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15. In the diagram at the right, draw vector arrows (straight lines with arrowheads) which indicate the following for an object which is moving in a clockwise circle.

- the net force at point A.
- the acceleration at point B.
- the velocity at point C.

See diagram for answers. The force and acceleration vectors are directed inwards towards the center of the circle and the velocity vector is directed tangent to the circle. |

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Speed and Velocity || Acceleration || The Centripetal Force Requirement |

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16. The diagram at the right shows a satellite orbiting the Earth in an elliptical path in a clockwise direction.

- Draw a vector representing the velocity of the object at position A.
- raw a vector representing the force on the object at position D.
- Draw a vector representing the acceleration of the object at position C.
- At which of the four positions is the satellite moving fastest?

See diagram for answers. The force and acceleration vectors are directed towards the planet being orbited (which is at a foci of the ellipse) and the velocity vector is directed tangent to the path. |

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Circular Motion Principles for Satellites || Mathematics of Satellite Motion |

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17. A 1.20-kg bucket is held by a string and whirled in a vertical circle. The radius of the circle is 1.30 m. The speed of the bucket is 3.70 m/s at the top of the loop and 7.10 m/s at the bottom of the loop. On the diagrams below, construct a free-body diagram (label all forces according to type) for the bucket at both the top and the bottom of the loop. Fill in the blanks and clearly indicate the magnitude of the individual force values on your free-body diagram.

In each case, the
acceleration is found using the equation: a = v Top: F Using these equations
and the calculated values of F |

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Newton's Second Law - Revisited |

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18. Construct a free-body diagram showing the direction and types of forces acting upon the car at the top and the bottom of a loop. Be sure to label the forces according to type.

Anna experiences a downward acceleration of 24.0 m/s^{2}
at the top of the loop and an upwards acceleration of 12.0
m/s^{2} at the bottom of the loop. Use Newton's
second law to
fill in all the blanks and to ultimately determine the normal force
acting upon Anna's 800. kg roller coaster car. **PSYW**

Top |
Bottom |

_{net }= 19200 N, down |
_{net }= 9600 N, up |

_{grav }= 7840 N, down |
_{grav }= 7840 N, down |

_{norm }= 11360 N, down |
_{norm }= 17440 N, up |

This problem is very
similar to question #17 above. The net force is found by calculating
m•a; its direction is in the same direction as the Top: F The difference in
these two equations is attributed to the fact that at the top of the
loop, the F |

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Roller Coasters and Amusement Park Physics |

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19.
A fullback is running a sweep around the left side of the line. As he
rounds the turn, he is momentarily moving in circular motion,
sweeping out a quarter-circle with a radius of 4.17 meters. If the
83.5-kg fullback makes the turn with a speed of 5.21 m/s, then what
is his acceleration, the net force, the angle of lean (measured to
the vertical), and the total contact force with the ground?
**PSYW**

a = 6.51 m/s^{2}F

_{net}= 544 NF

_{grav}= 818 NF

_{norm}= 818 NF

_{frict}= 544 NF

_{contact}= 982 NAngle of Lean = 33.6 degrees

The acceleration is
found using the equation: a = v |

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Athletics |

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20.
Pete Zaria is riding the Cliff Hanger at Great America. Pete enters a
cylindrical barrel which makes 18.5 revolutions every minute. The
diameter of the barrel is 6.92 m. Pete's mass is 64.7 kg. Construct a
free-body diagram showing the forces acting upon Pete. Calculate the
acceleration and net force acting upon Pete Zaria. Finally, determine
the coefficient of friction between Pete and the walls that would be required to support Pete's weight.
**PSYW**

As usual, F Now the acceleration is: ^{2}/R
= (6.703 m/s)^{2}/(3.46 m) = 12.986 m/s^{2}Now the net force (and the normal force) can be calculated: _{net}
= F_{norm} = m•a = (64.7 kg)•(12.986 m/s/s) = 840.20 NNow that the normal force and the friction force have been found, the coefficient of friction can be determined: _{frict})/(F_{norm})
= (634 N)/(840.20 N) = 0.755 |

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Roller Coasters and Amusement Park Physics |

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21. Two objects attract
each other with a force of
gravity (F_{grav}) of 36 N. If the distance
separating the
objects is doubled, then what is the new force of gravitational
attraction? **PSYW**

22. Two objects attract
each other with a force of
gravity (F_{grav}) of 36 N. If the distance
separating the
objects is doubled and the masses one of the objects is tripled, then
what is the new force of gravitational attraction? **PSYW**

23. Two objects attract
each other with a force of
gravity (F_{grav}) of 36 N. If the distance
separating the
objects is increased by a factor of 4 and the masses of both objects
are tripled, then what is the new force of gravitational attraction?
**PSYW**

For Questions #21-#23,
Newton's universal gravitation equation must be used as a guide to
thinking about how an alteration in one variable would effect another
variable. From the equation, it can be deduced that a change in either
one of the masses would produce a proportional change in the force of
gravity. However, a change in the separation distance will produce the
inverse square effect upon the force of gravity. So increasing
separation distance by a factor of x, will decrease the force of
gravity by a factor of x 21. New force = 36 N
/ 4 = 22. New force = 36 N
• 3 / 4 = 23. New force = 36 N
•3 •3 / 16 = |

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24. Suppose that the acceleration of gravity on the surface of planet X is 12 m/s/s. Determine the acceleration of gravity at a location ...

- of 2 radii from the center of planet X.
- of 4 radii from the center of planet X.
- of 2 radii from the surface of planet X.
- on the surface of planet X if the planet mass were twice as large (same radius).
- on the surface of planet X if the planet mass were one-half as large (same radius).

Like the force of
gravity, the acceleration of gravity varies inversely with distance. So
increasing the separation distance by a factor of x, decreases the 24a. 24b. 24c. 24d. 24e. |

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The Value of g |

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25. Suppose that a planet
was located 12.0 times
further from the sun than the earth's distance from the sun.
Determine the period of the planet. **PSYW**

This question
explores the R-T relationship. As such, Kepler's third law - that the T _{earth}^{2}/R_{earth}^{3}
= T_{planet}^{2}/R_{planet}^{3}
(1 year) Rearranging this equation to solve for the period of the planet yields the equation: _{planet}^{2
}= (1 year)^{2 }• (12.0•R_{earth})^{3
}/R_{earth}^{3}
= (1 year)^{2} • (12.0)^{3 }=
1728 year^{2 }
T |

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Kepler's Three Laws |

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26. The acceleration of
gravity on the moon is
approximately one-sixth the value on earth's surface. If a person
weighs 60.0 N on the moon's surface, what is his/her approximate mass
on Earth? **PSYW**

If the _{grav}/g
= (60.0 N) / (1.63 m/s/s) = 36.7 kgSince mass is
unaffected by location (moon or earth or |

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The Value of g |

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27. Determine the force of
gravitational attraction
between a 52.0-kg mother and a 3.0-kg child if their separation distance
is 0.50 meters. **PSYW**

Use Newton's universal gravitation equation: _{grav}
= G • m_{1} • m_{2} /R^{2}
= (6.67•10^{-11} N•m^{2}/kg^{2})
• (52.0 kg) • (3.0 kg) /(0.50 m)^{2} = 4.2•10^{-8}
N |

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Newton's Law of Universal Gravitation |

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28. Use the following
information to determine the
orbital velocity at *treetop level* on the surface of
the moon.
**PSYW**

Mass of moon = 7.36
x 10^{22} kg

Radius of Moon = 1.74 x 10^{6 }m

Radius of Moon = 1.74 x 10

The orbital velocity equation is _{central}/R)
= SQRT[ (6.67•10^{-11} N•m^{2}/kg^{2})
• (7.36•10^{22} kg) / (1.74•10^{6}
m) ]
v = SQRT (2.82•10 |

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Mathematics of Satellite Motion |

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