Circular Motion and Gravitation Review
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Part E: ProblemSolving
29. A roller coaster car loaded with passengers, has a mass of 500 kg; the radius of curvature of the track at the bottom point of a dip is 12 m. The vehicle has a speed of 18 m/s at this point.

In the space below, draw a freebody diagram for the car (label forces according to type).

Calculate the acceleration and the net force acting upon the car. PSYW

Calculate the force exerted on the vehicle by the track? PSYW
Answers:
The freebody diagram (part a) is shown at the right.
The acceleration of the car can be computed as follows:
a = v^{2}/R = (18.0 m/s)^{2}/(12.0 m) = 27.0 m/s^{2} (part b)
The net force can be found in the usual manner:
F_{net} = m•a = (500. kg)•(27.0 m/s^{2}) = 13500 N (part b)
Since the center of the circle (see diagram) is above the riders, then both the net force and the acceleration vectors have an upward direction. The force of gravity is downwards, so the net force is equal to the upward force minus the downward force:
F_{net} = F_{norm}  F_{grav}
where F_{grav} = m • g = (500. kg) • (9.8 m/s/s) = 4900 N
Thus,
F_{norm} = F_{net} + F_{grav} = 13500 N + 4900 N = 18400 N (part c)

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30. What is the acceleration of a piece of dust on an oldfashioned record album, 15 cm from the center, if the record is spinning at 33.3 rpm?
Answer: 1.82 m/s/s
To find acceleration, the speed and the radius must be known. The radius is given; the speed can be computed as distance per time. The dust moves a distance equivalent to 33.3 circumferences in 60 seconds. So
v = 33.3 • 2 • pi • (0.15 m) / (60 s) = 0.523 m/s.
The acceleration can now be computed using the centripetal acceleration equation:
a = v^{2}/R = (0.523 m/s)^{2 }/ (0.15 m) = 1.82 m/s/s

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31. What is the tension in a 0.500 meter rope which carries a 2.50 kg bucket of water in a vertical circle with a velocity of 3.00 m/s when the bucket is at the bottom of its swing?
Answer: 69.5 N
Begin with a freebody diagram for the bucket at the bottom of its string. Since the bucket is at the bottom of the circle, the net force is upwards (inwards). Notice that the tension force must be greater than the force of gravity in order to have a net upwards (inwards) force. Also note that while F_{tens} is directed in the direction of the net force, F_{grav} is in the opposite direction. Thus, the equation can be written:
F_{net} = F_{tens}  F_{grav}
This equation can be rearranged to solve for tension; the expressions for F_{grav} (m•g) and F_{net} (m•v^{2}/R) can be substituted into it:
F_{tens} = F_{net} + F_{grav }= m•v^{2}/R + m•g = (2.50 kg)•(3.00 m/s)^{2}/(0.500 m) + (2.50 kg)•(9.8 m/s^{2})
F_{tens} = 69.5 N

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32. A 53.5kg rider on a roller coaster car is moving 10.3 m/s at the top of a loop which has a radius of curvature of 7.29 m. Determine the normal force experienced by the rider.
Answer: 254 N
At the top of the loop, both the gravitational and the normal force are directed inwards (as shown in the freebody diagram at the right). Thus, the F_{net} equation can be written as:
F_{net} = F_{grav} + F_{norm}
This equation can be rearranged and the expressions for F_{grav} (m•g) and F_{net} (m•v^{2}/R) can be substituted into it:
F_{norm} = F_{net}  F_{grav }= m • v^{2 }/ R  m • g
F_{norm} = (53.5 kg) • (10.3 m/s)^{2 }/ (7.29 m)  (53.5 kg) • (9.8 m/s^{2})
F_{norm} = 254 N

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33. Determine the minimum coefficient of friction required to keep a 920.kg car on a turn with a radius of 26.8 m. The car is traveling 29.9 m/s and the roadway is level.
Answer: 3.40
As usual, begin with a freebody diagram. The gravity force is balanced by (and equal to) the normal force and the force of friction is the net force. The solution then begins by equating m • a to F_{frict} and carrying out the customary substitutions and algebra steps (using the fact that a = v^{2 }/ R and F_{frict} = µ • F_{norm} and F_{grav} = m • g).
m•a = F_{frict}
m • v^{2}/R = µ • F_{norm}
m • v^{2}/R = µ • m • g
The mass now cancels and the equation can be rearranged to solve for mu:
µ = v^{2}/(g• R) = (29.9 m/s)^{2} / [ (9.8 m/s^{2}) • (26.8 m)] = 3.40
(This is not a likely mu value and as such the car does not have much of a chance negotiating the turn at this speed.)

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34. With what speed (in mi/hr) must you drive your car across the crest of a hill of radius 37.1m in order to feel weightless (1.00 m/s = 2.24 mi/hr)?
Answer: 42.8 mi/hr (19.1 m/s)
The feeling of weightlessness is experienced when a person is under the sole influence of gravity. So the freebody diagram for such a person would reveal only one force  gravity, pointing straight down. In such a case, the seat does not push upward upon the person. The car has passed over the crest of the hill at such a high speed that it is momentarily raised off the roadway and is unable to support the weight of its passengers. The force of gravity is the net force and the following equation can be written:
m • a = m • g
The mass can be canceled from both sides of the equation (sigh!) and the expression for acceleration can be substituted into the equation, yielding:
v^{2}/R = g
Rearrangement and substitution yields the solution:
v = SQRT(g • R) = [(9.8 m/s^{2}) • (37.1 m)] = SQRT[364 m^{2}/s^{2}] = 19.1 m/s
This value for speed can be converted to mi/hr using the equivalency: 2.24 mi/hr = 1m/s; the car is traveling with a speed of 42.8 mi/hr.

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35. What is the reading of a bathroom scale when a man of mass 72 kg stands on it while riding in an elevator accelerating upwards at 5.4 m/s^{2}?
Answer: 1.1 x 10^{3} N
The scale reading is a measure of the amount of normal force acting upon the person. As is the usual case, begin with a freebody diagram (shown at the right). Since the acceleration is upwards, the upwards normal force (F_{norm}) is greater than the downwards gravitational force (F_{grav}). From the diagram, the F_{net} equation can be written:
F_{net} = F_{norm}  F_{grav}
Algebraic rearrangement of this equation, combined with the understanding that F_{net} = m•a and F_{grav} = m•g yields the following solution:
F_{norm }= F_{net }+ F_{grav }= m•a + m•g = (72 kg) • (5.4 m/s^{2}) + (72 kg) • (9.8 m/s^{2})
F_{norm} = 1.1 x 10^{3} N

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36. An airplane pilot makes a vertical loop with a 562m radius. Determine the normal force acting upon the pilot's 62.6kg body at the top of the loop if his air speed is 64.1 m/s.
Answer: 3.96 x 10^{3} N
As usual, begin with a freebody diagram (shown at the right). From the diagram, the equation relating net force to the individual forces can be written:
F_{net} = F_{norm} + F_{grav}
Since F_{net} = m•a = m•v^{2}/R and F_{grav} = m•g, substitutions can be made and the equation can be rewritten:
m•v^{2}/R = F_{norm} + m•g
Rearranging to solve for F_{norm} yields:
F_{norm} = m•v^{2}/R  m•g = (62.6 kg) • (64.1 m/s)^{2 }/ (56.2 m)  (62.6 kg) • (9.8 m/s^{2})
F_{norm} = 3.96 x 10^{3} N

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37. Use the following information to help determine the acceleration of the planet Pluto about the sun.
Mass of Sun = 1.991 x 10^{30} kg
SunPluto Distance = 5.91 x 10^{12 }m
Answer: g = 6.56 • 10^{6} m/s^{2}
Pluto's acceleration is caused by the gravitational force of attraction to the sun. Thus, one way (there are a couple of ways) to calculate the a of Pluto is to determine the acceleration of gravity towards the sun at the location of Pluto's orbit. The equation, substitution (using textbook values) and solution are shown below:
g = G • M_{sun} / R^{2} = (6.67 • 10^{11} N•m^{2}/kg^{2}) • (1.991 • 10^{30} kg) / (4.50 • 10^{12} m)^{2}
g = 6.56 • 10^{6} m/s^{2}

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38. What is the period of rotation of Venus in earth years if it is a distance of 1.08 x 10^{11} meters from the sun? (The earthsun distance is 1.5 x 10^{11} m.)
Answer: 0.61 yr
This question explores the RT relationship. As such, Kepler's third law  that the T^{2}/R^{3} ratio is the same for all the planets  must be utilized.
T_{earth}^{2}/R_{earth}^{3} = T_{Venus}^{2}/R_{Venus}^{3}
(1 year)^{2}/(1.5 x 10^{11})^{3} = T_{Venus}^{2 }/(1.08 x 10^{11 }m)^{3}
Rearranging this equation to solve for the period of the planet yields the equation:
T_{Venus}^{2 }= (1 year)^{2 }• (1.08 x 10^{11 }m)^{3 }/(1.5 x 10^{11})^{3} = (1 year)^{2}) • (0.72)^{3 }= 0.373 year^{2 }
T_{Venus} = SQRT(0.373 year^{2 }) = 0.61 years

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39. What is the gravitational attraction between the proton and electron in an Hydrogen atom if they are 5.3 x 10^{11} meters apart? (m_{proton} = 1.67 x 10^{27} kg; m_{electron} = 9.11x10^{31} kg)
Answer: 3.6•10^{47} N
Use Newton's universal gravitation equation:
F_{grav} = G • m_{1} • m_{2} /R^{2}
F_{grav} = (6.67•10^{11} N•m^{2}/kg^{2}) • (1.67 x 10^{27} kg) • (9.11x10^{31} kg) /(5.3 x 10^{11} m)^{2}
F_{grav} = 3.6•10^{47} N

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40. Use the following information to help determine the orbital velocity and orbital period of a satellite at a location of 15000 miles above the surface of the earth.
Mass of Earth = 5.98 x 10^{24} kg
Radius of Earth = 6.37 x 10^{6 }m
1609 m = 1.00 mi
Answers:
The first step is to find the separation distance from the center of the earth  this is R in most of the equations. The altitude of the satellite is 15000 mi or 2.41•10^{7} m; this distance must be added to the radius of the earth to determine the separation distance. The radius of the earth is 6.38•10^{6} m; the separation distance is 3.05•10^{7} m. The orbital velocity can be computed using the orbital velocity equation:
v = SQRT(G•M_{central}/R) = SQRT[ (6.67•10^{11} N•m^{2}/kg^{2}) • (5.98•10^{24} kg) / (3.05•10^{7} m) ]
v = SQRT (1.31•10^{7} m^{2}/s^{2}) = 3615 m/s
The orbital period can now be calculated using the v = 2•pi•R/T equation or the Kepler's second law equation [ T^{2}/R^{3} = 4•pi^{2}/(GM_{earth})]. Starting with the first equation, algebra yields:
T = (2•pi•R)/v = [2 • pi • (3.05•10^{7} m) / (3615 m/s) ] = 5.30 • 10^{4} s = 14.7 hr

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