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Circular Motion

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**Speed/Velocity in a Circle**

Consider an object moving in a circle around a specific origin. The DISTANCE the object covers in ONE REVOLUTION is called the CIRCUMFERENCE. The TIME that it takes to cover this distance is called the PERIOD. Speed is the MAGNITUDE of the velocity. And while the speed may be constant, the VELOCITY is NOT. Since velocity is a vector with BOTH magnitude AND direction, we see that the direction of the velocity is ALWAYS changing. We call this velocity, TANGENTIAL velocity as its direction is draw TANGENT to the circle.

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**Centripetal Acceleration**

Suppose we had a circle with angle, q, between 2 radaii. You may recall: Dv v v vo q vo Centripetal means “center seeking” so that means that the acceleration points towards the CENTER of the circle

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**Drawing the Directions correctly**

So for an object traveling in a counter-clockwise path. The velocity would be drawn TANGENT to the circle and the acceleration would be drawn TOWARDS the CENTER. To find the MAGNITUDES of each we have:

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**Circular Motion and N.S.L**

Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true: Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE. NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D.

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Examples The blade of a windshield wiper moves through an angle of 90 degrees in 0.28 seconds. The tip of the blade moves on the arc of a circle that has a radius of 0.76m. What is the magnitude of the centripetal acceleration of the tip of the blade?

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Examples What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= m), when the penny is placed at the outer edge of the record? Top view FN Ff mg Side view

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**A 1500-kg car is traveling at 24 m/s through a flat 100-meter radius turn.**

How large is the frictional force required to keep the car moving in its circular path? What is the correspondingly minimal coefficient of friction between the road surface and the car's tires? How would this coefficient of friction be changed if a 3000-kg pickup truck were traveling through the same curve?

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**Answers Ff = Fc Ff = m(v2/r) Ff = (1500)(242/100) Ff = 8640 FN**

Ff = μ(mg) 8640 = μ(1500)(9.81) 0.587

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Vertical Circle Unlike horizontal circular motion, in vertical circular motion the speed, as well as the direction of the object, is constantly changing. Gravity is constantly either speeding up the object as it falls, or slowing the object down as it rises. We will begin by looking at two special positions which are usually analyzed in problems: the very top of a vertical circle and the very bottom of the circle.

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**net force to the center = N + mg N + mg = m(v2/r) N = m(v2/r) - mg**

Now, consider an example of a person riding a roller coaster through a circular section of the track, a "loop-the-loop” Let's look at the formulas needed to calculate the normal force, N, exerted on a object traveling on the inside surface of a vertical circle as it passes through the bottom and through the top of the ride While at the bottom net force to the center = N - mg N - mg = m(v2/r) N = m(v2/r) + mg At the top net force to the center = N + mg N + mg = m(v2/r) N = m(v2/r) - mg

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**While driving to work you pass over a "crest" in the road that has a radius of 30 meters.**

How fast would you need to be traveling to experience apparent "weightlessness" while passing over the crest? If we let the normal approach 0 to represent apparent weightlessness, then

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Examples The maximum tension that a 0.50 m string can tolerate is 14 N. A 0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle? T mg

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Examples At the bottom? T mg

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Conical Pendulums Our next example is also an object on the end of string but this time it is a conical pendulum. Notice, that its path also tracks out a horizontal circle in which gravity is always perpendicular to the object's path. T cos θ is balanced by the object's weight, mg. It is T sin θ that is the unbalanced central force that is supplying the centripetal force necessary to keep the block moving in its circular path: T sin θ = Fc = mac.

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Suppose a 75-gram ball is being whirled as a conical pendulum by a child. The ball is attached to a 50-cm string and tracks out a horizontal circle with a radius of 40 cm. What is the measure of angle θ? What is the tension in the rope? How fast is the ball traveling as it swings? What is the period of the stopper?

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**Answers T cos θ = mg T cos (53º) = (0.075)(9.81)**

T = 1.22 N T sin θ = m(v2/r) 1.22(sin 53º) = (0.075)(v2/0.40) v2 = 5.20 v = 2.28 m/sec r = L sin θ 0.40 = 0.50 sin θ θ = 53º v = 2πr/T 2.28 = 2π(0.40)/T T = 1.1s

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Banked Curves If instead, the curve is banked then there is a critical speed at which the coefficient of friction can equal zero and the car still travel through the curve without slipping out of its circular path.

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At this critical speed, there is no need for any friction between the car and the road's surface. If the speed of the car were to exceed vcritical then the car would drift up the incline. If the speed of the car is less than vcritical then the car would slip down the incline.

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Free body diagram of the forces acting on the car would show weight and a normal. Since the car is not sliding down the bank of the incline, but is instead traveling across the incline, components of the normal are examined. N sin θ is the unbalanced central force; that is, N sin θ = Fc = mac. This component of the normal is supplying the centripetal force necessary to keep the car moving through the banked curve Dividing the equations N sin θ = mv2/r N cos θ = mg Solving for v produces the desired result yields the equation. tan θ = v2/rg. vcritical = √ (rg tan θ).

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**At a local NASCAR racetrack, cars travel through a 316-meter radius curve that is banked at 31º**

At what speed would the race cars be traveling if they wanted to pass through this curve frictionlessly? Does your answer to the previous question depend on the mass of the car? If the cars actually travel through this turn in excess of 195 mph, or 87 m/sec, what would supply the additional centripetal force?

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Answers No, in the derivation for our equation for vcritical, the mass of the race car cancelled out. friction

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**Newton’s Law of Gravitation**

What causes YOU to be pulled down? THE EARTH….or more specifically…the EARTH’S MASS. Anything that has MASS has a gravitational pull towards it. What the proportionality above is saying is that for there to be a FORCE DUE TO GRAVITY on something there must be at least 2 masses involved, where one is larger than the other.

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N.L.o.G. As you move AWAY from the earth, your DISTANCE increases and your FORCE DUE TO GRAVITY decreases. This is a special INVERSE relationship called an Inverse-Square. The “r” stands for SEPARATION DISTANCE and is the distance between the CENTERS OF MASS of the 2 objects. We us the symbol “r” as it symbolizes the radius. Gravitation is closely related to circular motion as you will discover later.

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**N.L.o.G – Putting it all together**

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Try this! Let’s set the 2 equations equal to each other since they BOTH represent your weight or force due to gravity SOLVE FOR g!

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Examples Venus rotates slowly about its axis, the period being 243 days. The mass of Venus is 4.87 x 1024 kg. Determine the radius for a synchronous satellite in orbit around Venus. (assume circular orbit) Fg 1.54x109 m

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