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Chapter 5 – Circular Motion

Chapter 5 – Circular Motion. This one’s going to be quick. Uniform Circular Motion. Uniform Circular Motion = an object following a circular path AT CONSTANT SPEED. Why can we not say “at constant velocity”? Definition: Period = length of time required to travel around the circle once

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Chapter 5 – Circular Motion

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  1. Chapter 5 – Circular Motion This one’s going to be quick

  2. Uniform Circular Motion • Uniform Circular Motion = an object following a circular path AT CONSTANT SPEED. • Why can we not say “at constant velocity”? • Definition: • Period = length of time required to travel around the circle once • Symbol is “T” • What would the units be for T?

  3. UCM continued • What is the formula for the circumference of a circle? • Circumference = 2Πr, where r = radius of circle • What are the units for r? • Circumference is a distance and period (T) is a time, so we can define the speed v that an object has moving in a circle by • V = 2Πr/T

  4. Examples of UCM • Jerry the racecar driver on a circular track. • Child whirling a rock on a string overhead. • A satellite in orbit around the earth (sorta) • Others?

  5. UCM speed example • The wheel of a car has a radius of 0.29m and is rotating at 830 rpm on a tire-balancing machine. Determine the speed of the outer edge of the wheel. • So T = 0.072 sec and circumference = 2Π(0.29m) • V = 2Πr/T, so v = 25 m/s 830 rev 1 min = 13.8 rev/sec, so 1 rev = 0.072sec 1 min 60 sec

  6. Centripetal acceleration • As an object moves in a circle, it changes its direction, even if the speed remains the same. • Recall from the beginning of the year that acceleration = Δv/ Δt and you can have acceleration even if you are only changing direction V at time t1 θ C V at time t2

  7. Centripetal acceleration • Centripetal Acceleration is the acceleration towards the center of a circle and is what keeps the object moving in a circle. • Centripetal Acceleration is a vector, so it has magnitude and direction • Direction = always towards the center of the circle (so it changes constantly) • Magnitude = aC = v2/r

  8. Centripetal Acceleration: Example Problem • A salad spinner (that thing you put lettuce into and spin it around to dry it off) has a radius of 12 cm and is rotating at 2 rev/sec. what is the magnitude of the centripetal acceleration at the outer wall? • V = 2Πr/T; r = 0.12m and T = 0.5 sec (2 rev/sec means 1 rev every ½ sec), so V = 1.5m/s • AC = v2/r = (1.5m/s)2/(0.12m) = 18.9m/s2, which is slightly less than 2g

  9. Centripetal Acceleration: Example Problem 2 • The bobsled track at the 1994 Olympics had two turns with radii 33m and 24m. Find the centripetal acceleration if the speed was 34 m/s and express in multiples of g = 9.8 m/s2. • aC = (34 m/s)2/33m = 35 m/s2 = 3.6 g • aC = (34 m/s)2/24m = 48 m/s2 = 4.9 g R = 33m R = 24m

  10. Centripetal Force • Centripetal Force is what causes centripetal acceleration • If aC = v2/r and F = ma, then what do you think the formula for centripetal force is? • FC = mv2/r • Like aC, FC also always points to the center of the circle and changes direction constantly

  11. Centripetal force example • A model airplane has a mass 0.9 kg and moves at constant speed in a circle that is parallel to the ground. Find the tension T in a guideline (length = 17 m) for speeds of 19 and 38 m/s. • T1 = (0.9kg)(19m/s)2/17m = 19N • T2 = (0.9kg)(38m/s)2/17m = 76N • So, the second speed is twice the first. What is the difference in force?

  12. What provides the centripetal force in these situations? • The model airplane from the previous example? • A car driving around a circular track? • The bobsled example? • An airplane making a banked turn?

  13. A problem for you to solve • Compare the max speeds at which a car can safely negotiate an unbanked turn (radius = 50m) in dry weather (μs = 0.9) and in icy weather (μs = 0.1). • FC = μsFN = μsmg = mv2/r • V = √ μsgr • Dry: v = √(0.9)(9.8m/s2)/(50m) = 21 m/s • Icy: v = √(0.1)(9.8m/s2)/(50m) = 7 m/s

  14. How can we further improve safety on a curve in the road? • Add a bank to the curve. • With proper banking angle, a car could negotiate the curve even if there were no friction FN FNcos(θ) θ FN sin(θ) mg θ

  15. How can we further improve safety on a curve in the road? • Which force points in towards the center of the curve? • FN sin(θ) does, So FN sin(θ)= mv2/r And we can also see that FNcos(θ) = mg FN FNcos(θ) θ FN sin(θ) mg θ

  16. How can we further improve safety on a curve in the road? • So, for seemingly no good reason, we can divide one equation by the other: FN sin(θ) = mv2/r FN sin(θ) = mv2/r FNcos(θ) = mg FNcos(θ) = mg FN FNcos(θ) θ So tan(θ) = v2/rg, giving us the banking angle that allows safe driving with no friction. FN sin(θ) mg θ

  17. Example: Daytona 500 • At Daytona International Speedway, the turns have a max radius of 316m and are steeply banked with θ = 310. if there were no friction, at what speed could Junior drive around the curve? • From before, we see that tan(θ) = v2/rg, so • V = √rgtan(θ) = √(316m)(9.8m/s2)(tan(310)) • V = 43m/s (96mph)

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