Transcript Chapter 5 * Circular Motion

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.
830 rev 1 min
1 min
60 sec
= 13.8 rev/sec, so 1 rev = 0.072sec
• So T = 0.072 sec and circumference = 2Π(0.29m)
• V = 2Πr/T, so v = 25 m/s
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
C
θ
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
R = 33m
R = 24m
• 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
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 cos(θ)
θ
FN
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
FN cos(θ)
θ
FN
FN sin(θ)
mg
θ
And we can also see that
FNcos(θ) = mg
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
FNcos(θ) = mg
FN cos(θ)
θ
FN sin(θ) = mv2/r
FNcos(θ) = mg
FN
FN sin(θ)
mg
θ
So tan(θ) = v2/rg,
giving us the banking angle
that allows safe driving with
no friction.
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)