uniform circular motion. a b c answer: b v circular motion act 1 a ball is going around in a circle...
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Uniform Circular Motion
A
B
C
Answer: B
v
Circular Motion ACT 1
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: constant
Speed: constant
r
v
circumference 2
period
rv
T
period the time required for one complete rota on.tiT
For a particle in uniform circular motion, the velocity vector v remains constant in magnitude, but it continuously changes its direction.
Uniform Circular Motion
Acceleration in Uniform Circular Motion
vv1vv2
vvvv2
vv1RRRR
avR
2
centripetal acceleration aaave= vv / 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
2
(centripetal acceleration)v
ar
2 (tangential velocity)r
vT
Centripetal Acceleration ACT
Which motion has the largest centripetal acceleration?
PreflightConsider 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?
A) 1 B) 2 C) 3 D) 4
f
W
FN
correct
16
F = ma = mv2/R
a=v2/R R0%28%44%28%
“Fn = Normal Force, W = Weight, the force of gravity, f = centripetal force.”
“Gravity, Normal Force, Friction”
Common Incorrect Responses• Acceleration: F = 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
Net force may be provided by the tension in a string, the normal force, or friction, among other sources – as with any net force.
Circular Motion Requires 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
f
W
FN
16
A. Zero B. Pointing radially inward C. Pointing radially outward F = ma = mv2/R
a=v2/R R
correct22%67%11%
“Ever spin stuff in a bowl? It gravitates outward.”
“Centripetal force is always pointing inward"
“This is why many racetracks have banked roadways”
Dip ACTSuppose 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 B. FN = mg C. FN > mg
v
mg
FN
R
F = ma FN - mg = mv2/RFN = mg + mv2/R
a=v2/R
correct
20
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?
yF N mg
217.0
80.065.0
355.7
ms
net
net
F kgm
F N
784 355.7
1140
N N N
N N
Dip Example
2
net
net
F ma
vF m
r
2(80.0 ) 9.8
784
net
net
mF N mg N kgs
F N N
• 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:
(a)(a) the same as Bonnie’s (b)(b) twice Bonnie’s(c)(c) half Bonnie’s
Klyde Bonnie
Bonnie travels 2 R in 2 seconds vB = 2 R / 2 = 9.42 m/sKlyde travels 2 (R/2) in 2 seconds vK = 2 (R/2) / 2 = 4.71 m/s
1
2Klyde BonnieV V
Merry-Go-Round ACT
33
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 s = 0.82, what is the maximum speed the car can have on the curve without skidding?
x s
netx s
F f
F N
0yF
N w
N mg
2
s
vmg m
r
2(0.82)(45.0 m)(9.81 m/s ) 19.0 m/ssv rg
Question: How does this result depend on the mass of the car?
net sF mg2
net
vF m
r
Banked Curves
Revolving in a CircleAn 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?
2
net
vF m
r
netF TRadial:
0netF n w Vertical:m = 25 kgr = 2 mT = 100 Nv = ?
2vT m
r
2rTv
m (2 )(100 )
2.83 /25
rT m Nv m s
m kg
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.
w
N
(67 )(9.8 )
656.6
net
net
net
net
F w N
F mg N
NF kg Nkg
F N N
2
212 /
(67 )35
275.7
net
net
net
vF ma m
r
m sF kg
mF N
netF F
netF ma 656.6 275.7
656.6 275.7
380.9
N N N
N N N
N N
51•• Referring to Problem 50, at what speed must you go over the bump if people in your car are to feel “weightless?”
CentrifugeA 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 22(89.3 m/s)
93,800 m/s 9,560 g(0.0850 m)cp
va
r