motion in a plane chapter 8. centripetal acceleration centripetal acceleration – acceleration that...
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Motion in a Plane
Chapter 8
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Centripetal Acceleration
• Centripetal Acceleration – acceleration that points towards the center of a circle.– Also called Radial Acceleration (aR)
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vBall rolling in a straight line (inertia)
v Same ball, hooked to a string
aR
vaR aR = v2
r
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If you are on a carousel at constant speed, are you experiencing acceleration?
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If you twirl a yo-yo and let go of the string, what way will it fly?
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Period and Frequency
Period (T)– Time required for one complete (360o) revolution– Measured in seconds
Frequency– Number of revolutions per second– Measured in rev/s or Hertz (Hz)
T = 1 f
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Formulas
v = 2r v = r T
aR = v2 a = r
r
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A 150-kg ball is twirled at the end of a 0.600 m string. It makes 2.00 revolutions per second. Find the period, velocity, and acceleration.
(0.500 s, 7.54 m/s, 94.8 m/s2)
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The moon has a radius with the earth of about 384,000 km and a period of 27.3 days.
A.Calculate the acceleration of the moon toward the earth. (2.72 X 10-3 m/s2)
B.Calculate the previous answer in “g’s” (2.78 X 10-4 g)
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Centripetal Force – the “center seeking” force that pulls an object in a circular path.– Yo-yo– Planets– Merry-go-round– Car rounding a curve
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Centrifugal Force
A word about Centrifugal Force
• Doesn’t really exist.• “apparent outward
force”• Water in swinging cup
example Centripetal Force of string
Direction water wants to go
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Centripetal Motion
F = maR = mv2
rA 0.150 kg yo-yo is attached to a 0.600 m string
and twirled at 2 revolutions per minute.a. Calculate the velocity in m/s ()b.Calculate the centripetal force in the string
(14.2 N)
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Thor’s Hammer (mjolnir) has a mass of 10 kg and the handle and loop have a length of 50 cm. If he can swing the hammer at a speed of 3 m/s, what force is exerted on Thor’s hands?
(Ans: 180 N)
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Can Thor swing his hammer so that it is perfectly parallel to the ground?
FR
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What angle will the hammer take with the horizontal?
F R
mg
Let’s resolve the FR vector into it’s components:
FRx = FRsin
FRy = FRcos
Fy = 0 (the hammer is not rising or falling)
Fy = 0 = FRcos – mg
FRcos = mg
cos = mg/FR
= 57o
How about if he swings faster?
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A father places a 20.0 kg child on a 5.00 kg wagon and twirls her in a circle with a 2.00 m rope of tension 100 N. How many rpms does the wagon make ()? (14 rpm)
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A 0.150 kg ball is swung on a 1.10-m string in a vertical circle. What minimum speed must it have at the top of the circle to keep moving in a circle?
At the top of the circle, both the weight and the tension in the string contribute to the centripetal force
F = FT + mg
mg FT
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F = FT + mg
FR = FT + mgmv2 = FT + mg r(tricky part: assume FT = 0, just as the cord
goes slack, but before the ball falls)mv2 = mg rv2 = grv = 3.28 m/s
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Note: this equation is also the minumum velocity for orbit of a satellite
v = \/rg
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What is the tension in the cord at the bottom of the arc if the ball moves at twice the minimum speed? (v = 6.56 m/s)
mg
FT
At the bottom of the circle, the weight opposes the centripetal force.
F = FT – mgmv2 = FT - mg rFT = mv2 + mg rFT = 7.34 N
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Car Rounding a Turn
• Friction provides the centripetal force• Use the coefficient of static friction (s). The
wheels are turning, not sliding, across the surface
• Wheel lock = kinetic friction takes over. k is always less than s, so the car is much more likely to skid.
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A 1000-kg car rounds a curve (r=50 m) at a speed of 14 m/s. Will the car skid if the road is dry and s=0.60?
Ffr = FR
mg
FNLet’s first solve for the Normal Force
FN = mg = (1000 kg)(9.8 m/s2)
FN = 9800 N
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Fx = Ffr
FR = Ffr
mv2 = sFN
r(1000 kg)(14m/s)2 = (0.60)(9800 N) (50 m)3920 N < 5800 N
The car will make it. 3920 N are required, and the frcition provides 5800 N.
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Will the car make it if it is icy and the s = 0.25
Fx = Ffr
FR = Ffr
mv2 = sFN
r(1000 kg)(14m/s)2 = (0.25)(9800 N) (50 m)3920 N > 2450 N
The car will not make it. 3920 N are required, and the friction only provides 2450 N.
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What is the maximum speed a 1500 kg car can take a flat curve with a radius of 50 m (s = 0.80)
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BANKED CURVES• Banked to reduce the reliance on friction• Part of the Normal Force now contributes
to the centripetal force
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FR = Ffr + FNsin
(ideally, we bank the road so that no friction is required: Ffr = 0)
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Banked Curves: Example 1A 1000-kg car rounds a 50 m radius turn at 14
m/s. What angle should the road be banked so that no friction is required?
mg
FN
FN = mgcos
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Now we will simply work with the Normal Force to find the component that points to the center of the circle
mg
FN
First consider the y forces.
Fy = FNcos - mg
Since the car does not move up or down:
Fy = 0
0 = FNcos – mg
FNcos = mg
FN = mg/cos
FNcos
FNsin
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mv2 = FNsin rmv2 = mgsin r cosv2 = gtan rv2 = gtanrv2 = tangr
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tan = (14 m/s)2 = 0.40 (50 m)(9.8m/s2)
= 22o
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Fred Flintstone places a 1.00 kg rock in a 1.00 m long sling. The vine breaks at a tension of 200 N.
a. Calculate the angle below the horizontal plane that the rock will take. (2.81o)
b.Calculate the maximum linear velocity the rock can twirl. (14.1 m/s)
c. Calculate the angular velocity in rpm’s. (135 rpm)
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Circular Orbits• Orbits are freefall (not true weightlessness)• Orbital velocity must match the weight
mg = mv2
r g = v2 v = √ gr r
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A satellite wishes to orbits at a height of 200 miles above the earth’s surface.
a. Calculate the height above the center of the earth if Rearth = 6.37 X 106 m. (6.69 X 106 m)
b.Calculate the orbital velocity. (8098 m/s)c. Calculate the period in minutes. (86.5 min)
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Review of Angular KinematicsA motor spins a 2.0 kg block on an 80.0 cm arm at
200 rpm. The coefficient of kinetic friction is 0.60.a. Draw a free body diagram of the block.b. Calculate the tangential acceleration of the block
(due to friction). (-5.88 m/s2)c. Calculate the angular acceleration. (-7.35 rad/s2)d. Calculate the time until the block comes to a rest.
(2.8 s)e. Calculate the number of revolutions. (4.7 rev)