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Uniform Circular Motion (UCM) • The object travels in a circular path with a constant speed. • Its velocity is tangent to the circle and is changing due to the changing direction. • The unbalanced force is always directed toward the center of the circle. • The acceleration is always directed toward the center of the circle. v v F v F F v F F v v F F F v v Without a force toward the center the object will fly off on a line tangent to the circle. Something you should already know: The Circumference (C) of a circle: A C 2 r d By the way the area is given by: Ar 2 New Definition: T = Period = The time for one cycle. (measured in seconds) In the case of circular motion the period is the time for one cycle, which is the time for one complete revolution. time T # of revolutions What are the units for an angle: 1/4 π/2 90º Revolutions Degrees Radians π/3 θθ==1/6 60º = 3.14/3 ≈ 1 rad θ π 180º 1/2 3/4 270º 3π/2 000º 2π Linear Velocity vs. Angular Velocity V = Linear Velocity – The rate at which the object travels a distance. s 2 r v t T Units: m/s ω = Angular Velocity – The rate at which the object travels through an angle. t Units: deg/s, rad/s, Revolutions per minute (rpm) Two objects travel in a circular path. Object A has a radius of 0.8 meters and object B has a radius of 2 meters. They both complete a revolution in 5 seconds. Compare their linear and angular velocities. Object A 2 r 2 (.8) m v 1.0 T 5 s 2 rad 1.26 .8m t A 5 s Object B 2m B 2 r 2 (2) m v 2.5 T 5 s 2 rad 1.26 t vA vB 5 s A B Let’s do some conversions: ? 3.35 80rpm = ______ m/s radius = 40cm m 80rev 1 min 2 (.4) m 3.35 min 1rev 60s s 2.1 ? radians 120º = _____ 240 2 radians radians 120 deg 360 360 deg 2 radians 2.1 radians 3 An object traveling in a circle with a constant speed is still accelerating due to its changing direction. This acceleration is called the centripetal acceleration and is given by the following equation: 2 v ac r Units: m s 2 m m 2 s2 m 2 m s The steps to solving a problem: • Draw a force diagram • Apply Newton’s 2nd Law • Substitute the centripetal acceleration equation for the acceleration. F ma 2 v F m r Example 1 A 300g mass is swung on a string in a horizontal circle. (Neglect gravity) The mass has an angular velocity of 120rpm. Find the tension in the string. F ma 2 v FT m r (.3)(37.7) 2 FT 3 FT 142N 3m FT m=300g s 120(2 r ) v t t 120(2 (3)) m 37.7 60 s Example 2: A car traveling with a constant speed rounds a circular curve of radius 15 meters. The coefficient of friction between the road and the tires is 0.8. What is the maximum velocity the can have and still successfully negotiate the curve? F ma 2 v f m r 2 v F m r2 v mg m r Top view v2 g r v gr v gr 2 v (.8)(10)(15) m v 11 s 15m f v Example 3: A 500kg satellite is in a geosynchronous orbit. (It takes 24 hours to orbit the earth one time.) The satellite has a orbital radius of 42000km. What is the force of gravity on the satellite? F ma2 v Fg m r 30542 Fg 500 7 4.2 10 Fg 111N r = 42000km Fg 2r 2 (42000000) m v 3054 T 24(3600) s