the Ferris Wheel.  If a man (65.0 kg) and a girl (17.0 kg) were both in a Ferris wheel and not strapped in, then both would be thrown from the ride.

21. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 21 Circular Motion and Force #32  A test dummy (m = 62.5 kg) rides freely in a high-speed roller coaster with a radius of curvature of r = 52.00 m.  What type of a problem is this one?  A Ferris wheel problem.  If the coaster was moving at v T = 18.00 m/s at the top of the coaster, then what is the apparent weight of the test dummy at this point?  What is the maximum safe speed of the roller coaster? RUN

22. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 22 Circular Motion and Force #33  An airplane, whose engine failed, was gliding to the ground.  Once the engine restarted, the pilot “pulled up” in order to keep the plane from crashing into the ground.  The radius of curvature of the pull up is 175.0 m, and the plane’s speed at the lowest point of this curve is 78.0 m/s.  What is the apparent weight of the 58.2 kg pilot at this lowest point in flight?

23. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 23 Friction Types for Rolling and Skidding Tires  What type of friction acts between the road and a tire that is rolling without skidding?  What type of friction acts between the road and a tire that is skidding?  Watch the red dot on the tire below as it rolls and then slides.  When the tire is rolling, the red spot is stationary above the position on the road that it is in contact with; therefore, stronger static friction acts on the tire.  When the tire is skidding, the red spot slides relative to road’s surface; therefore, the weaker kinetic friction acts on the tire.

24. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 24 Skidding and Curves  When an ATV safely negotiates a curve, the centripetal acceleration produced by the inward acting static friction is able to accelerate (change the direction of) the ATV.  The tracks left behind by this ATV would look like the ones to the right.  When an ATV goes around the curve at too high a speed, the tires begin to skid.  The centripetal acceleration produced by the inward acting kinetic friction is not able to accelerate (change the direction of) the ATV quickly enough causing the ATV to go out of control.  The tracks left behind by this ATV would look like the ones to the right.  If you are a race car fan, then you have likely seen similar marks left behind by many cars on the surface of the race track.

25. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 25 Negotiating Curves  Watch the animation to the right and explain what has happened.  What force acted on the ATV that prevented it from slipping off the road during the first lap and in which direction was it acting?  The static friction directed inwards towards the center of curvature.  What forces were acting on the ATV when it slipped off the road?  Excess speed caused the tires to skid and to slide changing the static friction to a much smaller kinetic friction that was not able to make the ATV change direction.

26. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 26 Negotiating Unbanked Curves #36  When a car negotiates an unbanked curve, the only force causing the centripetal acceleration is the static friction acting inwards along the radius of curvature.  This static friction accelerates the car and causes it to turn.  Use Newton’s laws to derive an expression for the maximum speed the ATV can have to safely negotiate a curve of radius r when the coefficient of static friction between the tires and the road is  s.  The equation for static friction is as follows.  Since we are considering the maximum static friction, the static friction equation becomes as follows.  Is the size of the body of any importance in this equation? Why or why not?

27. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 27 Circular Motion and Force #37  A semi tractor approaches a curve at 20.0 m/s.  The coefficient of static friction between the truck’s tires and the road is 0.68.  The radius of curvature of the road is 45.0 m.  Will the truck skid on this road?  What is the radius of curvature if the maximum safe speed around this curve is 35.0 m/s?

28. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 28 Curves and Merry-Go-Rounds  A motorcycle runs on a curved (circular) track next to a boy riding a Merry-Go- Round.  What kind of friction acts between the tires and the track? Boy’s shoes and MGR?  What would happen if the motorcycle went to fast? MGR spins too fast?  As a result, a MGR can be thought of as a curve problem.

29. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 29 Circular Motion and Force #38  Two boys play on a merry-go-round as shown.  The white boy (m = 38.0 kg) is a distance of 1.35 m from the center traveling at a speed v = 3.33 m/s,  The black boy (m = 31.0 kg) is 1.20 m from the center raveling at a speed v = 2.96 m/s.  The coefficient of static frictions  s = 0.68.  Which boy will slide off the merry-go-round? Why?

30. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 30 Cresting Hills  A truck carrying a crate negotiates a hill as shown below.  What would happen to the crate if the truck went too fast?  What type of problem could you classify the truck cresting a hill as being? Why?

31. © 2001-2005 Shannon W. Helzer. All Rights Reserved. 31 Circular Motion and Force #20 and #21  A young child (m = 33.00 kg) rides in the bed of a pickup truck as the truck approaches a hill with a radius of curvature of 25.00 m. If the truck is traveling at a speed of 15.00 m/s, then what will be the apparent weight of the boy at the very top of the hill?  On the return trip, the driver of the vehicle is traveling at 16.00 m/s. Will the child be safe riding in the back of the vehicle? Hint: find the maximum safe speed and properly interpret your findings.  How would this maximum speed vary if it was a full grown adult riding in the back of the truck instead of a child.

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