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Chapter 5 – Circular MotionPowerPoint Presentation

Chapter 5 – Circular Motion

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### 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
- Symbol is “T”
- What would the units be for T?

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

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?

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

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

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

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

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

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

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?

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?

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

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

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

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

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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