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April 26, 2009 04:36 PM
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Are you kidding me? This reminds me of a homework question, in which a complete real-world answer with the information provided is impossible, but I think a few simplifications and clarifications can help:
First , the statement of the problem does not explicitly say, but it seems that it is acceptable to ignore drag and friction. (I explain my reasoning at the bottom so as not to get sidetracked).
Second, the problem is to find which one reaches the bottom first. For this, you need just perform a conservation of energy analysis, but the trick with this type of problem is not to ignore the effects of rotational inertia (actually, Galileo used a similar setup to calculate the acceleration objects experience due to gravity, but he got a wrong result because he forgot to include rotational inertia see p. 73).
Simply put, the potential energy the objects have at the top of the slope is converted entirely into kinetic energy at the bottom PLUS energy that is used up overcoming the rotational inertia of each of the object.
Potential energy , PE = mgh = (mass)(acceleration due to gravity)(height)
Kinetic energy, KE = 1/2 mv^2 = 1/2(mass)(final linear velocity)^2
That takes care of the linear part. For the rotational aspect, you have to consider the rotational analog of mass, the “moment of inertia,” I, as well as the rotational analog of linear velocity, the “angular velocity, ω. The rotational analog to the kinetic energy is then T = 1/2 Iω^2 = 1/2 (moment of inertia)(angular velocity)^2
Moment of inertia, like mass, will be shape-specific—it is sort of a way to think about how the mass distribution in 3D affects the angular acceleration of the object. It can be calculated from I = int(r^2 dm) , or looked up in a table. For example, I_solid disk = 1/2 mR^2 (where R is the outer diameter). For a thin hoop, it is I_hoop = mR^2, etc.
So to answer your question, you need to look at conservation of energy:
PE_top = KE_bottom + T
mgh = 1/2 mv^2 + 1/2 mR^2 ω^2 (for the solid disk)
The beauty of this approach is that the mass cancels out everywhere—and it should do so regardless of which moment of inertia you use. Further, the starting height is the same in all cases, as is the acceleration due to gravity, so the PE term is largely moot—a constant in each case. Your question therefore boils down to comparing the moments of inertia of these objects.
In the equation, you can see that the PE minus the rotational energy, T, gives you the final velocity (after some minor algebra). I.e. whichever object has a GREATER moment of inertia will have a LOWER final velocity (because more of the potential energy has gone into overcoming its rotational intertia).
In order of highest final velocity (order of lowest moment of inertia):
1) solid sphere
2) solid disk
3) hollow sphere
4) hoop
As already noted, the mass is irrelevant. It drops out of the conservation of energy equation.
Because the moment of inertia in each case goes as the radius squared, increased the radius will decrease the final velocity in every instance. If the radius of all objects is kept constant, the order in which they finish is unaltered, but the final velocity of each will increase or decrease depending on how the radius parameter is adjusted. You can consider the cases of the solid cylinder and hollow cylinder, and any other case of interest by following the same type of analysis.
*Whew* All of that for $1M ? : )
Notes:
* The reason for ignoring drag is that the disk and hoop are states to have negligible width (not a real-world case). In any event, drag depends in part upon the cross section of the object, and if the hoop and disk are rolling straight forward and their cross sections are negligible, they experience no drag. The spheres have the same cross section, but the solid one is more massive, and it therefore experience more drag, but it’s impossible to use this to add any meaningful information to the problem as it is stated, even if we are to assume that the objects are all made of the same material and therefore have identical surface characteristics.
*Friction is another dissipative force that can be ignored if we assume that the object roll *without slipping* down the incline. In this case, friction is responsible for allowing the rolling motion to occur, but unless they are at any point sliding down the incline, it doesn’t cause them to slow.
Source(s):
K.R. Symon Mechanics, 3rd ed., p. 211, Addison Wesley, Reading, MA, 1971.
E. Hecht, Physics, p. 282, Brooks / Cole Publishing Co., Albany, NY, 1996.
N. Speilberg and B.D. Anderson, Seven Ideas that Shook the Universe, Galilean Mechanics, p. 73, John Wiley, New York, 1995.
http://hyperphysics.phy-astr.gsu.edu/HBASE/mi.html#cmi
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The solid sphere would travel faster than a hollow sphere downhill, and the solid disk would travel faster than a thin hoop. Overall the solid disk would be the fastest because it's less surface area than the solid sphere, which cuts down the amount of friction involved.
For the cylinder question, the solid cylinder would be faster. The solid cylinder would still be slower than the solid disk because of how much extra surface area is touching the ground and friction would slow it down.
It would most likely be the solid disk, the solid sphere, thin hoop, the hollow sphere, the hollow cylinder and solid cylinder. (think the amount of surface area and weight of the solid cylinder would pretty much counteract any momentum it would pick up)
Making them from a heavier material would actually help as long as the surface they are rolling on is solid and it wouldn't affect them sinking into the surface. It would cause more momentum to be gained due to the higher weight.
If you increase or decrease the diameter, but keep them the same weight, it shouldn't make a difference. The weight difference is what would decide the speed difference.
Whew, that was a long answer.
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So the way a physicist thinks about something like this is to think of all the different forces affecting the object.
There is one force pulling the object down the hill, which is the force due to gravity.
There are two forces acting in an uphill direction, the force of friction with the ground, and the force of drag or air resistance.
So the total force on the object is:
F(total) = F(gravity) - F(friction) - F(drag)
It is also important to understand what force is. Force = Mass * Acceleration
So lets look at gravity:
Gravity provides constant acceleration in the downhill direction. This means that if there was no drag or friction, all of the objects would be accelerating at the same rate down the hill. However while they would be accelerating the same, the force due to gravity would be greater for an object with greater mass.
Now lets look at friction:
Friction depends on a lot of things. Here is a list:
Weight of object
Composition and texture of object
Composition and texture of slope
Angle of slope
All of these things affect the force of friction drastically, so without knowing them it is impossible to answer your question.
Now lets look at drag:
Drag is much simpler. For this question it basically only depends on the size and shape of the rolling object. An object with more surface area or with a rougher surface will have more drag.
So to sum this all up, there are two many factors that have to be considered to say definitively which shape will roll faster.
For example lets look at the hollow sphere vs. the solid sphere.
The hollow sphere would have a lower force due to gravity pulling it down the hill, but also a lower force due to drag, and a lower force due to friction pushing it back up the hill than a solid sphere. Without knowing all the information, it is impossible to say how much lower any of those parameters will be, so your question can't really be answered.
I realize that other people are going to give you definitive answers, however based on the information you have provided, I am pretty certain it is mathematically IMPOSSIBLE to answer ANY of your questions. I hope I have been able to communicate why it is impossible, although this is a pretty complicated subject which ties a lot of physics concepts together into one question. In fact there are entire physics textbooks which basically address only the concepts which are presented in this question.
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Answered Question
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Which rolls down a slope faster, given all have the same diameter: A solid sphere, a thin hollow sphere, a solid disk, or a thin hoop?
Is there any difference among these, and if so, which rolls faster? The disk and hoop can be considered to have a negligible width, and the thickness of the shell of the sphere and the hoop can be considered negligible, if that helps. In a real-world comparison, something like a bowling ball -vs- soccer ball -vs- bucket lid -vs- thin spoked wheel (all with same diameter) would be the comparison.
One more thought: What about a solid cylinder of some non-negligible width, and a hollow cylinder of non-negligible width? (Like the disk and hoop but extruded along their axis? Does additional length equate to rolling faster or slower?)
If all of the above were in a race all starting at exactly the same time, what order would they place?
Does the mass of each one make a difference? (If the solid sphere were made of lead, vs wood, for example?)
Finally, would increasing or decreasing the diameter make the objects roll faster, slower, or would they roll at the same speed?
One more thought: What about a solid cylinder of some non-negligible width, and a hollow cylinder of non-negligible width? (Like the disk and hoop but extruded along their axis? Does additional length equate to rolling faster or slower?)
If all of the above were in a race all starting at exactly the same time, what order would they place?
Does the mass of each one make a difference? (If the solid sphere were made of lead, vs wood, for example?)
Finally, would increasing or decreasing the diameter make the objects roll faster, slower, or would they roll at the same speed?
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| April 26, 2009 09:56 PM |
First , the statement of the problem does not explicitly say, but it seems that it is acceptable to ignore drag and friction. (I explain my reasoning at the bottom so as not to get sidetracked).
Second, the problem is to find which one reaches the bottom first. For this, you need just perform a conservation of energy analysis, but the trick with this type of problem is not to ignore the effects of rotational inertia (actually, Galileo used a similar setup to calculate the acceleration objects experience due to gravity, but he got a wrong result because he forgot to include rotational inertia see p. 73).
Simply put, the potential energy the objects have at the top of the slope is converted entirely into kinetic energy at the bottom PLUS energy that is used up overcoming the rotational inertia of each of the object.
Potential energy , PE = mgh = (mass)(acceleration due to gravity)(height)
Kinetic energy, KE = 1/2 mv^2 = 1/2(mass)(final linear velocity)^2
That takes care of the linear part. For the rotational aspect, you have to consider the rotational analog of mass, the “moment of inertia,” I, as well as the rotational analog of linear velocity, the “angular velocity, ω. The rotational analog to the kinetic energy is then T = 1/2 Iω^2 = 1/2 (moment of inertia)(angular velocity)^2
Moment of inertia, like mass, will be shape-specific—it is sort of a way to think about how the mass distribution in 3D affects the angular acceleration of the object. It can be calculated from I = int(r^2 dm) , or looked up in a table. For example, I_solid disk = 1/2 mR^2 (where R is the outer diameter). For a thin hoop, it is I_hoop = mR^2, etc.
So to answer your question, you need to look at conservation of energy:
PE_top = KE_bottom + T
mgh = 1/2 mv^2 + 1/2 mR^2 ω^2 (for the solid disk)
The beauty of this approach is that the mass cancels out everywhere—and it should do so regardless of which moment of inertia you use. Further, the starting height is the same in all cases, as is the acceleration due to gravity, so the PE term is largely moot—a constant in each case. Your question therefore boils down to comparing the moments of inertia of these objects.
In the equation, you can see that the PE minus the rotational energy, T, gives you the final velocity (after some minor algebra). I.e. whichever object has a GREATER moment of inertia will have a LOWER final velocity (because more of the potential energy has gone into overcoming its rotational intertia).
In order of highest final velocity (order of lowest moment of inertia):
1) solid sphere
2) solid disk
3) hollow sphere
4) hoop
As already noted, the mass is irrelevant. It drops out of the conservation of energy equation.
Because the moment of inertia in each case goes as the radius squared, increased the radius will decrease the final velocity in every instance. If the radius of all objects is kept constant, the order in which they finish is unaltered, but the final velocity of each will increase or decrease depending on how the radius parameter is adjusted. You can consider the cases of the solid cylinder and hollow cylinder, and any other case of interest by following the same type of analysis.
*Whew* All of that for $1M ? : )
Notes:
* The reason for ignoring drag is that the disk and hoop are states to have negligible width (not a real-world case). In any event, drag depends in part upon the cross section of the object, and if the hoop and disk are rolling straight forward and their cross sections are negligible, they experience no drag. The spheres have the same cross section, but the solid one is more massive, and it therefore experience more drag, but it’s impossible to use this to add any meaningful information to the problem as it is stated, even if we are to assume that the objects are all made of the same material and therefore have identical surface characteristics.
*Friction is another dissipative force that can be ignored if we assume that the object roll *without slipping* down the incline. In this case, friction is responsible for allowing the rolling motion to occur, but unless they are at any point sliding down the incline, it doesn’t cause them to slow.
Source(s):
K.R. Symon Mechanics, 3rd ed., p. 211, Addison Wesley, Reading, MA, 1971.
E. Hecht, Physics, p. 282, Brooks / Cole Publishing Co., Albany, NY, 1996.
N. Speilberg and B.D. Anderson, Seven Ideas that Shook the Universe, Galilean Mechanics, p. 73, John Wiley, New York, 1995.
http://hyperphysics.phy-astr.gsu.edu/HBASE/mi.html#cmi
| Asker's Rating: |
• Exactly the answer I was looking for; even inferred the unstated intent of ignoring friction. Ladies and gentlemen, we have a winner.
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Other Answers (2)
April 26, 2009 05:43 PM
Okay, I'm no physics expert, but I'm gonna give this my best shot. The solid sphere would travel faster than a hollow sphere downhill, and the solid disk would travel faster than a thin hoop. Overall the solid disk would be the fastest because it's less surface area than the solid sphere, which cuts down the amount of friction involved.
For the cylinder question, the solid cylinder would be faster. The solid cylinder would still be slower than the solid disk because of how much extra surface area is touching the ground and friction would slow it down.
It would most likely be the solid disk, the solid sphere, thin hoop, the hollow sphere, the hollow cylinder and solid cylinder. (think the amount of surface area and weight of the solid cylinder would pretty much counteract any momentum it would pick up)
Making them from a heavier material would actually help as long as the surface they are rolling on is solid and it wouldn't affect them sinking into the surface. It would cause more momentum to be gained due to the higher weight.
If you increase or decrease the diameter, but keep them the same weight, it shouldn't make a difference. The weight difference is what would decide the speed difference.
Whew, that was a long answer.
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April 26, 2009 11:49 PM
Well it can't really be a question of surface area, since the solid and hollow sphere would have the same surface areas, and it can't really be a matter of friction regarding the solid disk and the hoop, since they would have the same frictional contact patch (if friction were even a factor, which it isn't in the theoretical case where we're talking about rolling, not sliding, down the slope, and specified negligible width for both cases.)
Making them of heavier materials would have no effect, since the mass in fact cancels out of rotational moment of inertia.
And finally, increasing or decreasing the diameter does have an effect.
Again, I appreciate the time taken and especially some of the thought that went into this answer, but it misunderstood that the question was one about rotational moment of inertia. I do appreciate taking a shot at it though!
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Making them of heavier materials would have no effect, since the mass in fact cancels out of rotational moment of inertia.
And finally, increasing or decreasing the diameter does have an effect.
Again, I appreciate the time taken and especially some of the thought that went into this answer, but it misunderstood that the question was one about rotational moment of inertia. I do appreciate taking a shot at it though!
April 26, 2009 06:58 PM
First off let me say that there is not a simple answer to this question. So the way a physicist thinks about something like this is to think of all the different forces affecting the object.
There is one force pulling the object down the hill, which is the force due to gravity.
There are two forces acting in an uphill direction, the force of friction with the ground, and the force of drag or air resistance.
So the total force on the object is:
F(total) = F(gravity) - F(friction) - F(drag)
It is also important to understand what force is. Force = Mass * Acceleration
So lets look at gravity:
Gravity provides constant acceleration in the downhill direction. This means that if there was no drag or friction, all of the objects would be accelerating at the same rate down the hill. However while they would be accelerating the same, the force due to gravity would be greater for an object with greater mass.
Now lets look at friction:
Friction depends on a lot of things. Here is a list:
Weight of object
Composition and texture of object
Composition and texture of slope
Angle of slope
All of these things affect the force of friction drastically, so without knowing them it is impossible to answer your question.
Now lets look at drag:
Drag is much simpler. For this question it basically only depends on the size and shape of the rolling object. An object with more surface area or with a rougher surface will have more drag.
So to sum this all up, there are two many factors that have to be considered to say definitively which shape will roll faster.
For example lets look at the hollow sphere vs. the solid sphere.
The hollow sphere would have a lower force due to gravity pulling it down the hill, but also a lower force due to drag, and a lower force due to friction pushing it back up the hill than a solid sphere. Without knowing all the information, it is impossible to say how much lower any of those parameters will be, so your question can't really be answered.
I realize that other people are going to give you definitive answers, however based on the information you have provided, I am pretty certain it is mathematically IMPOSSIBLE to answer ANY of your questions. I hope I have been able to communicate why it is impossible, although this is a pretty complicated subject which ties a lot of physics concepts together into one question. In fact there are entire physics textbooks which basically address only the concepts which are presented in this question.
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April 26, 2009 11:40 PM
No, there really is a simple answer to this question. Really, there totally is. It is about items rolling, not sliding, down a plane. Friction is a factor in the real world case, but this is a theoretical question about rotational moment of inertia. You state that the hollow sphere would have a lower force due to gravity, and this is just not only wrong, it is first year high school physics class wrong. The force of gravity will be the same, provided we're performing the experiment on say, Earth. This is the classic "feather and bowling ball dropped in a vacuum chamber" experiment. -Both experience the same force and have no drag in a vacuum and both hit the ground at the same time. You are additionally simply dead wrong when you say it is mathematically impossible to answer this question. In fact this is a question that math is extraordinarily superbly suited to answering. And rather than entire physics books devoted to answering this type of question, it's usually just one chapter on rotational mechanics.
I do appreciate your taking the time to answer this, but I'm a little amazed at how wrong you can be, and how certain you sound of your answer. I would very humbly suggest that if you have no actual knowledge regarding this type of question, that it would be better to not answer at all, than to answer with not just incorrect, but *very* incorrect answers.
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I do appreciate your taking the time to answer this, but I'm a little amazed at how wrong you can be, and how certain you sound of your answer. I would very humbly suggest that if you have no actual knowledge regarding this type of question, that it would be better to not answer at all, than to answer with not just incorrect, but *very* incorrect answers.
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Since I've been wondering for years exactly which finishes first, and since I adore the simplicity of Mahalo answers, it seemed the perfect place to ask. If people don't want to chime in on this type of question, I can respect that. It probably *is* someone's homework somewhere, and I wouldn't condone just asking for the answers to one's homework on Mahalo, as it would defeat the purpose of learning the physics involved, which is after all, kind of the point of education. But without a readily available physics book handy, it seemed appropriate to throw it out there on Mahalo. I've always found it an interesting question and I presume it has some application to real world mechanics, like choosing the best wheel design for a racing bicycle or for an automotive wheel. Obviously other factors come into play for real world cases, but the theoretical case of the different shapes is what I was really interested in.
Finally, does extending the length of the solid disk or hoop into a solid cylinder or tube change the moment of inertia? Would a long enough solid tube outrace the solid sphere? (ignoring friction, and absolutely only talking about rolling, not sliding, down the inclined surface?
To answer your question about cylinders--no, as long as they're rolling down the slope, the axis of rotation is through the center of the axis of symmetry, and length is not an issue. E.g. for a solid cylinder rotating about it's axis of symmetry (i.e. the one it rotates about while rolling), the moment of inertia is 1/2 m r^2 (the same as the solid disk!). Only if it rotates about some other axis does the length come into play. (e.g. a cylinder rotating about an axis perpendicular to its axis of symmetry has I = 1/4 m r^2 + 1/12 m L^2). So, if the cylinder were tumbling down the slope, the length would matter, but otherwise, no.
If the radii of the cylinders were the same as that of the other objects, then the solid cylinder would tie with the solid disk and the hollow cylinder would tie with the hoop (assuming their walls had the same thickness). Nice.
Really, I think it's a great question--one that really demonstrates the beauty of symmetry requirements and a conservation of energy approach to physics. It almost seems counterintuitive that mass is irrelevant, but that's how it works out--the geometry turns out to be the thing that really matters.