Circular Motion
Notes
(Ref p234-239 HRW)
Examples of circular motion
Uniform Circular Motion
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Definition
•
Motion exhibited by a body following a circular path at
constant (or uniform) speed, with characteristics of …
Frequency (f) – “how often”
•
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# of cycles, rotations or revolutions per unit time.
SI units are cycles/sec or hertz (Hz)
•
•
Time taken for 1 complete rotational cycle or
revolution.
SI units are seconds.
Period (T) – “how long”
Frequency-Speed relationship
•
f = 1/T
Example
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Q: A merry-go-round takes 5 seconds to
make 1 complete revolution.
• What is the frequency (f) of rotation?
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A: Given T, find f = 1/T
• T = 5 seconds
• f = 1/T = 1/5
• f = 0.2 rev/sec (or cycles/sec)
Example 2
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Q: A CD rotates at 500 rpm near the
center.
• What is the period (T) of the rotation?
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A: Given f, find T = 1/f
• f = 500 rpm
• T = 1/f = 1/500 = 0.002 minutes
• T = 0.12 sec (= 0.002 x 60)
• T = 120 milliseconds (ms)
Linear (tangential/orbital) velocity
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v = distance traveled around a circle per
unit time (vav = dist/time)
• vav = circumference/period
• vav = 2(radius)/period
• vav = 2r/T
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Direction of velocity vector is tangential
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(perpendicular) to radius
Units are m/s
Example – tether ball
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What is the orbital speed (v) of a tennis ball
that is being swung on the end of a 1.5 m
(r) string and takes 1.25 seconds (T) for a
complete revolution?
Given: r = 1.5m, T = 1.25 s
Solve v = 2(radius)/period
• v = 2 (1.5)/1.25
• v = 7.54 m/s
Centripetal Acceleration (ac)
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ac = linear velocity2
radius of rotation
• ac = v2/r
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Direction of acceleration
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vector is inward along
radius (center seeking).
Units are m/s2
Acceleration = rate of change of velocity, so…
Example
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What is the centripetal acceleration (ac)
of the tennis ball in the previous
example?
Solve ac = v2/r (v = 7.54 m/s, r=1.5)
• ac = 7.542/1.5
• ac = 37.89 m/s2
Per Newton, if there is acceleration there
must be a non-zero…
Centripetal force (Fc)
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Fc = mass x ac (Newton’s 2nd Law)
Fc = mac = mv2/r
Direction of force vector is inward along
radius (center-seeking).
Units are Newtons (N)
Example – Merry Go Round
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A merry-go-round has a diameter of 10m
and a period of 10 seconds. There are 2
circles of horses, at 3 m radius and at
the outer edge. If a rider has a mass of
50 kg, calculate:
• Speed of the horses at the edge and 3 m
• Centripetal acceleration of each horse
• Centripetal force on each horse
Solution – Merry Go Round
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v(edge) = 2π r/T = 2π(5)/10 = 3.14 m/s
v(3 m) = 2π r/T = 2π(3)/10 = 1.88 m/s
ac (edge) = v2/r = 3.142/5 = 1.972 m/s2
ac (3m) = v^2/r = 1.882/3 = 1.178 m/s2
Fc(edge) = mac = 50(1.972) = 98.6N (inward)
Fc (3m) = mac = 50(1.178) = 58.9N (inward)
Circular Motion Summary
Torque ()
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Turning force or “moment” of a force
Product of the perpendicular component of a
force applied at a distance from the axis of
rotation (pivot or fulcrum) of an object.
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•
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 = Force┴ x lever arm
 = F┴ x L
Units are N-m.
Convention direction
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Counterclockwise (ccw) = +ve
Clockwise (cw) = -ve
pivot
Torque Examples
Torque – Change angle and lever
arm
Models
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2 models for analysis
• Single pivot (ex. see-saw)
• Double pivot (ex. plank,
stretcher, bridge)
Single Pivot (Seesaw) Model
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To achieve balance
• total ccw torque = total cw torque
• total means all objects contributing to the
respective torques.
• ccw = cw
d
W1
1
d2
W2
Center of Gravity (CoG)
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The point on an object through which all
the mass is deemed to be acting.
• The object is assumed to be uniform in
structure
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Important concept when considering
double pivot models
Pondering…What happens to the plane if the CoG moves backward?
Double Pivot (Bridge/Plank)
Model
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For balance,
• the sum of the CW torques = the sum of the
CCW torques
mass
FL
FR
Mass of box
Mass of plank
CCW
CW
Torque Practice
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Overheads…
Practice – where should the girl
sit?
Moment of Inertia or Rotational
Symbol
Inertia (I)
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Defn: The tendency of a body rotating about a
fixed axis to resist change in rotational motion.
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•
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Rotational equivalent of mass.
Dependent on the distribution of the mass of the
object (where is the mass located?)
Formula
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I = “shape” constant x mass x linear dimension
(squared)
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Where shape is a numeric multiplier dependent upon the
object shape
I = (k)mr2
•
see text (P103) for models
Units
• kg-m2
Models
Rotational Inertia Models (P103)
Note: cylinder = disk!
Example
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What is the moment of inertia of of a diskshaped wheel that has a mass of 15 kg
and a diameter of 2.8 m?
Solve: Refer to formulas on p 103
• Model = cylinder > I = ½ mr2
• I = ½ (15)(2.8/2)2
• I = 14.7 kg.m2
Angular Momentum (L)
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A measure of the how much a rotating body
resists stopping.
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Symbol
Rotational equivalent of linear momentum
L = mass x linear velocity x radius of rotation
•
L = mvr
Units
•
kg.m/s.m > kg.m2/s
Example of Angular Momentum
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Jupiter orbits the sun with a speed of
2079 m/s and orbit of 71,398,000 m from
the sun. If Jupiter’s mass is 1.9 x 1027 kg,
what is its angular momentum (L)?
Solve: L = mvr
• L = 1.9 x 1027 x 2,079 x 71,398,000
• L = 2.82 x 1038 kg.m2/s
Conservation of Angular
Momentum
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Linear momentum is conserved, so…
Angular momentum is conserved across an
event
Visualize an ice skater doing pirouettes…
• Skater has arms out (larger radius↑) – smaller
•
•
•
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spinning velocity↓
Skater pulls her arms in (smaller radius↓)– greater
spinning velocity↑.
L before = Lafter (Conservation of Ang Momentum)
(mvr)before = (mvr)after
v1r1 = v2r2 (mass (m) cancels)
Skater Example of Cons of
Angular Momentum
Before
v*R
After
V*r
v*r product does not change across an event
Example of Conservation of
Angular Momentum
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A physics student is spinning around in a chair @
1.5 m/s with his arms stretched out 0.6 m from the
center of his body, holding in each hand a 2 kg
mass. If he pulls in his arms to 0.2 m from the
center of his body, how fast does he now spin?
Solve: Conservation of Angular momentum
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Angular momentum (before) = angular momentum (after)
v1 x r1 = v2 x r2
1.5 x 0.6 = v2 x 0.2
v2 = 0.9/0.2
v2 = 4.5 m/s
Linear vs Rotational Model
Parameter
Linear
Rotational
Inertia
Mass
Moment of Inertia
Forces
Force
Torque
Momentum
Momentum
Angular momentum