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

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Presentation on theme: "Chapter 5 – Circular Motion This one’s going to be quick."— Presentation transcript:

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 = 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

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 = 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

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 C θ V at time t1 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 = a C = v 2 /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 = 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

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/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

10 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

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. 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?

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 (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

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 mg θ θ FNFN F N cos(θ) F N sin(θ)

15 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

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: 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

17 Example: Daytona 500 At Daytona International Speedway, the turns have a max radius of 316m and are steeply banked with θ = 31 0. 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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