Unit 6 Part 2 Circular Motion and Force

1. Unit 6 Part 2Circular Motion and Force

2. Circular Motion and Centripetal Acceleration • Let us take another look at our 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.

3. 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 circleat all times. • In this problem that force is the tension in the string.

4. Centripetal Force - Swinging Ball Examples • 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 a FBD • 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?

5. Centripetal Force - Swinging Ball Examples • 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? • TT = 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 or radius r must have at least the speed given by this equation in order to continue to travel in a vertical circle.

6. Circular Motion and Force WS 17 - 3 • 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.

7. 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.

8. Centripetal Force – Ferris Wheel Examples • On this slide we will derive the equations relating the Normal force at 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 a FBD. • 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?

9. Centripetal Force – Ferris Wheel Examples • 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.

10. Circular Motion and Force • 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 vT = 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

11. Circular Motion and Force • 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?

12. Physics GT Tires The Fast Ones 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.

13. 88 88 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.

14. 88 88 88 Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires Physics GT Tires The Fast Ones The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones Physics GT Tires The Fast Ones 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.

15. Negotiating Curves - Unbanked • 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?

16. Circular Motion and Force • 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?

17. Circular Motion and Force • Two boys play on a merry-go-round (r = 1.5 m) as shown. • The white boy (m = 38.0 kg)is a distance of 1.35 m from the center, and the black boy (m = 31.0 kg) is 1.20 m from the center. • The merry-go-round has an angular speed of 2.47 rad/s. • Which boy will slide off the merry-go-round? Why?