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

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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?

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UCM continued What is the formula for the circumference of a circle? Circumference = 2 Π r, 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 = 2 Π r/T

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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?

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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 = sec and circumference = 2 Π( 0.29m) V = 2 Π r/T, so v = 25 m/s 830 rev 1 min 60 sec = 13.8 rev/sec, so 1 rev = 0.072sec

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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 C θ V at time t1 V at time t2

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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 = a C = v 2 /r

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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 = 2 Π r/T; r = 0.12m and T = 0.5 sec (2 rev/sec means 1 rev every ½ sec), so V = 1.5m/s A C = v 2 /r = (1.5m/s) 2 /(0.12m) = 18.9m/s2, which is slightly less than 2g

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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/s 2. a C = (34 m/s) 2 /33m = 35 m/s 2 = 3.6 g a C = (34 m/s) 2 /24m = 48 m/s 2 = 4.9 g R = 33m R = 24m

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Centripetal Force Centripetal Force is what causes centripetal acceleration If a C = v 2 /r and F = ma, then what do you think the formula for centripetal force is? F C = mv 2 /r Like a C, F C also always points to the center of the circle and changes direction constantly

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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. T 1 = (0.9kg)(19m/s) 2 /17m = 19N T 2 = (0.9kg)(38m/s) 2 /17m = 76N So, the second speed is twice the first. What is the difference in force?

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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?

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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 (μ s = 0.9) and in icy weather (μ s = 0.1). FC = μ s F N = μ s mg = mv 2 /r V = √ μ s gr Dry: v = √(0.9)(9.8m/s 2 )/(50m) = 21 m/s Icy: v = √(0.1)(9.8m/s 2 )/(50m) = 7 m/s

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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 mg θ θ FNFN F N cos(θ) F N sin(θ)

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How can we further improve safety on a curve in the road? Which force points in towards the center of the curve? F N sin(θ) does, So F N sin(θ) = mv 2 /r mg θ θ FNFN F N cos(θ) F N sin(θ) And we can also see that F N cos(θ) = mg

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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: mg θ θ FNFN F N cos(θ) F N sin(θ) F N sin(θ) = mv 2 /r F N cos(θ) = mg So tan(θ) = v 2 /rg, giving us the banking angle that allows safe driving with no friction. F N sin(θ) = mv 2 /r F N cos(θ) = mg

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Example: Daytona 500 At Daytona International Speedway, the turns have a max radius of 316m and are steeply banked with θ = if there were no friction, at what speed could Junior drive around the curve? From before, we see that tan(θ) = v 2 /rg, so V = √rgtan(θ) = √(316m)(9.8m/s2)(tan(31 0 )) V = 43m/s (96mph)

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