<p>PHYSICS 20 - STA NOTE BOOKLET</p><p>Unit 3 – Circular and Energy Chapter 5 – Circular Motion Name: ______5.1 – Defining Circular Motion 5.2 – Centripetal Acceleration 5.3 – Horizontal and Vertical Acceleration 5.4 – Satellites and Orbits 5.1 – Defining Circular Motion</p><p>Uniform Circular Motion</p><p>• An object moving at a constant velocity</p><p>Constant velocity means TWO things</p><p>1. Constant speed</p><p>2. Constant direction</p><p>• An object that is undergoing uniform circular motion is moving around a circle at a constant speed.</p><p>• But, any object moving in a circle is constantly changing direction . . . . so it cannot have a constant velocity.</p><p>So, for an object moving in a circle, the direction of the velocity is always tangent to the circle . </p><p>The diagram depicts an object (bob) moving in a clockwise circle. o if the “bob” (or in this case rock) is constantly changing direction, by definition it is ACCERLERATING (acceleration = any change in velocity) o if it wasn’t accelerating, it would be travelling in a straight line</p><p> o the Δv is towards the centre the acceleration is in this direction o called CENTRIPETAL ACCELERATION meaning ‘centre seeking’ o According to Newton’s 2nd Law, what is happening when something is ACCELERATING? o since acceleration is towards the centre, force must also be towards the centre called Centripetal Force Do not confuse CENTRIPETAL with CENTRIFUGAL! A person in a car going around a circle might say a ‘centrifugal force’ was pushing them across the seat and out of the circle. In fact, that’s just INERTIA! The car is trying to keep you in the circle, that’s CENTRIPETAL!</p><p>Centripetal Acceleration</p><p> o circular motion is also called periodic motion (periodically it gets back to where it started and keeps going)</p><p> o Two units of time are important in circular motion</p><p>Period</p><p> o time to go through one revolution</p><p> o “seconds per cycle”</p><p> o measured in seconds (s)</p><p>Frequency</p><p> o how often something repeats itself</p><p> o cycles per second</p><p> o measured in hertz (Hz) Example 1: </p><p>A pendulum swing at a frequency of 0.90 Hz. Calculate the period of the pendulum’s swing.</p><p>Example 2:</p><p>You are thinking of buying a new computer. One specification you are looking for is that the hard drive spins at 7200 rpm.</p><p> a) Explain what 7200 rpm means.</p><p> b) Determine the frequency in Hz.</p><p> c) Determine its period. Circular Motion</p><p> o objects travelling in a circle exhibit constant SPEED or Uniform Circular Motion</p><p> o formula for speed </p><p> o but our “d” is circumference </p><p> o creating…</p><p> o but… the time to complete one trip around the circle is called PERIOD and has the symbol T</p><p> o thus… Ex 1: What is the circular speed required if an object completes one orbit of a circle with a radius of 4.00 m in a time of 5.00 s?</p><p>Ex 2: Determine the length of a student’s arm if she can swing a pail around five times in a circle at 2.72 m/s in 7.5s.</p><p>Variation of formula if given frequency instead of T</p><p> so substitute</p><p>Ex 3: If an object travelling in a circle of 5 m radius has a frequency of 6 Hz, what is the circular speed of the object? 5.2 - Centripetal Acceleration</p><p>Centripetal Acceleration</p><p> o in order for VELOCITY to change there must be some acceleration</p><p> o Ex: What is centripetal acceleration of a student driving at 60 km/h in a traffic circle that is 120m across? Variation of the formula</p><p> o try deriving the formula using the alternative speed formula using frequency Ex: What is the centripetal acceleration of a horse running around a track with a radius of 37 metres, once every 12 s?</p><p>Ex: What is centripetal acceleration of a stone being whirled in a circle at the end of 1.75 m string with a frequency of 2.25 Hz?</p><p>Centripetal Force</p><p> o like all acceleration, F=ma, in this case</p><p> o Using our three different formulas for ac, we make three formulas for Fc Ex: A 100 kg astronaut in training is spinning in a centrifuge. Determine the centripetal force acting on the person if the centrifuge has a 8.80 m radius and spins at 10.0 m/s.</p><p>How many “g’s” is this?</p><p>Ex: Same astronaut, same centrifuge but up the speed to 15.0 m/s.</p><p>How many “g’s” is this? Ex: A 5.0 kg mass rotates in a circle with a 6.0 m radius. If the centripetal force applied is 47.4 N, what is the time for one revolution?</p><p>Ex 11: Determine the centripetal force acting on a 100 kg man in a 8.80m radius centrifuge if he is spinning at 15 rpm. 5.3 Horizontal and Vertical Circular Motion</p><p>Horizontal Circular Motion</p><p> o most common example – cars going around corners</p><p> o very important to civil engineers</p><p>Ex 1: Determine the maximum speed at which a 1500 kg car can round a curve that has a radius of 40.0 m if the static coefficient of friction is 0.60.</p><p>Ex 2: An Edmonton Oiler (m = 100 kg) carves a turn with a radius of 7.17 m while skating and feels his skates begin to slip. What is his speed of the coefficient of static friction between the skates and ice is 0.80?</p><p>Ex 3: Automotive manufacturers test the handling ability of a new car by driving a prototype on a test track in a large circle (r=100 m) at increasing speeds until the car begins to skid. A prototype car (m=1200 kg) is tested and found to skid at 95 km/h. Determine the coefficient of static friction between the car and tires.</p><p>Ex 4: A 600.0 g toy radio controlled car can make a turn at a speed of 3.0 m/s on the kitchen floor where the coefficient of friction is 0.900. What is the radius of the turn? Vertical Circular Motion</p><p>• create unique problems because weight (Fg) becomes a factor</p><p>• Fg (weight) never changes but Fc may vary from point to point</p><p>The weight of the object ( Fg ) and the tension in the string ( FA ) act in the same direction.</p><p>The centripetal force is always the net force that is acting toward the center.</p><p>The weight of the object ( Fg ) and the tension in the string ( FA ) act in the same direction.</p><p>The centripetal force is always the net force that is acting toward the center.</p><p>Ex 5: A 1.8 kg object is swung from the end of a 0.50 m string in a vertical circle. If the time of revolution is 1.2 s, what is the tension in the string at the top of the circle? The weight of the object ( Fg ) and the tension in the string ( FA ) act in opposite directions.</p><p>Again, the centripetal force is the net force that is acting toward the center</p><p>* not on formula sheet!</p><p>Ex 6: Same scenario as example 5, what is the tension in the string at the bottom of the circle?</p><p>Minimum Speed</p><p>• minimum speed occurs when the tension on the object (FT or Fa) is 0, therefore Fc = Fg</p><p>Ex 7: A 700.0 kg roller coaster full of people goes in a vertical loop that has a diameter of 50.0 m. What is the minimum speed the roller coaster must maintain at the top of the vertical loop to stay on the track? Ex 8: A bucket of water with a mass of 1.5 kg is spun in a vertical circle on a rope. The radius of the circle is 0.75 m and the speed of the bucket is 3.00 m/s. What is the tension on the rope in position C, as shown?</p><p>Ex 9: A 1.03 x 103 kg car goes over a hill as shown. If the radius of the curve is 40.0 m, how fast must the car travel so that it exerts no force on the road at the crest? </p>