Transcript Circular Motion - cloudfront.net

Circular Motion
Focused Learning Target
• Given Circular Motion and Torque Problems , I will be able to
calculate the centripetal acceleration , centripetal force ,
rotational speed and torque using the following equations:
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ac= v2 / r
Fc = mac = m v2 /r
Torque T = radius r X Force
Torque T = Radius X Force sinθ
Homework :
Chapter 7
P 255- #48,49,50,55
Chapter 8
• P 295 - # 22, 24, 26,27,28
Read and Summarize
• Read p 219-220, p 226
• Read p 96 Princeton Review
Vocabulary
1. Uniform Circular Motion
2. Centripetal acceleration –
3. Equation for centripetal acceleration
4. Discuss Centripetal force p229
1. Uniform Motion
• Type of circular motion which occurs when an object moves at
constant speed in a circular path .
• Circular motion which Object’s speed around its path remains
constant with changing velocity or changing direction
2. Centripetal Acceleration
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Acceleration in a uniform circular motion
Center seeking acceleration
Constant speed but changing direction constantly
must be directed radially inward
Acceleration vector points toward the center
Depends on the object’s speed, radius r
3. Equation of Centripetal Acceleration
:
• a c= v2 / r
4. Centripetal Force
• Net inward force which causes an inward acceleration
• Fc = mac = m v2 /r
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Directed toward the center of the circular path
Always perpendicular to the direction of motion
Could be the
A. tension of the string ,
B. friction or
C. caused by gravity
5.Critical Point – CRITICAL
VELOCITY
• Critical point occurs at the highest point of the circular motion
.
• The velocity at the highest point of a vertical uniform circular
motion is Critical Velocity.
• Critical point or critical velocity happens when the tension on
the string is 0 N . This is the speed necessary to maintain
circular motion.
Examples
• 6. An object of mass 5 kg at a constant speed of 6 m/s in a
circular path of radius 2m . Find the object’s acceleration and
the net force responsible for this motion .
• ac= v2 / r = ( 6 m /s )2 / 2m = 18 m/ s2
• Fc = mac = 5kg ( 18m/s2) = 90 N
7. A 10 kg mass is attached to a string that has a breaking point
of 200 N . If the mass is whirled in a horizontal circle of radius 80
cm , what is the maximum speed can it have?
Tension on the string provides the centripetal force
FT = Fc = m ac = m v2 / r
v = √ Fc r /m
v = √ ( 200 N ) ( 0.8 m ) / 10 kg
v = 4m/s
8. An athlete who weighs 800 N is running around a curve at a
speed of 5m/s in an arc whose radius of curvature r, is 5 m . Find
the centripetal force acting on him . What provides the
centripetal force? What could happen to him if r were smaller ?
Fc = m ac = m V2 /r
m = Fw / g = 800 N / 10 m/s2 = 80 kg
Fc = 80 kg ( 5m/s) 2 / 5m = 400 N
Friction provides the centripetal force .
The athlete will slip .
•
• 9. A roller coaster car enters the circular loop portion of the ride. At
the very top of the circle( where the people in the car are upside
down) the speed of the car is 15 m/s and the acceleration points
straight down. If the diameter of the loop is 40m and the total mass
of the car (plus the passengers) is 1200 kg , find the magnitude of
the normal force exerted by the track on the car at this point.
• Normal Force and
m = 1200kg d =40m
Weight provides
v = 15m/s
r = 20m
Centripetal force
Fc = FN + Fw = mac = m v2 / r = 1200 kg ( 15m/s)2/ 20m = 13,500 N
FN = Fc – Fw = 13,500 N – mg= 13,500 N- 1200kg (10m/s2) = 1500 N
• 10. How would the normal force change in example 9 if the
car was at the bottom of the circle ?
FN
Centripetal Force is the
combination of Normal force and
weight .
Fw
• Fc = FN- Fw
• FN = Fc + Fw = 13, 500 N + 12,000N = 25,500 N
CW:
1.A 50 cm rope is tied to the handle of the bucket which is then
whirled in a vertical circle. The mass of the bucket is 2 kg.
a. What is the speed of the bucket if the tension of the rope at
the lowest point of its path is 40 N ?
b. What is the critical speed if the rope would become slack
when the bucket reaches the highest point in the circle ?
2. An object moves at a constant speed in a circular path of
radius r at a rate of 2 revolution per second . What is its
acceleration in terms of r and π ?
3. An object m =2 kg at the end of the string is whirled around
in the vertical circle. The circular motion has a radius of 0.5 m. If
the speed of the object is 4m/s at the bottom of the circle
a. What is the tension of the string at that point ?
b. What is the minimum speed necessary for the object to
obtain circular motion ?
• 1 a 2.24 m/s b. 3.16 m/s
• 2. 16π2r/s2
• 3. FT = 44 N , 2.24 m/s
1. A. Fc = FT – FW = mV2/r
Fc = 40 N – 2kg ( 10m/s2)= 40 N -20N =20N
v = √(Fc r /m )
v = √(20 N X 0.5 m) / 2kg
V = 2.24 m/s
B. Fc = FT + FW = m v2 / r
Fc = 0 + 2kg (10m/s2) = 0 + 20 N = 20 N
v = √( Fc r / m)
v = √( 20 N X 0.5 m ) / 2 kg
v = 2.24 m/s
2. ac = V2 / r =
1 revolution = 2πr
ac= ( 2 X 2πr/ s) 2 /r = 16 π2r2 /s2 r = 16π2r/s2
3. Fc = FT – FW = mv2 /r
FT = mV2/r + Fw
FT = ( 2kg ( 4m/s)2 / 0.5 m ) + 2kg ( 10m/s2)
FT = 84 N
Fc = mg FT = 0
Fc = 2 kg ( 10m/s2)= 20 N
v =√ Fc r /m
v = √ 20 N X 0.5 m / 2kg
= 2.24 m/s
Experiment : Roller Coaster
• PURPOSE
• To allow the marble to complete the 1 and 2 loop track from
start to finish .
• To calculate the speed of the marble form start to finish on 1
and 2 loop track
• To calculate the potential energy and the kinetic energy of the
marble for each track
• To calculate the centripetal acceleration and centripetal force
for 1 loop and 2 loop track.
MATERIALS
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foam track
Marbles
Stopwatch
Meterstick
Calculator
Triple beam balance
PROCEDURE
EQUATIONS USED
Data Table
loop
1
2
Mass,
Marble,
kg
Highest
height ,
m
Total
Energy ,
Joules
V marble Ac, m/s2
, m/s
Fc ,
Newtons
And
Torque ,
Nm
CALCULATIONS
ANALYSIS AND CONCLUSIONS
• COMPARE THE 1 LOOP AND 2 LOOP TRACK ROLLER COASTER
• COMPARE YOUR ROLLER COASTER WITH 2 OTHER GROUPs’
ROLER COASTER WITH DIFFERENT STARTING HEIGHT .
COMPARE THEIR RESULTS AND YOURS.
HOW DOES THE START HEIGHT AFFECTS ALL OF THE
VARIABLES ?
11. Torque
• Read 264 and p 95-96
11. Definition of Torque
12. Factors affecting Torque
13. Moment arm or lever arm
14. Equation of torque
15. Symbol of torque
16. Unit of Torque
11. Torque
• Is the effectiveness of a force in producing rotational
acceleration.
12. Factors affecting Torque
• A. Magnitude of the force
• B. perpendicular distance from the axis of rotation
13. Moment or lever arm
• Perpendicular distance, m
14. Equation of Torque
• Torque T = radius r X Force
• Torque T = Radius X Force sinθ
15. Torque is tau τ
16 . Unit of Torque is N-m
EXAMPLES
• 17. A student pulls down with a force of 40 N on a rope that
winds around a pulley of radius 5 cm . What is the torque of
this force ?
•
F = 40 N
5
cm
• τ= r X F = 0.05 m X 40 N = 2 Nm
• 18. What is the net torque on the cylinder which is pinned at
the center of the two forces , F1 upward force , 12 cm
rightward from the center and F2 upward force ,8 cm leftward
from the center.
• Counterclockwise +
F1 =100N
• Clockwise F2 = 80 N
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τ= τ1 –τ2
τ= 100N (0 .12m)
- 80N ( 0.08 m)
τ= +5.6 N
8cm
°
12cm
Equilibrium
• Read p 265 -266 p 98
19. What is translational Equilibrium ?
20. What is rotational equilibrium ?
21. What are concurrent forces ?
22. What is mechanical equilibrium ?
23. What is static equilibrium ?
Solving Torque Problems at
Equilibrium
1.
2.
3.
4.
5.
6.
7.
8.
9.
Identify all Forces along x
Identify all Forces along y
Set up the equation EFx = 0 at equilibrium
Set up the equation EFy = 0 at equilibrium
Choose a pivot point to set up equation for ET= 0
Torque at the pivot point is zero.
Torque = perpendicular Force X lever arm
Clockwise Torque is negative
Counterclockwise Torque is positive .
Examples on Torque
• 24. A uniform bar of mass m and the length L extends
horizontally from a wall. A supporting wire connects the wall
to the bar’s midpoint , making an angle of 55o with the bar. A
sign of mass M hangs from the end of the bar .
• If the system is in static equilibrium and the wall has friction ,
determine the tension in the wire and the strength of the
force exerted on the bar by the wall if M =8kg and m=2 kg .
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55o
m
M
ΣFx =0 Fcx – FTX =0
Σfy= 0 Fcy + FTY -mg – Mg
Στ=0
Fcx – FT Cos 55o= 0
Fcy + FT Sin 550 –mg-Mg =0
FTy ( L/2) – mg(L/2) – Mg(L) =0
Fty (L/2)
= gL/2 (m + 2M)
FTY
= g(m+2M)
FT Sin 55o
= 10m/s2 ( 2kg + 2 (8kg))
FT
= 180 N / 0.819= 220 N
Fcx = FT Cos 550 = 220 N ( 0.57) = 126 N
Fcy = - 220 Sin 55 + 2kg(10m/s2) + 8kg (10m/s2) = - 80.2 N
Fc = √1262 + (-80.2 )2 = 149.36 N
• 25. A 45 kg boy is sitting on a seesaw 0.6 m from the balance
point as shown below. How far , on the other side should a60
kg girl sit so that seesaw will remain in balance ?
0.6m
X
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Counter clockwise +
Clockwise –
Choose a pivot point or fulcrum ; Torque on the pivot point =0
Στ=0
+τboy – τgirl = 0
Fwboy ( 0.6 m ) - Fwgirl (X) = 0
45 kg(10m/s2) (0.6m) = 60kg(10m/s2) X
X = 270 Nm/ 600kg
X = 0.45 m
• 26. A balanced stick is shown below . The distance from the
fulcrum is shown for each mass except the 10 g mass. What is
the approximate position of the 10g mass, based on the
diagram ?
• Pivot point = center of the bar ; torque of the weight of the
bar=0
•
40 cm
40 cm
X
20 cm
30g
40g
5cm
20g
10g
50g
• Στ=0
• 0.030 kg (10m/s2) .4m + 0.04kg (10m/s2) .2m + 0.02kg(10m/s2)
• 0.05m – 0.01kg(10m/s2) X – 0.05kg(10m/s2)(.4m)
• .012 + .008 +.001 = .01 X + .02
x = 0.1m or 10 cm
27. What is the torque about the pendulum’s suspension point
produced by the weight of the bob , given that the length of the
pendulum ,L is 50 cm and m = 0.6kg?
50o
L
τ= F sinθ X r
τ= mg sin 50o X L
τ= 0.6 kg (10m/s2) sin 50o X 0.5 m
τ= 2.3 Nm
CW : TORQUE
• 1. A solid cylinder consisting of an outer
radius R1 and an inner radius R2 is pivoted on
a frictionless axle as shown below. A string is
wound around the outer radius and is pulled
to the right with the force F1=3N. A second
string is wound around the inner radius and is
pulled down with the force F2=5N . If R1 =
0.75 m and R2= 0.35m , what is the net
torque acting on the cylinder?
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F1 =3N
F2= 5N
• 2. In an effort to tighten a bolt , a force F is applied as shown
in the figure above . If the distance from the end of the
wrench to the center of the bolt is 20 cm and F = 20 N ,
• What is the magnitude of the torque produced by F?
•
F
• 3. A uniform meter stick of mass 1 kg is
hanging from a thread attached at the stick’s
midpoint. One block of mass m =3kg hangs
from the left end of the stick and another
block , of unknown mass M, hangs below the
80 cm mark on the meter stick. If the stick
remains at rest in the horizontal position
shown above, what is M ?
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m
M
1. Στ = τ1 + τ2
Στ = - F1 ( 0.75m) +F2 ( 0.35m)
= - 3N ( 0.75m) + 5N (0.35m)
= -2.25Nm + 1.75 Nm
= -0.5 Nm
2. τ= 20 N X .2m = - 4 Nm
3. 3kg(10m/s2) (0.5m) – M(10m/s2)(0.3m )=0
M= 5 kg
Newton’s Law of Gravitation
27. gravitation force -Any two objects in the universe exert an
attractive force on each other
28. G – is the Universal Gravitational Constant =
• 6.67 X 10-11 Nm2/kg2
29. Newton’s Law of Gravitation : Any two objects in the
universe exert an attractive force on each other called
gravitational force whose strength is proportional to the product
of the object’s masses and inversely proportional to the square
of the distance between them as measured from center to
center .
30.
FG = Gm1m2 / r2
31.
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m1
m2
F1-2
F2-1
r
Examples
32. Given that the radius of the earth is 6.37 X 10 6 m ,
determine the mass of the earth.
M
m
r
FG = Gm1m2 / r2
FG= G Mm/r2
mg= GMm/r2
M= gr2/G = (10m/s2)(6.37 X 10 6 m)2
---------------------------------6.67 X 10-11 Nm2/kg2
M=6.1X 1024kg
• 33. Communications satellites are often parked in
geosynchoronous orbits above the Earth’s surface. These
satellites have orbit periods that are equal to earth’s rotation
period, so they remain above the same position on Earth’s
surface. Determine the altitude that a satellite must have to
be in geosynchoronous orbit above a fixed position on earth’s
equator. ( the mass of the earth is 5.98 X 1024 kg )
• Let m – satellite’s mass
• M- the mass of the earth
• R – distance between the center of the earth and the center
of the satellite
FG = Fc = mv2 / r
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M
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R
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m
Fc
FG = G Mm/ R2
FG = Fc = mv2 / R
mv2 / R = G Mm/ R2
V2 = GM /R
( 2πR/T )2 = GM/R
4π2R2/T2 = GM/R
4π2R3 /T2 = GM
R = √ GMT2 / 4π2
• Cube root
• R = √ ( 6.67 X 10-11)( 5.98 X 1024) (24 X 60 X 60 ) 2 / 4π2
• R = 4.23 X 107m
• If the radius of the earth is 6.37 X 106 m , the altitude of the
satellite above the earth is
H = 4.23 X 107m-6.37 X 106 m = 3.59 X 10 7 m
CW
• 1. An artificial satellite of mass m travels at a constant speed
in a circular orbit of radius R around the earth (mass M). What
is the speed of the satellite?
• The centripetal force on the satellite is provided by the Earth’s
gravitational pull.
• mV2/R = G mM/R2
• V = √ G M /R 2
• 2. A moon of Jupiter has a nearly circular orbit of radius R and
an orbit period of T. What is the mass of Jupiter ?
• mV2/R = G mM/R2
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v2 = GM/R
M = v2 R /G
M = (2πR/T)2 R / G
M = 4 π2 R3 /G T2