Download Examples of Circular Motion Uniform Circular Motion Angular

Examples of Circular Motion
PHYSICS 220
Lecture 07
Circular Motion
Lecture 7
Purdue University, Physics 220
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Lecture 7
Uniform Circular Motion
• The direction of the
velocity is continually
changing
– The vector is always
tangent to the circle
•Can be made effective I-D if use !
2
Angular Variables
• The motion of objects moving in circular (or nearly
circular) paths, is often described by angles measured in
radians rather than degrees.
• The angle ! in radians, is defined as:
s
!=
r
• Uniform circular
motion assumes
constant speed
• Displacement in x and y -- 2-D
Purdue University, Physics 220
• If s = r the angle is 1 rad
• If s = 2"r (the circumference of the circle) the angle is
2" rad. (In other words, 360° = 2" rad.)
Lecture 7
Purdue University, Physics 220
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Circular Motion
Circular to Linear
• Displacement #s = r #! (! in radians)
• Speed |v| = |#s/#t| = r |#!/#t| = r|$|
• Angular displacement #! = !2-!1
– How far it has rotated
|v| = 2"rf
|v| = 2"r/T
• Angular velocity $av = #!/#t
"$
– How fast it is rotating
! = lim
"t #0 "t
– Units: radians/second
(2" = 1 revolution)
• Direction of v is tangent to circle
• Period = 1/frequency
– T = 1/f = 2" / $
– Time to complete 1 revolution
Lecture 7
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Examples
The speed could be obtained by |v| = 2"r/T
1
= 1.2 ! 10"3 min/rev=0.072 s
830rev/min
v=
2! r 2! (0.29m)
=
= 25m / s
T
0.072s
• A CD spins with an angular frequency 20 radians/second.
What is the linear speed 6 cm from the center of the CD?
v = r $ = 0.06 % 20 = 1.2 m/s
Lecture 7
Purdue University, Physics 220
Purdue University, Physics 220
Demo 1D - 08
• The wheel of a car has a radius of 0.29 m and is being
rotated at 830 revolution per minute (rpm) on a tirebalancing machine. Determine the speed (in m/s) at which
the outer edge of the wheel is moving:
T=
Lecture 7
7
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Centripetal Acceleration
S = r!
|v| = r$
|a| = v$
|a| = r$$ = r$2
|a| = vv/r = v2/r
•Magnitude of the velocity vector is constant, but direction is
constantly changing
•At any instant of time, the direction of the instantaneous
velocity is tangent to the path
•Therefore: nonzero acceleration
Lecture 7
Purdue University, Physics 220
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Uniform Circular Motion
Roller Coaster Example
Circular motion with constant speed
R
a
v
Recall:
v=$R
v2
ar =
= ! 2R
R
centripetal
acceleration
• Instantaneous velocity is tangent to circle
• Instantaneous acceleration is radially inward
• There must be a force to provide the acceleration
Lecture 7
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What is the minimum speed you must have at the
top of a 20 meter diameter roller coaster loop, to
keep the wheels on the track.
y-direction: F = ma
-N – mg = m a
N
Let N = 0, just touching
-mg = m a
-mg = -m v2/R
g = v2 / R
v = sqrt(g*R) = 10 m/s
Lecture 7
mg
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Unbanked Curve
Unbanked Curve
What force accelerates a car around a turn on a level
road at constant speed?
A) it is not accelerating
B) the road on the tires
C) the tires on the road
D) the engine on the tires
What is the maximum velocity a car can go around an
unbanked curve in a circle without slipping?
fs ! µs N
y : N ! mg = 0 N = mg
x : Fc = f s = µ s N = µ s mg =
mv
r
2
v = µ s gr
The maximum
velocity to go
around an unbanked curve
depends only on µs
(for a given r)
Dry road: µs=0.9
Icy road: µs=0.1
Lecture 7
Purdue University, Physics 220
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Lecture 7
Banked Curve
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Banked Curve
A car drives around a curve with radius 410 m at a
speed of 32 m/s. The road is banked at 5.0°. The
mass of the car is 1400 kg.
!
x
(1)
! = 50
r = 410m
v = 32m / s
y-direction
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
!F y = may = 0
N cos" # mg # f sin " = 0
y
N
x-direction
!F x = max = ma
v2
N sin " + f cos" = ma = m
r
Lecture 7
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Lecture 7
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(2)
W
f
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Banked Curve
Banked Curve
2 equations and 2 unknown we can solve for N in
(1) and substitute in (2)
A car drives around a curve with radius 410 m at a
speed of 32 m/s. The road is banked at 5.0°. The
mass of the car is 1400 kg.
f sin ! + mg
v2
N =
N sin ! + f cos ! = m
r
cos !
mv 2
" f sin ! + mg %
$#
'& sin ! + f cos! =
cos!
r
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
Like an airplane
f =0
mv 2
f (sin ! + cos ! ) =
cos! ( mg sin !
r
" v2
%
f = m $ cos! ( g sin ! ' = 2300N
# r
&
2
Lecture 7
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Purdue University, Physics 220
v2
cos! = g sin !
r
v = gr tan ! = 19m / s
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Lecture 7
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