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Circular Motion described, using The Physics Classroom for animations and diagrams.

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- 1. Characteristics of Circular Motion <ul><li>Q&A using animations and questions from The Physics Classroom at </li></ul><ul><li>http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l1c.html </li></ul>
- 2. Acceleration Characteristics for Circular Motion <ul><li>An object moving in uniform circular motion is moving in a circle with a uniform or constant speed. </li></ul>
- 3. Explain why there is acceleration in circular motion if the SPEED is Constant. <ul><li>An accelerating object is an object which is changing its velocity. </li></ul><ul><li>Since velocity is a vector which has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity. </li></ul><ul><li>For this reason, it can be boldly declared that an object moving in a circle at constant speed is indeed accelerating. </li></ul><ul><li>It is accelerating because its velocity is changing its directions. </li></ul>
- 4. In which direction does the velocity change (Acceleration) vector point? <ul><li>The velocity change vector is directed towards the center. </li></ul><ul><li>An object moving in a circle at a constant speed from A to B experiences a velocity change and therefore an acceleration. </li></ul><ul><li>This acceleration is directed towards the center of the circle . </li></ul>
- 5. Briefly explain why, using geometric proof. <ul><li>In this time, the velocity has changed from v i to v f . The process of subtracting v i from v f is shown in the vector diagram; this process yields the change in velocity. </li></ul><ul><li>There is a velocity change for an object moving in a circle with a constant speed. </li></ul><ul><li>Note that this velocity change vector is directed towards the center. </li></ul><ul><li>In the case of an object moving in a circle about point C, the object has moved from point A to point B. </li></ul>
- 6. Give two real world demonstrations of this inward acceleration. <ul><li>If a glass with a lit candle is held at the end of an outstretched arm as you spin in a circle at a constant rate (such that the flame experiences an acceleration), then the candle flame will no longer extend vertically upwards. </li></ul><ul><li>Instead the flame deflects from its upright position, signifying that there is an acceleration when the flame moves in a circular path at constant speed. </li></ul><ul><li>The deflection of the flame will be in the direction of the acceleration. </li></ul><ul><li>When you do this experiment, you find that the flame deflects towards the center of the circle, thus indicating that not only is there an acceleration; but that there is an inward acceleration. </li></ul>
- 7. Give two real world demonstrations of this inward acceleration. <ul><li>a. Use a cork accelerometer (a cork submerged in a sealed flask of water) held in an outstretched arm and move in a circle at a constant rate of turning. </li></ul><ul><li>As the cork-water combination spun in the circle, the cork leaned towards the center of the circle. </li></ul><ul><li>Once more, there is proof that an object moving in circular motion at constant speed experiences an acceleration which directed towards the center of the circle. </li></ul>
- 8. Do the “C heck Your Understanding ” #1-8 on the site. <ul><li>Go to the website to use the interactive questions. </li></ul>
- 9. An object moving in a circle is experiencing acceleration. Even if moving around the perimeter of the circle with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is directed TOWARDS THE CENTER of the circle.
- 10. Newton's second law of motion says that … <ul><li>an object which experiences an acceleration must also be experiencing a net force </li></ul><ul><li>the direction of the net force is in the same direction as the acceleration. </li></ul>
- 11. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration.
- 12. This is sometimes referred to as the CENTRIPETAL FORCE REQUIREMENT.
- 13. The word "centripetal" means CENTER-SEEKING. For objects moving in circular motion, there is a net force acting towards the center which causes the object to seek the center.
- 14. Newton's first law of motion - the law of inertia states that…. <ul><li>"... objects in motion tend to stay in motion with the same speed and the same direction unless acted upon by an unbalanced force." </li></ul>
- 15. Objects will tend to naturally travel in straight lines; an unbalanced force is required to cause it to turn. The presence of THE UNBALANCED FORCE is required for objects to move in circles.
- 16. There is an outward force and an outward acceleration acting on your body when you sit in the passenger seat of a car making a left hand turn. <ul><li>True or False? </li></ul>
- 17. Explain your answer. <ul><li>It is the inertia of your body - the tendency to resist acceleration - which causes it to continue in its forward motion. </li></ul><ul><li>While the car is accelerating inward, you continue in a straight line. </li></ul><ul><li>If you are sitting on the passenger side of the car, then eventually the outside door of the car will hit you as the car turns inward. </li></ul><ul><li>In reality, you are continuing in your straight-line inertial path tangent to the circle while the car is accelerating out from under you. </li></ul>
- 18. <ul><li>The sensation of an outward force and an outward acceleration is a false sensation. </li></ul><ul><li>There is no physical object capable of pushing you outwards. </li></ul><ul><li>You are merely experiencing the tendency of your body to continue in its path tangent to the circular path along which the car is turning. </li></ul>
- 19. What is a centripetal force ? <ul><li>Some physical force pushing or pulling the object towards the center of the circle. </li></ul><ul><li>The word "centripetal" is merely an adjective used to describe the direction of the force. </li></ul>
- 20. Is centripetal force a new type of force? <ul><li>Yes or No ? </li></ul><ul><li>NO! </li></ul><ul><li>We are not introducing a new type of force but rather describing the direction of the net force acting upon the object which moves in the circle. </li></ul><ul><li>Whatever the object, if it moves in a circle, there is some force acting upon it to cause it to deviate from its straight-line path, accelerate inwards and move along a circular path. </li></ul>
- 21. Give three examples of centripetal force. <ul><li>As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion. </li></ul>
- 22. Give three examples of centripetal force. <ul><li>As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion. </li></ul>
- 23. Give three examples of centripetal force. <ul><li>As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion. </li></ul>
- 24. The centripetal force for uniform circular motion alters the direction of the object without altering its speed. <ul><li>True or False? </li></ul>
- 25. Here’s the situation: <ul><li>You are carrying a tennis ball upon a flat, level board. </li></ul><ul><li>Once the tennis ball and the board are in motion, they will continue in motion in the same direction at the same speed unless acted upon by an unbalanced force. </li></ul><ul><li>This is in accord with Newton's first law of motion. </li></ul>
- 26. Here’s the question: <ul><li>If you apply an unbalanced force to the flat board, then the flat board will accelerate. </li></ul><ul><li>If the force is continually directed towards a point at the center of the circle, then the flat board will round the corner in a circular-like path. </li></ul><ul><li>What will happen to the ball as the flat board is moved in a circle? </li></ul>
- 27. The answer: <ul><li>Go to the animation at this site. View the animation on the LEFT side of the page. </li></ul><ul><li>http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/cf.html . </li></ul>
- 28. An altered situation: <ul><li>Now a block is secured to the board in such a manner that the block applies an unbalanced force to the ball that is directed towards the center of the circle. </li></ul><ul><li>The block provides a normal force directed inward. </li></ul><ul><li>What will happen to the ball as the flat board is moved in a circle? </li></ul>
- 29. The Answer: <ul><li>Go to the animation at this site. View the animation on the LEFT side of the page. </li></ul><ul><li>http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/cf.html </li></ul>
- 30. Circular Motion <ul><li>Centripetal: “towards the center”. </li></ul><ul><li>Radial: Along the radius of a circle. </li></ul><ul><li>Centripetal is often used interchangeably with radial. </li></ul>
- 31. Circular Motion vs. Linear Motion <ul><li>Moving around a fixed point remaining the same distance from that point at all times. </li></ul><ul><li>Moving in a straight line away from a fixed point, increasing the distance from that point at all times. </li></ul>
- 32. AXES Circular Motion vs. Linear Motion <ul><li>Radial axis which points to the center of the circle. </li></ul><ul><li>Tangential axis which points in the direction of motion. </li></ul><ul><li>Depending on point of view, a y-axis may be used as well. </li></ul><ul><li>X-axis which usually points horizontally. </li></ul><ul><li>Y-axis which usually points vertically. </li></ul>
- 33. ACCELERATION Circular Motion vs. Linear Motion <ul><li>Radial Acceleration always points to the center of the circle. </li></ul><ul><li>Tangential Acceleration depends on whether velocity is increasing, decreasing or constant. </li></ul><ul><li>X-axis acceleration depends on whether velocity is increasing, decreasing or constant. </li></ul><ul><li>Y-axis acceleration depends on whether velocity is increasing, decreasing or constant. </li></ul><ul><li>Acceleration on one of the axes is always zero. </li></ul>
- 34. FORCES Circular Motion vs. Linear Motion <ul><li>There MUST be at least one force pointing towards the center of the circular path. </li></ul><ul><li>WHY? </li></ul><ul><ul><li>Since there is always centripetal acceleration, there must be a centripetal force. </li></ul></ul><ul><ul><li>WHO said that? </li></ul></ul><ul><ul><ul><li>Newton (his second law) </li></ul></ul></ul><ul><li>There can be forces in any direction. </li></ul><ul><li>There must be a force pointing in the direction of acceleration. </li></ul>
- 35. IMPORTANT NOTE <ul><li>An object WILL NOT move in a circular path unless there is a force acting on the radial axis pointing in towards the center of the circle. </li></ul>
- 36. In which direction will the object move without the centripetal force? <ul><li>Along the tangential axis </li></ul><ul><li>In the direction of motion </li></ul><ul><li>In a straight line. </li></ul>
- 37. MOTION DIAGRAMS Circular Motion vs. Linear Motion <ul><li>On the radial axis, motion diagrams only consist of the acceleration and net force direction; velocity is not needed. </li></ul><ul><li>On the tangential or y-axis, motion diagrams are normal, if needed. </li></ul><ul><li>Draw velocity and acceleration arrows to represent increasing, decreasing or constant speed. </li></ul>
- 38. FREE BODY DIAGRAMS Circular Motion vs. Linear Motion <ul><li>Draw forces the same as always. </li></ul><ul><li>Make sure there is a force on the radial axis pointing towards the center of the circle. </li></ul><ul><li>Axes must include the radial (positive towards the center) and either tangential or y-axis. </li></ul><ul><li>Forces are represented with arrows pointing in the direction of the push or pull. </li></ul><ul><li>X- and y-axes point in the direction of the most forces, usually with x in the direction of motion. </li></ul>
- 39. Equations Circular Motion vs. Linear Motion <ul><li>Newton’s Second Law: </li></ul><ul><ul><ul><li>F = ma </li></ul></ul></ul><ul><li>Acceleration: </li></ul><ul><ul><ul><li>a = v 2 /R </li></ul></ul></ul><ul><li>Newton’s Second Law: </li></ul><ul><ul><ul><li>F = ma </li></ul></ul></ul><ul><li>Acceleration: </li></ul><ul><ul><ul><li>a = (v f – v o )/t </li></ul></ul></ul><ul><li>(You will recognize this equation as v f = v 0 + at) </li></ul>
- 40. WHAT IS A CENTRIFUGAL FORCE? <ul><li>Centrifugal forces are a MYTH! They do not really exist. </li></ul><ul><li>It is the effect of Newton’s First Law acting on an object while the object is being moved in a circular path. </li></ul>
- 41. What are three mathematical quantities which will be of primary interest to us? <ul><li>force, speed, and acceleration. </li></ul>
- 42. Velocity is determined using the equation
- 43. The acceleration of an object moving in a circle can be determined by the equations
- 44. The net force is related to the acceleration of the object (as is always the case) and is given by the equations
- 45. This set of circular motion equations can be used in two ways: <ul><li>a. as a recipe for algebraic problem solving in order to solve for an unknown quantity. </li></ul><ul><li>b. as a "guide to thinking" about how an alteration in one quantity would effect a second quantity </li></ul><ul><li>. </li></ul>
- 46. How can the equations for circular motion be used as a guide to thinking? <ul><li>the equation relating the net force ( F net ) to the speed ( v ) of an object moving in uniform circular motion is F net = mv 2 /R </li></ul><ul><li>This equation shows that the net force required for an object to move in a circle is directly proportional to the square of the speed of the object. For a constant mass and radius, the F net is proportional to the speed 2 . </li></ul>
- 47. <ul><li>The factor by which the net force is altered is the square of the factor by which the speed is altered. Subsequently, if the speed of the object is doubled, the net force required for that object's circular motion is quadrupled. And if the speed of the object is halved (decreased by a factor of 2), the net force required is decreased by a factor of 4. </li></ul>
- 48. <ul><li>The process of solving a circular motion problem is much like any other problem in physics class. </li></ul><ul><li>The process involves: </li></ul><ul><ul><li>a careful reading of the problem </li></ul></ul><ul><ul><li>the identification of the known and required information in variable form </li></ul></ul><ul><ul><li>the selection of the relevant equation(s) </li></ul></ul><ul><ul><li>substitution of known values into the equation </li></ul></ul><ul><ul><li>finally algebraic manipulation of the equation to determine the answer </li></ul></ul>
- 49. 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. <ul><li>To determine the acceleration of the car, use the equation a = (v 2 )/R. The solution is as follows: </li></ul><ul><li>a = (v 2 )/R a = ((10.0 m/s) 2 )/(25.0 m) </li></ul><ul><li>a = (100 m 2 /s 2 )/(25.0 m) </li></ul><ul><li>a = 4 m/s 2 </li></ul><ul><li>To determine the net force acting upon the car, use the equation Fnet = m*a. The solution is as follows. </li></ul><ul><li>F net = m*a F net = (900 kg)*(4 m/s 2 ) </li></ul><ul><li>F net = 3600 N </li></ul>
- 50. 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 of the halfback. <ul><li>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: </li></ul><ul><li>v = d/t v = (0.25 * 2 * pi * R)/t </li></ul><ul><li>v = (0.25 * 2 * 3.14 * 12.0 m)/(2.1 s) </li></ul><ul><li>v = 8.97 m/s </li></ul>
- 51. 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 acceleration of and net force acting upon the halfback. <ul><li>To determine the acceleration of the halfback, use the equation a = (v 2 )/R. The solution is as follows: </li></ul><ul><li>a = (v 2 )/R a = ((8.97 m/s) 2 )/(12.0 m) </li></ul><ul><li>a = (80.5 m 2 /s 2 )/(12.0 m) </li></ul><ul><li>a = 6.71 m/s 2 </li></ul><ul><li>To determine the net force acting upon the halfback, use the equation Fnet = m*a. The solution is as follows. </li></ul><ul><li>F net = m*a F net = (95.0 kg)*(6.71 m/s 2 ) </li></ul><ul><li>F net = 637 N </li></ul>
- 52. Anna fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle? <ul><li>Answer the question and then check at The Physics Classroom! </li></ul>
- 53. A Lincoln Continental and a Yugo are making a turn. The Lincoln is four times more massive than the Yugo. If they make the turn at the same speed, then how do the centripetal forces acting upon the two cars compare. Explain. <ul><li>Answer the question and then check at The Physics Classroom! </li></ul>
- 54. The Gravitron at Six Flags is a ride in which occupants line the perimeter of a cylinder and spin in a circle at a high rate of turning. When the cylinder begins spinning very rapidly, the floor is removed from under the riders' feet. What effect does a doubling in speed have upon the centripetal force? Explain. <ul><li>Answer the question and then check at The Physics Classroom! </li></ul>
- 55. Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters. <ul><li>Answer the question and then check at The Physics Classroom! </li></ul>

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