Transcript Circular Motion - Garnet Valley School District

Circular Motion
Period and Frequency
• A CD rotates in a player at a constant speed of
240 rpm.
– How frequently does it complete a revolution:
• In minutes?
rev
240
min
• In seconds?
rev
4
sec
– How long does it take to complete one revolution?
T = 0.25sec
Period and Frequency
• Period (T): The time it takes for an object in circular
motion to make one complete revolution (or cycle)
– Unit: s/rev
–
s/cycle
–
seconds
• Frequency (f): The number of revolutions, or cycles, in
a given period of time
– Equation: f
– Unit: 1/s
– s-1
– Hertz (Hz)
= 1/ T
Speed of an Object traveling in
a Circle
• A car traveling in a circle with a radius of 40 m.
– How far does it travel per cycle?
•
Distance = 2p r = 2p 40 = 251.2 m
• The car completes 5 revolutions per minute.
– What is its period? (How long does it take to complete a rev?)
•
1 Min
60 Sec
T =
=
= 12 s
5 cycles
5 cycles
• What is the car’s speed?
–
251.2
distance 2pr
=
Speed =
=
= 20.9 m/s
12
time
T
Each Revolution takes 2 seconds
1
What was the distance traveled by
box 1?
2
5m
What was the distance traveled by
box 2?
What was the speed of box 1?
10 m
What was the speed of box 2?
Centripetal Acceleration
• Velocity is a Vector
– Magnitude (i.e. size)
– Direction
After 7 Sec
After 6 Sec
v
v
v
After 5 Sec
v
After 4 Sec
v
After 3 Sec
v
v
After 2 Sec
– Speed is constant, direction changing with time
–  Velocity is changing with time, thus accelerating
Centripetal Acceleration
• Centripetal Acceleration: The
acceleration of an object traveling in a
circle at a constant speed
– Speed is constant
– Direction is changing
– Acceleration vector points inward (center
seeking)
v
ac =
v
2
r
ac
Centripetal Acceleration
• Why does the acceleration vector point
inward?
Dv
vi
-vi
vvff
v f - vi
Dv
=
a=
Dt
Dt
v f - v i = v f + (-v i )
-v
vii
Dv
-vi vf
Centripetal Force
• ac, centripetal acceleration (m/s2)
– “Center seeking”
– Necessary to keep the object traveling in a circle
• Fc, centripetal force (N)
– Newton’s 2nd Law
FNET = ma
Fc = mac
• Newton’s 2nd Law for Circles
v
Fc = m
r
2
• NOTE: CENTRIFUGAL FORCE IS AN
IMAGINARY FORCE TO EXPLAIN INERTIA!
DOES NOT EXIST!
Centripetal Force
• The centripetal acceleration is caused by
another force such as:
– Friction
– Tension
– Normal Force
– Gravity
Centripetal Force
• Steps to solve centripetal force problems:
– Draw a free body diagram
– Identify the force causing the centripetal force
• The force that is keeping the object going in a circle
– Set the causal force = centripetal force eq.
– Solve for unknown
A 0.50 kg box is attached to string on a frictionless horizontal table. The box
revolves in a circle of radius 2.8 m. If the box completes 1 revolution
every 2.0 seconds, what is the tension in the string?
FN
r
FT
Fg
The centripetal force is caused by… the tension in the string
FT = FC
v2
FT = m
r
We know m & r, but
need v
2pr 2p (2.8)
v=
=
= 8.80
T
2
8.80 2
FT = 0.5
2.8
FT = 13.8N
A 1200 kg car approaches a circular curve with a radius of
45.0 m. If the coefficient of friction between the tires and road
is 1.20, what is the maximum speed at which the car can
negotiate the curve?
FN
Ff
r
Fg
The centripetal force is caused by… friction
Ff = FC
v2
mg =
r
v
Ff = m
r
2
v
mFN = m
r
2
(Car is going into
the page and
turning right)
v
mmg = m
r
v = rmg = 45.0(1.20)(9.8)
m
v = 23
s
2