#$&*
course phy 201
12/9 7:45 pm
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
032. Moment of inertia
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Question: `q001. Note that this assignment contains 3 questions.
The moment of inertia of a concentrated mass m lying at a distance r from the axis of rotation is m r^2. Moments of inertia are additive--that is, if an object with a moment of inertia about some axis is added to another object with its moment of inertia about the same axis, the moment of inertia of the system about that axis is found by simply adding the moments of inertia of the two objects.
Suppose that a uniform steel disk has moment of inertia .0713 kg m^2 about an axis through its center and perpendicular to its plane. If a magnet with mass 50 grams is attached to the disk at a point 30 cm from the axis, what will be the moment of inertia of the new system?
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Your solution:
Moment of inertia for disc: 0.0713 kg m^2; moment of inertia for magnet: 50 grams * 90 cm; 0.05 kg * 0.9 meters; 0.045 kg m^2; total inertia: 0.1163 kg m^2;
confidence rating #$&*:32; 3
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Given Solution:
A mass of m = .05 kg at distance r = .30 meters from the axis of rotation has moment of inertia I = m r^2 = .05 kg * (.30 m)^2 = .0045 kg m^2.
The moment of inertia of the new system will therefore be the sum .0713 kg m^2 + .0045 kg m^2 = .0758 kg m^2 of the moments of inertia of its components, the disk and the magnet.
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Self-critique (if necessary):
I didn’t change the units before the square, I also didn’t calculate cm^2 to m^2, but cm to m, which resulted in the moment of inertia being a decimal off.
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Self-critique rating: 3
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Question: `q002. A uniform rod with mass 5 kg is 3 meters long. Masses of .5 kg are added at the ends and at .5 meter intervals along the rod. What is the moment of inertia of the resulting system about the center of the rod?
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Your solution:
Moment of inertia for rod: 1/12m * l^2; moment of inertia for rod: 3.75 kg m^2; 0.5 kg masses added at 0.5 increments; 2 masses at 0.5 meters from center, 2 masses at 1 meter from center, 2 masses at 1.5 meters from center; moment of inertia for 0.5 distance mass: 0.5 kg * (0.5 m)^2; 0.125 kg m^2; total inertia added: 0.25 kg m^2; moment of inertia for 1 distance mass: 0.5 kg * 1m^2; 0.5 kg m^2; total inertia added: 1 kg m^2; moment of inertia for 1.5 distance mass: 0.5 kg * (1.5 m)^2; 1.125 kg m^2; total inertia added: 2.25 kg m^2; total inertia added by masses: 3 kg m^2; total moment of inertia for system: 6.75 kg m^2;
confidence rating #$&*:32; 2
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Given Solution:
The rod itself, being rotated about its center, has moment of inertia 1/12 M L^2 = 1/12 * 5 kg * (3 m)^2 = 3.75 kg m^2.
The added masses are at distances 1.5 meters (the two masses masses on the ends), 1.0 meters (the two masses .5 m from the ends), .5 meters (the two masses 1 m from the ends) and 0 meters (the mass at the middle of the rod) from the center of the rod, which is the axis of rotation.
At 1.5 m from the center a .5 kg mass will have moment of inertia m r^2 = .5 kg * (1.5 m)^2 = 1.125 kg m^2; there are two such masses and their total moment of inertia is 2.25 kg m^2.
The two masses lying at 1 m from the center each have moment inertia m r^2 = .5 kg * (1 m)^2 = .5 kg m^2, so the total of the two masses is double is, or 1 kg m^2.
The two masses lying at .5 m from the center each have moment of inertia m r^2 = .5 kg ( .5 m)^2 = .125 kg m^2, so their total is double this, or .25 kg m^2.
The mass lying at the center has r = 0 so m r^2 = 0; it therefore makes no contribution to the moment of inertia.
The total moment of inertia of the added masses is therefore 2.25 kg m^2 + 1 kg m^2 + .25 kg m^2 = 3.5 kg m^2. Adding this to the he moment of inertia of the rod itself, total moment of inertia is 3.75 kg m^2 + 3.5 kg m^2 = 7.25 kg m^2.
We note that the added masses, even including the one at the center which doesn't contribute to the moment of inertia, total only 3.5 kg, which is less than the mass of the rod; however these masses contribute as much to the moment of inertia of the system as the more massive uniform rod.
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Self-critique (if necessary):
I had the basic idea down fairly well, however I screwed up where the masses were one the rod.
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Self-critique rating: 3
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Question: `q003. A uniform disk of mass 8 kg and radius .4 meters rotates about an axis through its center and perpendicular to its plane. A uniform rod with mass 10 kg, whose length is equal to the diameter of the disk, is attached to the disk with its center coinciding with the center of the disk. The system is subjected to a torque of .8 m N. What will be its acceleration and how to long will it take the system to complete its first rotation, assuming it starts from rest?
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Your solution:
Alpha = tau/I; moment of inertia for disk: 1/2m * r^2; 0.64 kg m^2; moment of inertia for rod: 1/12m * l^2; 0.53 kg m^2; total moment of inertia: 1.17 kg m^2; alpha = 0.8 m N/1.17 kg m^2; alpha = 0.684 radians/s^2; distance to cover: 2pi radians, 6.28 radians; v0 = 0 radians/second; vf^2 = v0^2 + 2a* ds; vf^2 = 1.368 radians/s^2 * 6.28 radians; vf^2 = 8.59 radians^2/s^2; vf = 2.93 radians/second; omega = roughly 1.5 radians/second; time to complete rotation: 6.28 radians/ 1.5 r/s; roughly 4.2 seconds;
confidence rating #$&*:32; 3
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Given Solution:
The moment of inertia of the disk is 1/2 M R^2 = 1/2 * 8 kg * (.4 m)^2 = .64 kg m^2. The rod will be rotating about its center so its moment of inertia will be 1/12 M L^2 = 1/12 * 10 kg * (.8 m)^2 = .53 kg m^2 (approx).
( Note that the rod, despite its greater mass and length equal to the diameter of the disk, has less moment of inertia. This can happen because the mass of the disk is concentrated more near the rim than near the center (there is more mass in the outermost cm of the disk than in the innermost cm), while the mass of the rod is concentrated the same from cm to cm. ).
The total moment of inertia of the system is thus .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2. The acceleration of the system when subject to a .8 m N torque will therefore be
• `alpha = `tau / I = .8 m N / (1.17 kg m^2) = .7 rad/s^2, approx..
To find the time required to complete one revolution from rest we note that the initial angular velocity is 0, the angular displacement is 1 revolution or 2 `pi radians, and the angular acceleration is .7 rad/s^2. By analogy with `ds = v0 `dt + 1/2 a `dt^2, which for v0=0 is `ds = 1/2 a `dt^2, we write in terms of the angular quantities `d`theta = 1/2 `alpha `dt^2 so that
`dt = +- `sqrt( 2 `d`theta / `alpha )
= +- `sqrt( 2 * 2 `pi rad / (.7 rad/s^2))
= +-`sqrt( 12.56 rad / (.7 rad/s^2) ) = +-4.2 sec.
We choose the positive value of `dt, obtaining `dt = +4.2 sec..
INSTRUCTOR'S FURTHER CLARIFICATION
You have the acceleration and you know how far the system travels from rest.
If the system was accelerating along a line these quantities would be a, `ds and v0. The third equation of uniformly accelerated motion (`ds = v0 `dt + 1/2 a `dt^2) would apply, with `dt as the unknown. Since v0 = 0 the equation becomes `ds = 1/2 a `dt^2.
That equation isn't appropriate here because this is a rotating system. So instead of using `ds we use `dTheta, which is the symbol for angular displacement, and instead of a we use alpha, the symbol for angular acceleration.
Our equation therefore translates to
`dTheta = 1/2 alpha `dt^2.
Now using alpha = .7 rad/s^2 and `dTheta = 2 pi radians, we solve for `dt.
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Self-critique (if necessary):OK
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Self-critique rating: OK
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Question: `q004. To a uniform rod of length 30 cm and mass 80 grams, initially at rest and constrained to rotate about an axis through its center, we add a 20-gram domino at one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from the axis.
Show that the rod will balance about its center.
If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular velocity?
What centripetal force will be required to keep the domino moving around its circular path? Answer the same for the magnet.
What will be the kinetic energy of the domino? Answer the same for the magnet.
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Your Solution:
Torque = m * l; Torque of domino: 20 grams * 30 cm: 600 gram cm; Torque of magnet: 50 grams * 6 cm; 300 gram cm; torque on one end is half the torque on the other. This rod would NOT be balanced unless the magnet was TWELVE cm from the axis (50grams * 12 cm = 600 gram cm); I will proceed with the rest of the problem under the assumption that the magnet is 12 cm from the axis as opposed to 6;
@&
This is fine, but note that the 20 gram mass is not 30 cm from the center of the 30 cm rod, it's 15 cm from the center of the rod, so the magnitude of that torque would be 300 gram cm.
I say 'would be' in the sense that if torque was measured in gram cm, this would be so. However torque is measured in units of force * distance, not mass * distance. So technically the torques have magnitudes
20 grams * 980 cm/s^2 * 15 cm
and
50 grams * 980 cm/s^2 * 6 cm.
The torques still have identical magnitudes and opposite directions.
I point this out because if you calculate a torque in gram * cm, then try to use this to calculate the angular acceleration of a system, it's not going to work.
*@
Moment of inertia for rod; rod by itself: 0.0006 kg m^2; domino: 0.0018 kg m^2; magnet: 0.05 kg * (0.12 m)^2; 0.0072 kg m^2; total moment of inertia for system: 0.0096 kg m^2; tau: 0.03 m N; alpha = tau / inertia; alpha = 0.03 m N/0.0096 kg m^2; 3.125 radians/s^2; velocity after 5 seconds: 15.625 radians/second;
Find centripetal force for domino; acceleration: 3.125 radians/s^2; 2pi radians = one full revolution; for domino, 2pi radians = 188.4 cm; 6.28 radians = 1.88 meters; 3.34 radians = 1 meter; 3.125 radians/s^2 * 1 meter/3.34 radians; 0.936 m/s^2; mass of domino: 20 grams; force: 18.72 grams m/s^2;
@&
The domino moves around a circle of radius 15 cm. At 15.6 rad / sec, its velocity will be about 240 cm/sec.
Thus its centripetal acceleration is
a_cent = v^2 / r = (234 cm/s)^2 / (15 cm) = 3500 cm/s^2, approximately, which is also 35 m/s^2.
The magnet moves around a circle of radius 6 cm so its centripetal acceleration will be less.
Centripetal forces will be equal to centripetal accelerations multiplied by masses.
*@
Centripetal force for magnet: acceleration: 3.125 radians/second; 6.28 radians = 24 cm * pi; 6.28 radians = 75.38 cm; 6.28 radians = 0.754 meters; 8.33 radians = 1 meter; 3.125 radinas/s^2 * 1 meter/8.33 radians = 0.375 m/s^2; centripetal force = 50 grams * 0.375 m/s^2; 158.75 grams m/s^2;
KE = ½ m * v^2; acceleration of domino: 0.936 m/s^2; time: 5 seconds; vf: 4.68 m/s; KE = ½ 20 grams * (4.68 m/s)^2; KE = 10 grams * 21.9 m^2/s^2; 219 grams m^2/s^2;
For magnet:
Acceleration: 0.375 m/s^2; velocity: 0.375 m/s^2 * 5 seconds; 1.875 m/s^2; KE = ½ m * v^2; KE = 25 grams * 3.51 m^2/s^2; KE = 87.75 grams m^2/s^2;
confidence rating #$&*:
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@&
The angular velocity of the domino is about 15.6 rad / sec to its speed is about .936 m/s. It doesn't have an acceleraiton of .936 m/s^2.
You can calculate its KE based on its .936 m/s speed.
*@
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Question: `q004. To a uniform rod of length 30 cm and mass 80 grams, initially at rest and constrained to rotate about an axis through its center, we add a 20-gram domino at one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from the axis.
Show that the rod will balance about its center.
If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular velocity?
What centripetal force will be required to keep the domino moving around its circular path? Answer the same for the magnet.
What will be the kinetic energy of the domino? Answer the same for the magnet.
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Your Solution:
Torque = m * l; Torque of domino: 20 grams * 30 cm: 600 gram cm; Torque of magnet: 50 grams * 6 cm; 300 gram cm; torque on one end is half the torque on the other. This rod would NOT be balanced unless the magnet was TWELVE cm from the axis (50grams * 12 cm = 600 gram cm); I will proceed with the rest of the problem under the assumption that the magnet is 12 cm from the axis as opposed to 6;
@&
This is fine, but note that the 20 gram mass is not 30 cm from the center of the 30 cm rod, it's 15 cm from the center of the rod, so the magnitude of that torque would be 300 gram cm.
I say 'would be' in the sense that if torque was measured in gram cm, this would be so. However torque is measured in units of force * distance, not mass * distance. So technically the torques have magnitudes
20 grams * 980 cm/s^2 * 15 cm
and
50 grams * 980 cm/s^2 * 6 cm.
The torques still have identical magnitudes and opposite directions.
I point this out because if you calculate a torque in gram * cm, then try to use this to calculate the angular acceleration of a system, it's not going to work.
*@
Moment of inertia for rod; rod by itself: 0.0006 kg m^2; domino: 0.0018 kg m^2; magnet: 0.05 kg * (0.12 m)^2; 0.0072 kg m^2; total moment of inertia for system: 0.0096 kg m^2; tau: 0.03 m N; alpha = tau / inertia; alpha = 0.03 m N/0.0096 kg m^2; 3.125 radians/s^2; velocity after 5 seconds: 15.625 radians/second;
Find centripetal force for domino; acceleration: 3.125 radians/s^2; 2pi radians = one full revolution; for domino, 2pi radians = 188.4 cm; 6.28 radians = 1.88 meters; 3.34 radians = 1 meter; 3.125 radians/s^2 * 1 meter/3.34 radians; 0.936 m/s^2; mass of domino: 20 grams; force: 18.72 grams m/s^2;
@&
The domino moves around a circle of radius 15 cm. At 15.6 rad / sec, its velocity will be about 240 cm/sec.
Thus its centripetal acceleration is
a_cent = v^2 / r = (234 cm/s)^2 / (15 cm) = 3500 cm/s^2, approximately, which is also 35 m/s^2.
The magnet moves around a circle of radius 6 cm so its centripetal acceleration will be less.
Centripetal forces will be equal to centripetal accelerations multiplied by masses.
*@
Centripetal force for magnet: acceleration: 3.125 radians/second; 6.28 radians = 24 cm * pi; 6.28 radians = 75.38 cm; 6.28 radians = 0.754 meters; 8.33 radians = 1 meter; 3.125 radinas/s^2 * 1 meter/8.33 radians = 0.375 m/s^2; centripetal force = 50 grams * 0.375 m/s^2; 158.75 grams m/s^2;
KE = ½ m * v^2; acceleration of domino: 0.936 m/s^2; time: 5 seconds; vf: 4.68 m/s; KE = ½ 20 grams * (4.68 m/s)^2; KE = 10 grams * 21.9 m^2/s^2; 219 grams m^2/s^2;
For magnet:
Acceleration: 0.375 m/s^2; velocity: 0.375 m/s^2 * 5 seconds; 1.875 m/s^2; KE = ½ m * v^2; KE = 25 grams * 3.51 m^2/s^2; KE = 87.75 grams m^2/s^2;
confidence rating #$&*:
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@&
The angular velocity of the domino is about 15.6 rad / sec to its speed is about .936 m/s. It doesn't have an acceleraiton of .936 m/s^2.
You can calculate its KE based on its .936 m/s speed.
*@
#*&!
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Question: `q004. To a uniform rod of length 30 cm and mass 80 grams, initially at rest and constrained to rotate about an axis through its center, we add a 20-gram domino at one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from the axis.
Show that the rod will balance about its center.
If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular velocity?
What centripetal force will be required to keep the domino moving around its circular path? Answer the same for the magnet.
What will be the kinetic energy of the domino? Answer the same for the magnet.
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Your Solution:
Torque = m * l; Torque of domino: 20 grams * 30 cm: 600 gram cm; Torque of magnet: 50 grams * 6 cm; 300 gram cm; torque on one end is half the torque on the other. This rod would NOT be balanced unless the magnet was TWELVE cm from the axis (50grams * 12 cm = 600 gram cm); I will proceed with the rest of the problem under the assumption that the magnet is 12 cm from the axis as opposed to 6;
@&
This is fine, but note that the 20 gram mass is not 30 cm from the center of the 30 cm rod, it's 15 cm from the center of the rod, so the magnitude of that torque would be 300 gram cm.
I say 'would be' in the sense that if torque was measured in gram cm, this would be so. However torque is measured in units of force * distance, not mass * distance. So technically the torques have magnitudes
20 grams * 980 cm/s^2 * 15 cm
and
50 grams * 980 cm/s^2 * 6 cm.
The torques still have identical magnitudes and opposite directions.
I point this out because if you calculate a torque in gram * cm, then try to use this to calculate the angular acceleration of a system, it's not going to work.
*@
Moment of inertia for rod; rod by itself: 0.0006 kg m^2; domino: 0.0018 kg m^2; magnet: 0.05 kg * (0.12 m)^2; 0.0072 kg m^2; total moment of inertia for system: 0.0096 kg m^2; tau: 0.03 m N; alpha = tau / inertia; alpha = 0.03 m N/0.0096 kg m^2; 3.125 radians/s^2; velocity after 5 seconds: 15.625 radians/second;
Find centripetal force for domino; acceleration: 3.125 radians/s^2; 2pi radians = one full revolution; for domino, 2pi radians = 188.4 cm; 6.28 radians = 1.88 meters; 3.34 radians = 1 meter; 3.125 radians/s^2 * 1 meter/3.34 radians; 0.936 m/s^2; mass of domino: 20 grams; force: 18.72 grams m/s^2;
@&
The domino moves around a circle of radius 15 cm. At 15.6 rad / sec, its velocity will be about 240 cm/sec.
Thus its centripetal acceleration is
a_cent = v^2 / r = (234 cm/s)^2 / (15 cm) = 3500 cm/s^2, approximately, which is also 35 m/s^2.
The magnet moves around a circle of radius 6 cm so its centripetal acceleration will be less.
Centripetal forces will be equal to centripetal accelerations multiplied by masses.
*@
Centripetal force for magnet: acceleration: 3.125 radians/second; 6.28 radians = 24 cm * pi; 6.28 radians = 75.38 cm; 6.28 radians = 0.754 meters; 8.33 radians = 1 meter; 3.125 radinas/s^2 * 1 meter/8.33 radians = 0.375 m/s^2; centripetal force = 50 grams * 0.375 m/s^2; 158.75 grams m/s^2;
KE = ½ m * v^2; acceleration of domino: 0.936 m/s^2; time: 5 seconds; vf: 4.68 m/s; KE = ½ 20 grams * (4.68 m/s)^2; KE = 10 grams * 21.9 m^2/s^2; 219 grams m^2/s^2;
For magnet:
Acceleration: 0.375 m/s^2; velocity: 0.375 m/s^2 * 5 seconds; 1.875 m/s^2; KE = ½ m * v^2; KE = 25 grams * 3.51 m^2/s^2; KE = 87.75 grams m^2/s^2;
confidence rating #$&*:
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@&
The angular velocity of the domino is about 15.6 rad / sec to its speed is about .936 m/s. It doesn't have an acceleraiton of .936 m/s^2.
You can calculate its KE based on its .936 m/s speed.
*@
#*&!#*&!
&#Good responses. See my notes and let me know if you have questions. &#