1. Section 6.3 Apparent Forces in Circular Motion (cont.)
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2. Centrifugal Force?
If you are a passenger in a car that turns a corner quickly, it is the force of the car door, pushing inward toward the center of the curve, that causes you to turn the corner.
What you feel is your body trying to move ahead in a straight line as outside forces (the door) act to turn you in a circle.
A centrifugal force will never appear on a free-body diagram and never be included in Newton’s laws.
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3. Apparent Weight in Uniform, Vertical Circular Motion
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4. Apparent Weight in Unif. Vert. Circular Motion
The force you feel, your apparent weight, is the magnitude of the contact force that supports you.
When the passenger on the roller coaster is at the bottom of the loop:
The net force points upward, so n > w.
Her apparent weight is wapp= n, so her apparent weight is greater than her true weight.
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5. Apparent Weight in Unif. Vert. Circular Motion
Newton’s second law for the passenger at the bottom of the circle gives that
so her apparent weight is greater than her true weight, w.
Newton’s second law for the passenger at the top of the circle gives that
Note that her apparent weight can exceed the true weight,
but might not depending on the speed.
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6. Apparent Weight in Unif. Vert. Circular Motion
As the car goes slower there comes a point where n becomes zero:
The speed for which n = 0 is called the critical speed vc. Because for n to be zero we must have , the critical speed is
The critical speed is the slowest speed at which the car can complete the circle.
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7. QuickCheck 6.10
A physics textbook swings back and forth as a pendulum. Which is the correct free-body diagram when the book is at the bottom and moving to the right?
Answer: C
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8. QuickCheck 6.10
A physics textbook swings back and forth as a pendulum. Which is the correct free-body diagram when the book is at the bottom and moving to the right?
Centripetal acceleration requires an upward force. C.
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9. QuickCheck 6.11
A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant?
Answer: A
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10. QuickCheck 6.11
A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant?
Now the centripetal acceleration points down. A.
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11. QuickCheck 6.12
A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected.
Answer: E
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12. QuickCheck 6.12
A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected.
The track is above the car, so the normal force of the track pushes down. E.
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13. Centrifuges
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14. Try It Yourself: Human Centrifuge
If you spin your arm rapidly in a vertical circle, the motion will produce an effect like that in a centrifuge. The motion will assist outbound blood flow in your arteries and retard inbound blood flow in your veins. There will be a buildup of fluid in your hand that you will be able to see (and feel!) quite easily.
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15. Example 6.10 Analyzing the ultracentrifuge
An 18-cm-diameter ultracentrifuge produces an extraordinarily large centripetal acceleration of 250,000g, where g is the free-fall acceleration due to gravity.
What is its frequency in rpm?
What is the apparent weight of a sample with a mass of kg?
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16. Example 6.10 Analyzing the ultracentrifuge
prepare The acceleration in SI units is
The radius is half the diameter, or r = 9.0 cm = m.
solve We can rearrange Equation 6.5 to find the frequency given the centripetal acceleration:
Converting to rpm, we find
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17. Example 6.10 Analyzing the ultracentrifuge (cont.)
The acceleration is so high that every force is negligible except for the force that provides the centripetal acceleration. The net force is simply equal to the inward force, which is also the sample’s apparent weight: A 3 g sample has apparent weight around 1700 pounds! assess Because the acceleration is 250,000g, the apparent weight is 250,000 times the actual weight. This makes sense, as does the fact that we calculated a very high frequency, which is necessary to give the large acceleration.
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18. QuickCheck 6.13
A coin sits on a turntable as the table steadily rotates counterclockwise. What force or forces act in the plane of the turntable?
Answer: A
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19. QuickCheck 6.13
A coin sits on a turntable as the table steadily rotates counterclockwise. What force or forces act in the plane of the turntable? A.
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20. Example Problem
A car of mass 1500 kg goes over a hill at a speed of 20 m/s. The shape of the hill is approximately circular, with a radius of 60 m, as in the figure. When the car is at the highest point of the hill,
What is the force of gravity on the car?
What is the normal force of the road on the car at this point?
Answer: A: the force of gravity on the car is independent of any motion and is simply m*g = 1500*9.8 = 1.5x10^4 N.
B: the car is going through uniform circular motion at the top of the hill, and the net force on the car thus equals m*v^2/r. The net force on the car = 1500*20^2/60 = 10,000 N downward. Gravity supplies 15,000 N downward (part (a)), so the normal force must be 5000 N upward for a net force of 10,000 N down.
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21. Section 6.4 Circular Orbits and Weightlessness
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22. Orbital Motion
The force of gravity on a projectile is directed toward the center of the earth.
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23. Orbital Motion
If the launch speed of a projectile is sufficiently large, there comes a point at which the curve of the trajectory and the curve of the earth are parallel.
Such a closed trajectory is called an orbit.
An orbiting projectile is in free fall.
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24. Orbital Motion
The force of gravity is the force that causes the centripetal acceleration of an orbiting object:
An object moving in a circle of radius r at speed vorbit will have this centripetal acceleration if
That is, if an object moves parallel to the surface with the speed
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25. Orbital Motion
The orbital speed of a projectile just skimming the surface of a smooth, airless earth is
We can use vorbit to calculate the period of the satellite’s orbit:
This gives T = 2.2 hours.
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26. Weightlessness in Orbit
Astronauts and their spacecraft are in free fall.
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27. QuickCheck 6.19
Astronauts on the International Space Station are weightless because
There’s no gravity in outer space.
The net force on them is zero.
The centrifugal force balances the gravitational force.
g is very small, although not zero.
They are in free fall.
Answer: E
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28. QuickCheck 6.19
Astronauts on the International Space Station are weightless because
There’s no gravity in outer space.
The net force on them is zero.
The centrifugal force balances the gravitational force.
g is very small, although not zero.
They are in free fall.
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29. The Orbit of the Moon
The moon, like all satellites, is simply “falling” around the earth.
If we use the distance to the moon, r = 3.84 × 108 m, in:
we get a period of approximately 11 hours instead of one month.
This is because the magnitude of the force of gravity, and thus the size of g, decreases with increasing distance from the earth.
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