This ones 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 ID: 528773
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Slide1
Chapter 5 – Circular Motion
This one’s going to be quickSlide2
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?Slide3
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/TSlide4
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?Slide5
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
Πr/T, so v = 25 m/s
830 rev
1 min
1 min
60 sec
= 13.8 rev/sec, so 1 rev = 0.072secSlide6
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 =
Δ
v/
Δ
t and you can have acceleration even if you are only changing direction
C
θ
V at time t1
V at time t2Slide7
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
/rSlide8
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 2gSlide9
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 gaC = (34 m/s)2/24m = 48 m/s
2 = 4.9 g
R = 33m
R = 24mSlide10
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 = mv2/rLike aC, FC also always points to the center of the circle and changes direction constantlySlide11
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 = 76NSo, the second speed is twice the first. What is the difference in force?Slide12
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?Slide13
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 =
μsFN = μsmg = mv2/r
V = √ μsgrDry: v = √
(0.9)(9.8m/s2)/(50m) = 21 m/sIcy: v = √(0.1)(9.8m/s
2)/(50m) = 7 m/sSlide14
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
F
N
F
N
cos
(
θ
)
F
N
sin(
θ
)Slide15
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) = mv2/r
mg
θ
θ
F
N
F
N
cos
(
θ
)
F
N
sin(
θ
)
And we can also see that
F
N
cos
(
θ
) = mgSlide16
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
F
N
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
) = mgSlide17
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
) = v2/rg, soV = √rgtan(θ) = √
(316m)(9.8m/s2)(tan(310))V = 43m/s (96mph)