Round and Round we Go….

1. Circular Motion Round and Round we Go….

2. Circular Motion Motion that repeat itself is called periodic motion. Projectile motion, since it doesn't repeat, is not periodic. The motion of a swinging ball is an example of circular motion. Let's analyze the circular motion. Consider a planet around the Sun.

3. When there is no external force, an object will travel in a straight line (Newton's First Law of Motion). In order for an object to travel in a circle, there has to be a force that makes it travel in a circle.

4. An object in circular motion always tries to move in a straight line (Law of Inertia). However, there is a force that actstoward the center of the motion. This force is called thecentripetal force. It is the centripetal force that makes an object travel in a circular path. This force could be friction or a gravitational force, but we call it a centripetal force.

5. When the centripetal force is too strong, the ball will accelerate toward the center of the circle. When the centripetal force is too weak, the ball will get out of the orbit. An object will maintain a circular motion only when the centripetal force is well balanced.

6. Into Orbit? If an object is throw horizontally at a very fast speed, it will actually go into orbit! Actually, the high speed just allows the object to fall around the earth instead of directly into it.

7. The velocity at which this occurs is 8km/s. This is equivalent to 18,000 mi/hr!!! When pieces of space junk fall to earth, their high speed causes them to burn up and disintegrate. Space craft fall or “orbit” around the earth at a higher altitude to avoid the friction from the atmosphere.

8. It would be incorrect to say there is no gravity as the space shuttle orbits the earth. In fact, there must be gravity so that the shuttle continuously falls “around” the earth. Otherwise they would fly off into space. The astronauts only feel weightless because the are continually in free fall.

9. Circular Motion Which moves faster on a merry go round, the horses on the inside, or outside? The answer depends if you’re asking for linear or rotational speed.

10. Each horse completes on revolution per amount of time. They would have the same rotational speed. RPM = revolutions per minute

11. However, since the objects on the outside travel a farther distance (larger circumference), they have a larger linear speed. Tangential speed = linear speed.

13. Circumference = 2*pi*Radius

14. Rotational Example: You are on a rotating platform that has a diameter of 5m. It has a rotational speed of 20 RPM. What is your linear speed at the edge?

15. You make 20 revolutions per minute. How big is each revolution? C = 2πr = πd C = (3.14) 5m = 15.7 m x 20 for 20 revolutions in 1 minute =314 m Ave v = d/t v = 314m / 60s v = 5.23 m/s

16. Equations as a Recipe for Problem-Solving The process of solving a circular motion problem is much like any other problem in physics class.The process involves a careful reading of the problem, the identification of the given and unknown in variable form,the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer.

17. Sample Problem #1 A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car. Given Information: m = 900 kg v = 10.0 m/s R = 25.0 m Unknown Information: a = ???? Fnet = ???? To determine the acceleration of the car, use the equationa = (v2)/R. The solution is as follows:

18. Sample Problem #1 a = (v2)/R a = ((10.0 m/s)2)/(25.0 m) a = (100 m2/s2)/(25.0 m) a = 4 m/s2 To determine the net force acting upon the car, use the equation Fnet = m*a. The solution is as follows. Fnet = m*a Fnet = (900 kg)*(4 m/s2) Fnet = 3600 N

19. Sample Problem #2 A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback. Known Information: m = 95.0 kg R = 12.0 m Traveled 1/4-th or .25 of the circumference in 2.1 s

20. Sample Problem #2 Unknown Information: v = ???? a = ???? Fnet = ???? To determine the speed of the halfback, use the equation v = d/t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows: v = d/t v = (0.25 * 2 * pi * R)/t v = (0.25 * 2 * 3.14 * 12.0 m)/(2.1 s) v = 8.97 m/s

21. Sample Problem #2 Requested Information: v = 8.97 m/s a = ???? Fnet = ???? To determine the acceleration of the halfback, use the equation a = (v2)/R. The solution is as follows: a = (v2)/R a = ((8.97 m/s)2)/(12.0 m) a = (80.5 m2/s2)/(12.0 m) a = 6.71 m/s2

22. Sample Problem #2 Requested Information: v = 8.97 m/s a = 6.71 m/s2 Fnet = ???? To determine the net force acting upon the halfback, use the equation Fnet = m*a. The solution is as follows. Fnet = m*a Fnet = (95.0 kg)*(6.71 m/s2) Fnet = 637 N

23. Sample Problem #3 A 900-kg car makes a 180-degree turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car. Known Information: m = 900 kg v = 10.0 m/s R = 25.0 m Requested Information: Ffrict = ??? mu = ????("mu" - coefficient of friction)

24. Sample Problem #3 a = (v2)/R Fnet = m*a force of friction is3600 N Substituting 3600 N forFfrictand 9000 N forFnormyields a coefficient of friction of0.400

25. Sample Problem #4 The coefficient of friction acting upon a 900-kg car is 0.850. The car is making a 180-degree turn around a curve with a radius of 35.0 m. Determine the maximum speed with which the car can make the turn. Fnet = 7650 N maximum speed of16.4 m/s

26. QUESTION:A ball with a mass of 250 g has a centripetal acceleration of 5 m/s2. What is the centripetal force acting on this ball?

27. QUESTION: A ball with a speed of 10 m/s is in a circular motion. If the circle has a radius of 20 m, what would the period of the ball be?

28. QUESTION:The Moon's orbit around earth is nearly circular. The orbit has a radius of about 385,000 km and a period of 27.3 days. Determine the acceleration of the moon toward Earth.

29. Uniform Circular Motion Quiz 1.Knowing what you know about uniform circular motion, and physics, when a NASCAR races around a circular track at 160 mph what keeps it from flying off the sides of the track? Gravitational Force. Centripetal Force. Frictional Force. Electromagnetic Force. Strong Nuclear Force.

30. 2.You attached a rock to a rope, and were swinging it around in a circle very quickly. If you could keep the velocity of the rock constant as you pulled the rope in, what would happen to the force transmitted to you by the rope? Increase linearly. Decrease linearly. Increase as the square of the distance. Decrease as the square of the distance. Remain the Same.

31. 3.While playing on the International Space Station (ISS), so you can neglect the effects of gravity, you attached a rock to a rope, and were swinging it around in a circle very quickly, when the rope suddenly broke. What did the path of the rock look like before it hit the ISS window a few feet away? An arc. A parabola. A hyperbola. A line. It dropped to the ground (and didn't hit the window).