Transcript Uniform Circular Motion

Uniform Circular Motion
Circular Motion ACT 1
B
A
C
v
Answer: B
A ball is going around in a circle attached to a string. If the
string breaks at the instant shown, which path will the ball
follow?
Circular Motion
Position: r  constant
Speed:
v  constant
Uniform Circular Motion
T period  the time required for one complete rotation.
circumference 2 r
v

period
T
For a particle in uniform
circular motion, the velocity
vector v remains constant in
magnitude, but it continuously
changes its direction.
Acceleration in Uniform
Circular Motion
v
v2
R
R
v1
v2
v1
v2
a
R
centripetal acceleration
aave= v / t
Acceleration toward center
Acceleration is due to change in direction, not
speed. Object turns “toward” center
 must be a force on object toward center.
Centripetal Acceleration
v2
a 
(centripetal acceleration)
r
2 r
v
(tangential velocity)
T
Centripetal Acceleration ACT
Which motion has the largest centripetal acceleration?
Preflight
Consider the following situation: You are driving a car with
constant speed around a horizontal circular track. On a piece
of paper, draw a Free Body Diagram (FBD) for the car. How
many forces are acting on the car?
FN
A) 1
B) 2
C) 3
D) 4
0%
28%
44%
28%
f
correct
R
W
SF = ma = mv2/R
“Gravity, Normal Force, Friction”
“Fn = Normal Force, W = Weight, the force of
gravity, f = centripetal force.”
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Common Incorrect Responses
• Acceleration: SF = ma
– Centripetal Acceleration
• Force of Motion (Inertia not a force)
– Forward Force,
– Force of velocity
– Speed
• Centrifugal Force (No such thing!)
– Centripetal (really acceleration)
– Inward force (really friction)
• Internal Forces (don’t count, cancel)
– Car
– Engine
Circular Motion Requires Net Force
Net force may be provided by the tension in a
string, the normal force, or friction, among other
sources – as with any net force.
Preflights
Consider the following situation: You are driving a car with
constant speed around a horizontal circular track. On a
piece of paper, draw a Free Body Diagram (FBD) for the
car. The net force on the car is FN
f
22% A. Zero
67% B. Pointing
11% C. Pointing
radially inward
radially outward
R
W
correct
SF = ma = mv2/R
“This is why many racetracks
have banked roadways”
“Centripetal force is always pointing inward"
“Ever spin stuff in a bowl? It gravitates outward.”
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Dip ACT
Suppose you are driving through a valley whose bottom has
a circular shape. If your mass is m, what is the magnitude
of the normal force FN exerted on you by the car seat as
you drive past the bottom of the hill
A. FN < mg
a=v2/R
B. FN = mg
R
correct
C. FN > mg
FN
SF = ma
FN - mg = mv2/R
FN = mg + mv2/R
v
mg
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Dip Example
While driving on a country road at a
constant speed of 17.0 m/s, you
encounter a dip in the road. The dip
can be approximated by a circular arc
with a radius of 65.0 m.
What is the normal force exerted by
the car seat on an 80.0 kg passenger
at the bottom of the dip?
F
y
 N  mg

Fnet  N  mg  N  (80.0kg ) 9.8 m
s2

N  784 N  355.7 N
N  1140 N
Fnet  N  784 N
Fnet  ma
v
Fnet  m
r
17.0 s 

 80.0kg 
m
2
Fnet
Fnet  355.7 N
65.0m
2
Merry-Go-Round ACT
• Bonnie sits on the outer rim of a merry-go-round with
radius 3 meters, and Klyde sits midway between the center
and the rim. The merry-go-round makes one complete
revolution every two seconds.
– Klyde’s speed is:
Klyde
Bonnie
(a) the same as Bonnie’s
(b) twice Bonnie’s
(c) half Bonnie’s
VKlyde
1
 VBonnie
2
Bonnie travels 2  R in 2 seconds
vB = 2  R / 2 = 9.42 m/s
Klyde travels 2  (R/2) in 2 seconds vK = 2  (R/2) / 2 = 4.71 m/s
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Rounding a Corner
A 1,200 kg car rounds a corner of radius r = 45.0 m. If the coefficient of
friction between the tires and the road is ms = 0.82, what is the maximum
speed the car can have on the curve without skidding?
F
x
 fs
Fnetx  m s N
v
 ms mg  m
r
2
Fnet  ms mg
v2
Fnet  m
r
F
y
0
N w
N  mg
v  ms rg  (0.82)(45.0 m)(9.81 m/s 2 )  19.0 m/s
Question: How does this result depend on the mass of the car?
Banked Curves
Revolving in a Circle
An energetic father places his 20 kg child in a 5.0 kg cart to
which is attached a 2.0 m long rope. He then holds the end
of the rope and spins the cart and child in a circle, keeping
the rope parallel to the ground. If the tension in the rope is
100 N, what is the cart’s tangential speed?
m = 25 kg
r=2m
T = 100 N
v=?
Vertical:
Radial:
v2
T m
r
rT
 v2
m
Fnet  0  n  w
Fnet  T
v2
Fnet  m
r
rT
(2m)(100 N )
v

 2.83m / s
m
25kg
50 •• Driving in your car with a constant speed of 12 m/s, you encounter a bump in the
road that has a circular cross-section, as indicated in Figure 6–30. If the radius of
curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as
you pass over the top of the bump.
N
Fnet   F
Fnet  w  N
w
Fnet  mg  N
Fnet  (67 kg )(9.8 N
kg
)N
Fnet  656.6 N  N
Fnet  ma
Fnet  ma  m
Fnet
2
v
r
12m / s 

 (67kg )
Fnet  275.7 N
35m
2
656.6 N  N  275.7 N
656.6 N  275.7 N  N
N  380.9 N
51•• Referring to Problem 50, at what speed must you go over the bump if people in
your car are to feel “weightless?”
Centrifuge
A centrifuge rotates at a rate such that the
bottom of a test tube travels at a speed of 89.3
m/s. The bottom of the test tube is 8.50 cm
from the axis of rotation. What is the centripetal
acceleration acp at the bottom of the test tube in
m/s and in g (where 1 g = 9.81 m/s2)?
2
2
v
(89.3 m/s)
acp  
 93,800 m/s2  9,560 g
r (0.0850 m)