# Chapter 5 – Circular Motion - PowerPoint PPT Presentation

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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

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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)