Circular Motion ANNOTATED.notebook
December 19, 2011
Unit 4 Vocabulary Terms
Circular Motion
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Circular Motion ANNOTATED.notebook
December 19, 2011
Radius ­ R
The radius is defined as half the diameter of a circle or the distance from the edge of a circle to its center. The radius is a unit of distance and can be measured in any unit of length. The symbol for the radius is “R”
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Circular Motion ANNOTATED.notebook
December 19, 2011
Period ­ T
The period is defined as the amount of time required for one object to complete one revolution around another object. The common unit for the period is the second. The symbol for the period is “T”.
e.g. The period for the earth to revolve around the sun is 365 days.
e.g. A tether ball may complete one revolution around a pole in 3 seconds.
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Circular Motion ANNOTATED.notebook
December 19, 2011
Frequency ­ f
The frequency is the number of revolutions or cycles occurring in one second of time. The symbol for frequency is “f ”. The common unit of frequency is cycles/sec also known as a Hertz (Hz).
e.g. The tether ball can complete 1/3 of a revolution or cycle in one second. f= 0.333 Hz
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Circular Motion ANNOTATED.notebook
December 19, 2011
Relation of Frequency to the Period
The frequency is related to the period by the following expression:
Period = 1 / Frequency T = __1__
f
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Circular Motion ANNOTATED.notebook
December 19, 2011
Circular Velocity ­ Vc
Remember that the velocity is equal to the change in displacement over the change in time. For a circle, we just replace displacement with circumference and time with the period.
vc = 2πR circumference
T period
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Circular Motion ANNOTATED.notebook
December 19, 2011
Centripetal Acceleration ­ ac
The acceleration an object experience as it travels in a circle is called the centripetal acceleration. Although the object is traveling at a constant speed, its direction is continuously changing thus its velocity is also constantly changing. As a result, the object experiences an acceleration towards the center of the circle (centripetal).
ac = vc2
R
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Circular Motion ANNOTATED.notebook
December 19, 2011
Centripetal Force ­ Fc
Centripetal force is the force resulting from a mass experiencing centripetal acceleration towards the center of a circle.
Fc = mac
Fc = m vc2
R
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Circular Motion ANNOTATED.notebook
December 19, 2011
There is NO Centrifugal Force!
When released, an object will not fly backwards from the center of the circle. Instead, the object will fly away tangentially from the edge of the circle. Centrifugal means “center fleeing”. This is common misconception in circular motion. 9
Circular Motion ANNOTATED.notebook
December 19, 2011
Solving Problems in Circular Motion
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Circular Motion ANNOTATED.notebook
December 19, 2011
An object moves in a counterclockwise horizontal circle as shown in the diagram (view is from above). The period of the object's motion is 8.14 seconds, and the radius of the path is 19.3 meters. Each of the positions shown on the diagram (not to scale) is listed below. The navigational angle indicated refers to the position of the object at each point (A through D), according to the diagram shown. 0o
270o
90o
180o
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Circular Motion ANNOTATED.notebook
December 19, 2011
An object moves in a counterclockwise horizontal circle as shown in the diagram (view is from above). The period of the object's motion is 8.14 seconds, and the radius of the path is 19.3 meters. Each of the positions shown on the diagram (not to scale) is listed below. The navigational angle indicated refers to the position of the object at each point (A through D), according to the diagram shown. 0o
90o
270o
Fill in the table with the
appropriate values!
Point
Position in Nav Units
180o
Velocity in Nav Units
Acceleration in Nav Units
A
B
C
D
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Circular Motion ANNOTATED.notebook
December 19, 2011
2. A car of mass 1266 kg moves through a semicircular turn of radius 26.3 m with a constant velocity of 9.33 m/s. The road has a coefficient of friction μ = 0.333. A. What friction force is necessary to hold this turn at this velocity? 4190 N
B. What is the maximum friction force available? 4130 N
C. Does the car skid as it moves through the turn at this velocity? YES
D. What is the maximum velocity through this turn? 9.26 m/s
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Circular Motion ANNOTATED.notebook
December 19, 2011
3. A car of mass 1266 kg drives over a semicircular hilltop of radius 26.3 m and through a semicircular valley of radius 27.4 m with a constant velocity of 7.32 m/s. A. What is the net centripetal force on the car at the top of the hill? 2580 N
B. What is the normal force on the car at the top of the hill? 9830 N
C. What is the maximum velocity with which the car can go over the hill and have the wheels stay on the road? 16.1 m/s
D. What is the net centripetal force on the car at the bottom of the valley? 2480 N
E. What is the normal force on the car at the bottom of the valley? 14900 N
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Circular Motion ANNOTATED.notebook
December 19, 2011
4. A ball of mass 206 grams on the end of string of length 82.6 cm is revolved in a vertical circle with a constant speed such that the period of revolution is 0.848 seconds. Point A is at the top of the path, point B is midway between the top and the bottom, point C is the bottom, and poind D is such that θ1 = 56.6°. Watch the units. A. What is the velocity of the ball in the circular path? 6.12 m/s
B. What is the ball's centripetal acceleration? 45.3 m/s
2
C. What is the tension in the string at point A in the path? 7.32 N
D. What is the tension in the string at point B in the path? 9.34 N
E. What is the tension in the string at point C in the path? 11.4 N
F. What is the tension in the string at point D in the path? 11.0 N
G. What is the minimum speed necessary for the ball to stay in the circular path? 2.85 m/s
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Circular Motion ANNOTATED.notebook
December 19, 2011
5. The tethered flyer shown is a 1/48 scale model of a helicopter with a mass of 126 grams on the end of a string of length 82.6 cm. When the model is turned on, it revolves in a circular path. The string makes an angle ϕ = 57.4° with the vertical. A. What is the radius of the model's path? 0.696 m
B. What is the tension in the string? 2.29 N
C. What is the velocity of the model? 3.27 m/s
D. What is the period of the model's motion? 1.34 s
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Circular Motion ANNOTATED.notebook
December 19, 2011
6. Use the Solar System Data sheet to calculate the following quantities for the planets listed.
A. What is the velocity of Earth in its orbit? 29600 m/s
B. What is the centripetal acceleration of Neptune in its orbit? 6.57e­06 m/s
2
C. What is the net centripetal force on Neptune in its orbit? 6.77e+20 N
D. What is the gravitational force between Saturn and the sun? 3.66e+22 N
E. What is the surface gravity on Mercury? 3.31 m/s
2
F. What is the Kepler constant for objects orbiting Uranus? 1.49e+14 m /s
3
2
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Circular Motion ANNOTATED.notebook
December 19, 2011
7. You are in charge of placing a satellite in orbit around an extrasolar planet. The planet has a mass of 2.67e+26 kg and a radius of 6.90e+07 m. The satellite has a mass of 2630 kg and is to be placed in an orbit with an altitude of 448 km. A. What is the orbital radius of the satellite? 6.94e+07 m
B. What is the orbital velocity of the satellite? 1.60e+04 m/s
C. What is the orbital period of the satellite? 27200 s
D. What is the centripetal acceleration of the satellite? 3.69 m/s2
E. What is the acceleration due to gravity at this altitude? 3.69 m/s2
F. What is the net centripetal force on the satellite? 9710 N
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Circular Motion ANNOTATED.notebook
December 19, 2011
8. The moon Carme orbits the planet Jupiter with an orbital radius of 22600000 km and an orbital period of 692 days. Use the orbital information of the moon to determine the following quantities.
A. What is the orbital radius of Carme in meters? 2.26e+10 m
B. What is the orbital period of Carme in seconds? 5.98e+07 s
C. What is the velocity of Carme in its orbit? 2380 m/s
D. What is the centripetal acceleration of Carme in its orbit? 0.000250 m/s
2
E. What is the mass of Jupiter based on the orbital motion of Carme ? 1.91e+27 kg
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Circular Motion ANNOTATED.notebook
December 19, 2011
9. A hypothetical planet has a radius 2.0 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface? 2.45m/s (downward)
2
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Circular Motion ANNOTATED.notebook
December 19, 2011
10. What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/18 of its value at the Earth's surface? 2.72e+07m
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Circular Motion ANNOTATED.notebook
December 19, 2011
11. Calculate the force of gravity on a spacecraft 12800 km (2 earth radii) above the Earth's surface if its mass is 1800 kg. 1950 N (toward the earth)
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Circular Motion ANNOTATED.notebook
December 19, 2011
Eccentricity
Focal Diatance "c"
Perigee
Apogee
Major Axis "a"
Eccentricity = =
Major Axis
e = 0.032
c
Focal Diatance
e = 0.50
a
e = 0.95
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