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course Phy 121
11/23 10pmThis is a practice for test 2.
I appreciate your feedback.
Hope you enjoyed the holiday.
Jeremy" "
Problem Number 1
A rigid rod rotating about the center of the circle holds an object in a vertical circle of radius 1.629 meters. Given the the object has mass .9399 kg and moves at .6 m/s, what is its the centripetal force acting on it?
What is the force exerted by the rod at each of the following three points:
The highest point of its path,
the lowest point, and
a point whose altitude is identical to that of the center of the circle.
aCent = v^2/r = (.6m/s)^2/1.629meters = .221 m/s^2
Fcent = aCent *m = .221m/s^2 * .9399kg = .208 N
weight = .9399 kg * 9.8m/s^2 = 9.21 N
At the top:
net force = rod force + weight
rod force = net force-weight
rod force = -.208N --9.21N = 9.00 N (approx)
At the bottom:
net force = rod force + weight
rod force = net force-weight
rod force = .208N --9.21N = 9.42 N (approx)
At point identical to the center:
Rod force horizontal = .208 N
Rod force vertical = 9.21N
Magnitude of rod force = c in a^2+b^2 = c^2
.208N ^2 + 9.21N ^2 = c^2
c = `sqrt(.208N ^2 + 9.21N ^2)
c= +-9.21N
Since we are concerned with magnitude, use 9.21N
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Problem Number 2
Given that G is approximately equal to 6.67 * 10^-11 kg m ^ 2/s ^ 2:
Give the strength of the gravitational attraction felt by a human being of mass 53 kg to a rock sphere with radius 2.9 km and density 3.7 times that of water (water's density is 1000 kg/m ^ 2), assuming that the person's entire mass is right at the surface of the sphere.
Give the attraction if the sphere was compressed to a radius of 290 meters, and if it was compressed to radius 29 meters.
To what radius would the sphere have to be compressed in order to exert a force equal to the weight of this individual on the surface of the Earth?
Rock sphere has a volume of 4/3 `pi *r^3
V= of 4/3 `pi *2900m^3
V= 1.02 * 10^11 m^3
1.02 * 10^11 m^3 * 3.7 * 1000kg/m^3 = 3.7799 * 10^14 kg
Sphere has a mass of 3.7799 * 10^14 kg
F = Gm1m2/r^2
F= 6.67*10^-11 Nm^2/kg^2 *53kg * 3.7799 * 10^14 kg/2900 m^2
F= .159 N
For the sphere with a radius of 290 m
F= 6.67*10^-11 Nm^2/kg^2 *53kg * 3.7799 * 10^14 kg/290 m^2
F = 15.9 N
For the sphere with a radius of 29 m
F= 6.67*10^-11 Nm^2/kg^2 *53kg * 3.7799 * 10^14 kg/29 m^2
F = 1590 N
To match the weight on the surface of the earth.
F= 53kg *9.8m/s^2 = 519 kg m/s^2 = 519 N (approx)
F = Gm1m2/r^2
r = `sqrt(Gm1m2/F)
r = `sqrt( 6.67*10^-11 Nm^2/kg^2 *53kg * 3.7799 * 10^14 kg/ 519N
r= +- 50.7 m
Use positive value
Radius would be approx 50.7m
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Problem Number 3
What is the moment of inertia of a disk that accelerates from .8 radians/second to 5.2 radians/second in 8.199 seconds when subject to a torque of 5.4 meter Newtons?
a = `dv/`dt
a =( 5.2 radians/sec -.8 radians/sec) / 8.199 sec
a = 4.4 radians/sec / 8.199sec
a = .537 radians/sec^2
Moment of Inertia = Torque/accelration
Moment of Inertia = 5.4 meter Newtons/ .537 radians/sec^2 = 5.4 kg m^2/sec^2/ .537 radians/sec^2
Moment of Inertia =10.1 kg m^2
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Problem Number 4
A person of mass 79 kg begins climbing a very high tower. The tower begins at the surface of the Earth, at a distance of 6400 km from the center, and rises to a position 3600 kilometers further from the center.
For each of the first three 1200 kilometer segments, determine the average of the initial and final gravitational forces encountered while climbing the segment.
Give the total work required for each segment, based on the average of the initial and final forces for the segment.
At an average power output of .62 watt/kg for 8 hours per day, how many days would be required to make the 3600 kilometer climb?
At the surface, force required = 79 kg * 9.8m/s^2 = 774.2 N
At surface + 1200 km force required = 79kg * 9.8m/s^2 /[(6400+ 1200)/6400] ^ 2 =549N
Average force = (774.2N + 549N)/2 = 661.6 N
work = 661.6N * 1,200,000 m = 7.94 *10^8 J approx
At surface + 2400 km force required = 79kg * 9.8m/s^2 /[(6400+ 2400)/6400] ^ 2 =409N
Average force = (549N +409 )/2 =479 N
work =479 N * 1,200,000 m = 5.75*10^8 J approx
At surface + 2400 km force required = 79kg * 9.8m/s^2 /[(6400+ 3600)/6400] ^ 2 =317N
Average force = (409N+317 N )/2 =363 N
work =363 N * 1,200,000 m = 4.36 *10^8 J
Total work over all three segments is 7.94 *10^8 J + 5.75*10^8 J + 4.36 *10^8J = 1.805 * 10^9 J
.62 watts/kg *79kg = 48.98watts = 48.98 J/sec
48.98J/s * 3600s/hr * 8hours/day = 1410624J/day
1.805 * 10^9 J / 1410624J/day = 1280 days approx.
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Problem Number 5
What will be the surface density of paint covering a sphere of radius 17.25 meters if when a sphere of radius 9.5 meters is covered with the same amount of paint the surface densyt of the paint is 9.400001 gallons / m^2?
The ratio of 17.25 meters to 9.5 meters is 1.82 (approx)
1.82^2 = 3.297 (approx)
Since the area is 3.297 times greater, the density will be 1/3.297 of the original
9.40001 gallons/m^2 * 1/3.297 = 2.85 gallons/m^2 (approx)
Surface density will be approximately 2.85 gallons/m^2
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Problem Number 6
An object makes a complete revolution around a circle in 4 seconds.
Through how many radians per second is the object moving?
How fast is the object traveling if it move on a circle of radius 18.99 meters?
.Cicumference = 2`pi radians = 6.28 radians (approx)
6.28radians/ 4secc = 1.57 radians per second
since on this circle 1 radian is equal to 18.99 meters
1.57 radians/sec * 18.99 meters/radian = 29.7 meters/sec
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Problem Number 7
Imagine that you are orbiting a neutron star whose mass is 31 * 10^30 kilograms at a distance of 20 kilometers from its center.
Determine the average force gradient (force change per unit of distance), in Newtons per meter, for one kilogram masses between 20 and 20+1 kilometers from the center of the neutron star.
Use this gradient to estimate the difference you would therefore expect between one kilogram of your left shoulder and one kilogram of your right shoulder, assuming that one shoulder is .45 meter further from the star than the other.
Determine the average velocity gradient, in (m/s) per meter, between orbits at 20 km and at 20+1 km from the center.
Determine how long it would therefore take one shoulder to move ahead of the other by one complete revolution.
First distance:
. F = G m M / r ^ 2,
F = 6.67 * 10^-11 N m ^ 2/kg ^ 2 * 1kg * 31 * 10^30 kg / (20000m)^2 = 5.17 * 10^12 N
Second distance:
F = 6.67 * 10^-11 N m ^ 2/kg ^ 2 * 1kg * 31 * 10^30kg / (21000m)^2 = 4.69 * 10^12 N
Change in force = 4.69 * 10^12 N - 5.17 * 10^12 N = -4.8 * 10^11N
We just need magnitude here.
4.8 * 10^11N/ 1000m = 4.8 *10^8 N/m
For the shoulders .41 meters apart:
.41meters * 4.8 *10^8 N/m = 1.97 *10^8 N
at 20km orbital velocity = `sqrt(G M / r)
v= `sqrt(6.67 * 10^-11 N m ^ 2/kg ^ 2*31 * 10^30kg * / 20000m)
v = +- 322000000 m/s (use positive)
at 21km orbital velocity = `sqrt(G M / r)
v= `sqrt(6.67 * 10^-11 N m ^ 2/kg ^ 2*31 * 10^30kg * / 21000m)
v = +- 314000000 m/s (use positive)
322000000 m/s - 314000000 m/s = 8,000,000m/s
Ave vel gradiant = 8,000,000m/s / 1000m = 8000m/s /m
Velocity difference for two shoulders is .41 m * 8000m/s/m = 3280m/s
One shoulder must travel 2`pi*.41meters further than the other or 2.58 m
dt=ds/v = 7.87* 10^-4 sec.
This is a brutal problem and not just because that's a lot of force on a human body!
???The last 2 steps really had me confused. The intro problem uses a distance corresponding to the two orbits. I used the distance between the two shoulders. Are my last 2 steps appropriate???
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Problem Number 8
How long does it take an object moving around a circular track to sweep out an angle of 13.99 radians while accelerating uniformly at .4 radians/second ^ 2? Its initial angular velocity is 7 radians/second.
What is its angular velocity after having swept out the 13.99 radians?
vf^2= v0^2 + 2a'ds
vf^2 = (7 radians/s)^2 + 2* .4radians/sec^2 * 13.99radians
vf = `sqrt(60.192radians^2/sec^2)
vf = +- 7.76 radians/sec (approx)
Angular velocity will be 7.76 radians/sec
vAVe = (vf+v0)/2 = 7.38 radians/sec
`dt= 'ds/vAve
`dt= 13.99 radians/ 7.38 radians/sec
`dt = 1.89 sec (approx)
It will take approximately 1.89 sec."
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You nailed it.
The .208 N is the tension in the rod. The vertical force is the result of the flexing of the rod which, unless there's a lot of flexion, is not associated with the tension.
Good luck on the test. Let me know if you have additional questions.
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And do enjoy the rest of your holiday.
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