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CIRCULAR MOTION

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NEWTON’S 1 ST LAW In order to understand how and why objects travel in circles, we need to look back at Newton’s 1 st Law Objects in motion (and at a constant velocity) will stay in motion (at a constant velocity) and objects at rest will stay at rest unless acted upon by an outside, unbalanced force. Remember this is often called the “Law of Inertia” Objects naturally want to stay at rest or keep moving at a constant velocity IN A STRAIGHT LINE If they are not moving in a straight line, it’s because SOMETHING is forcing them to change.

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BASIC PROBLEM Imagine the motion of a ball on a string being swung in a horizontal circular motion above a person’s head. Is the ball being swung at a constant speed accelerating? YES!! Velocity is changing. Velocity is a vector. There is a magnitude and a direction. The direction is changing so it’s velocity is changing! Acceleration will be caused by A change in speed A change in direction A change in both speed and direction

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Look at the below picture (snapshot of a swinging ball – bird’s eye view) You can see the magnitude of the velocity vector does not change, but direction does.

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WHICH DIRECTION IS ACCELERATION? If you have a toy car that travels only in a straight line, what would you have to do to make it travel in a circle? You would have to push the car INTO a circular path. Imagine doing this with little “nudges.” You would continuously tap the side of the car Since the tapping will always be directed to the center, this is the direction of acceleration. ANSWER: Inwards towards the center.

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HOW CAN THIS ACCELERATION BE CALCULATED? We can’t use a =(vf-vt)/t A c = v 2 R The variable A c stands for centripetal acceleration. Centripetal means “center-seeking”. The variable v is the velocity of the object that is moving in the circular path. The variable R is the radius of the circular path.

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EXAMPLES 1.Find the centripetal acceleration of a cat that travels around a circular turn of radius 50 m if it travels at a speed of 25 m/s. (12.5 m/s 2 inward towards the center) Ac =v 2 /r Ac = 25 2 /50 Ac = 12.5 m/s 2 inward

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EXAMPLES 2. Find the velocity of a ball on a string that is being swung in a horizontal circle above a man’s head. If the circular path has a radius of 1m and the centripetal acceleration is 100 m/s 2, what is the velocity of the ball if the string snaps? (what direction will the ball travel?) Ac = v2/r 100 = v2/1 V2 = 100 V = 10 m/s in a straight line tangent to the circle (because of intertia)

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TWO ADDITIONAL CONCEPTS Period of motion We will use the variable T to represent period. The period is the length of time that it takes to complete one full circular revolution. A large period refers to a slow moving object A small period refers to a fast moving object We will use the equation d = rt. Since speed is constant, this equation is appropriate. The distance is the circumference. Since circumference = C = 2πR How fast will it be able to do this? Well, since the speed is a constant, we will use the variable “v”. Therefore, d = rt or 2πR = vT. Rearranging the equation gives

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EXAMPLE 3. A man swings a ball in on a string in a horizontal circle above his head. If the ball is traveling at a speed of 5 m/s and the circle has a radius of 50 cm, find the time needed for the ball to complete one full revolution. [.628 sec]

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EXAMPLE A man holds his son by the arms and starts to spin him around in a circle of radius 1.5m. If the man spins at a constant velocity, and if he completes 5 full revolutions in 10 seconds, find the period of the boys motion. [2 sec] the speed of the little boy’s feet. [4.71 m/s] the speed of the little boy’s belt, which is only.75 m from the center of the circle. [2.36 m/s]

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2 ND CONCEPT Frequency of motion We could say that a fast moving ball completes a full circle more frequently than a ball that is moving very slow. The frequency depends on how fast the object moves around the circle. We will measure frequency in “revolutions per second” using the equation.

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EXAMPLE 5. On a dare, a softball player pus her forehead on the end of a bat and begins to spin around in a circle. Amazingly, she maintains a constant velocity and competes 10 full turns in 12 seconds. Find: The period of the motion. [1.2 sec (per full turn)] The frequency of the motion. [0.833 turns per second]

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The Equations of UNIFORM Circular Motion PERIOD (SEC): FREQUENCY (HZ): TANGENTIAL VELOCITY (M/S): CENTRIPETAL ACCELERATION (M/S 2 ): (because the velocity is constant.) If you see RADIANS in the Holt Physics book, convert to REVS by using the following conversion:

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HORIZONTAL CIRCLES In all examples, the positive direction will be chosen as “INTO THE CENTER” of the circle.

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A BALL ON A STRING SWINGING IN A HORIZONTAL CIRCLE (ON A FRICTIONLESS TABLE) R T m T supplies the centripetal force that keeps the mass in its circular orbit

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A CAR TRAVELING AROUND A CIRCULAR TRACK (OR CURVE) Friction supplies the centripetal force that keeps the car from slipping off the track (curve) R FfFf m Continued on next page

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FBD for …. A car traveling around a circular track (or curve) mg R FfFf FNFN The car “wants” to go this way (because of its inertia, it wants to continue in the “straight line” path) The frictional force supplies the centripetal force necessary to keep the car in the circular path. acac y x

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NEW FORMULAS FOR FRICTION Recall: F = ma from Newton’s 2 nd Law. The new formula we’ll be using is F c = ma c (with c meaning centripetal) Using our previous formula: A c = v 2 R We will rearrange F = ma to A c = F c /m then substitute it into the above formula. Result: F c = mv 2 /R The next slide provides a summary of formula substitutions

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This is the maximum velocity that a car can travel around the circular path without slipping. You can also use the following formula to find the minimum coefficient of friction for no slip at a certain velocity. mg R FfFf FNFN acac

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A CAR TRAVELING AROUND A CIRCULAR TRACK WITH A BANKED TURN Continued on next page

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VERTICAL CIRCLES Vertical circles still use the same formulas as horizontal circles. period (sec): T = 2πR = 1frequency (Hz): f = 1 V fT velocity (m/s): v = 2πR T centripetal acceleration (m/s 2 ): A c = v 2 = 4 π 2 R R T 2 centripetal force (N): Fc = mv 2 R

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HOWEVER…. Vertical circle problems also include the effects of GRAVITY!!

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INTRO TO VERTICAL CIRCLES All of the problems we looked at so far have dealt with objects moving in horizontal circles, like a car going around a turn. In that case, the Centripetal Force is being supplied by the friction between the tires and the road pushing the car into the circle. In some cases, like a ball on the end of a string, the force was from tension in the rope pulling the ball into the circle. In both cases, SOMETHING has to provide the centripetal force in order to make the object move in a circle. FBD: Since centripetal force must keep the object IN the circle, we only cared about the forces that point INTO the circle notice how weight and normal force do not point in? Thus they were never part of our problems! mg R FfFf FNFN acac

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Imagine a scenario where a man is in the middle of the circle and he’s constantly trying to push the car OUT of its circular path. He’s not very strong, so he can’t make the car leave the circle, but since he’s trying, we have to consider his pushing force in our FBD. Again, the friction is supplying the centripetal force, BUT the pushing force is working against friction. These must be subtracted in order to find centripetal force. F c = F f – F push So…when multiple forces point into or out of the circle, we need to find the net force. The sum of the forces pointing into the circle MINUS the sum of the forces pointing out of the circle provides centripetal force. F c = Ins - OUTs mg FfFf FNFN F push

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Figure B: VERTICAL CIRCLE Look at the FBD above. The ball has 4 key points along its path. At every point along the path, tension points inward. At every point along the path, weight (Fg) points downward. Remember, F c = Ins – OUTs At the top of the path, there are two “INs” At the bottom of the path, there are two “OUTs”

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EXAMPLE 1 : A BALL ON A STRING SWINGING IN VERTICAL CIRCLES What supplies the centripetal force when the ball is at the top of its path? Tension & the balls weight (together ) What supplies the centripetal force when the ball is at the bottom of its path? Tension (with the ball weight acting against it ) What supplies the centripetal force when the ball is at the side (3 O’Clock) of its circular path? Tension alone (b/c the weight is not pointing into the circle) R W v W T T Forces pointing in are positive

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EXAMPLE 1 : A BALL ON A STRING SWINGING IN VERTICAL CIRCLES Assuming a constant speed At top: Fc = T + W mv 2 = T + mg R At bottom: Fc = T – W mv 2 = T – mg R R W v W T T If moving too fast, the string may break due to too much tension If moving too slow, the tension will become zero at the top (the ball will fall

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EXAMPLE #2: A BIKE OR CAR GOING AROUND A VERTICAL – CIRCLE LOOP How is this different from a ball on a string being swung in vertical circles? It’s the same problem, but instead of tension supplying the centripetal force, the normal force (from the track) is the supplier. Since the normal force is what the road “feels” (and thus the force that it pushes back with), the normal force is also the “APPARENT WEIGHT” of the bike/car.

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A BIKE OR CAR GOING AROUND A VERTICAL- CIRCLE LOOP Assuming a constant speed: R W v W FNFN At top: At bottom: If moving too slow, the car/bike will lose contact at the top and fall. Before it separates from the track, F N goes to zero! A certain speed must be maintained to avoid this. 0 Forces pointing in are positive. FNFN

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PRACTICE PROBLEMS 1. A 0.5 kg ball is swung on a 1 m rope in a vertical circle with a constant velocity of 5 m/s. a. Find the position of the ball (top, bottom, left, right) when the tension is the greatest. b. Find the position of the ball when the tension is the least. c. Draw a FBD d. Find maximum tension in the string.

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2. A 0.2 kg ball is being swung on a 0.5 m rope in a vertical circle. The rope can withstand up to 50 N of force. a.Find the maximum speed of the ball before it breaks the rope. b.Find the minimum speed of the ball (critical velocity) so that it doesn’t fall out of the circle at the top. 3. A 35 kg boy is swinging on a rope 7 m long. He passes through the lowest position with a speed of 3 m/s. What is the tension in the rope at that moment?

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A 5 kg ball is swung on a 1 m long rope. The rope may be assumed to be weightless. If the ball travels at a constant velocity and takes 0.5 seconds to complete a full revolution, find the tension in the rope at the top and bottom of the swing.

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“WORKED-OUT” PROBLEMS At top: At bottom:

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2) For the same 5 kg ball swung on the 1 m long rope, what is slowest period that will allow the ball to maintain its circular path?

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“WORKED-OUT” PROBLEMS

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A man on a motorcycle (4800 N total weight) attempts to complete a vertical circular loop of radius 3m. How fast must he be driving in order to complete the loop (Give your answer in both m/s & mph) ? If he just exceeds this amount, what will be his apparent weight at the bottom of the loop?

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“WORKED-OUT” PROBLEMS The rider pulled 2 g’s! (twice his weight)

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