Satisfying explanation of Aristotle's Wheel Paradox.
$begingroup$
The paradox:
We have a circle and there is another circle with smaller radius. They are co-centeric.
If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.
Unsatisfying Solutions I found:
- "Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
- It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)
Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.
geometry paradoxes equivariant-maps
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
$endgroup$
add a comment |
$begingroup$
The paradox:
We have a circle and there is another circle with smaller radius. They are co-centeric.
If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.
Unsatisfying Solutions I found:
- "Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
- It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)
Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.
geometry paradoxes equivariant-maps
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
$endgroup$
1
$begingroup$
It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance.
$endgroup$
– Aretino
Jun 10 '18 at 13:55
add a comment |
$begingroup$
The paradox:
We have a circle and there is another circle with smaller radius. They are co-centeric.
If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.
Unsatisfying Solutions I found:
- "Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
- It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)
Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.
geometry paradoxes equivariant-maps
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
$endgroup$
The paradox:
We have a circle and there is another circle with smaller radius. They are co-centeric.
If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.
Unsatisfying Solutions I found:
- "Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
- It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)
Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.
geometry paradoxes equivariant-maps
geometry paradoxes equivariant-maps
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
asked Jun 10 '18 at 13:42
user2312512851user2312512851
1,197521
1,197521
1
$begingroup$
It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance.
$endgroup$
– Aretino
Jun 10 '18 at 13:55
add a comment |
1
$begingroup$
It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance.
$endgroup$
– Aretino
Jun 10 '18 at 13:55
1
1
$begingroup$
It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance.
$endgroup$
– Aretino
Jun 10 '18 at 13:55
$begingroup$
It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance.
$endgroup$
– Aretino
Jun 10 '18 at 13:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $vec V$ of the center $O$ and the velocity $vecomegatimesvec{OP}$ of rotation about the center, where $vecomega$ is angular velocity (perpendicular to the plane of the wheel).
A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have
$$
vec v_C=vec V+vecomegatimesvec{OC}
quadtext{and}quad
vec v_{C'}=vec V+vecomegatimesvec{OC'}
$$
If $vec v_C=0$ then $vec V=-vecomegatimesvec{OC}$ and
$$
vec v_{C'}=-vecomegatimesvec{OC}+vecomegatimesvec{OC'}
=vecomegatimes(vec{OC'}-vec{OC})=vecomegatimes(vec{CC'}).
$$
This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
$endgroup$
add a comment |
$begingroup$
Starting with three assumptions
Both wheels stay concentric all the way,
Both make full circle without sliding,
At the end the distance is equal to the perimeter
you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.
Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.
answered Jun 10 '18 at 13:47
RégisRégis
678310
$endgroup$
1
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
add a comment |
$begingroup$
Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:
Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.
Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.
In conclusion, the smaller the smaller circle is, the greater it slides.
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
$endgroup$
add a comment |
Your Answer
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3 Answers
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3 Answers
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$begingroup$
The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $vec V$ of the center $O$ and the velocity $vecomegatimesvec{OP}$ of rotation about the center, where $vecomega$ is angular velocity (perpendicular to the plane of the wheel).
A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have
$$
vec v_C=vec V+vecomegatimesvec{OC}
quadtext{and}quad
vec v_{C'}=vec V+vecomegatimesvec{OC'}
$$
If $vec v_C=0$ then $vec V=-vecomegatimesvec{OC}$ and
$$
vec v_{C'}=-vecomegatimesvec{OC}+vecomegatimesvec{OC'}
=vecomegatimes(vec{OC'}-vec{OC})=vecomegatimes(vec{CC'}).
$$
This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
$endgroup$
add a comment |
$begingroup$
The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $vec V$ of the center $O$ and the velocity $vecomegatimesvec{OP}$ of rotation about the center, where $vecomega$ is angular velocity (perpendicular to the plane of the wheel).
A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have
$$
vec v_C=vec V+vecomegatimesvec{OC}
quadtext{and}quad
vec v_{C'}=vec V+vecomegatimesvec{OC'}
$$
If $vec v_C=0$ then $vec V=-vecomegatimesvec{OC}$ and
$$
vec v_{C'}=-vecomegatimesvec{OC}+vecomegatimesvec{OC'}
=vecomegatimes(vec{OC'}-vec{OC})=vecomegatimes(vec{CC'}).
$$
This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
$endgroup$
add a comment |
$begingroup$
The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $vec V$ of the center $O$ and the velocity $vecomegatimesvec{OP}$ of rotation about the center, where $vecomega$ is angular velocity (perpendicular to the plane of the wheel).
A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have
$$
vec v_C=vec V+vecomegatimesvec{OC}
quadtext{and}quad
vec v_{C'}=vec V+vecomegatimesvec{OC'}
$$
If $vec v_C=0$ then $vec V=-vecomegatimesvec{OC}$ and
$$
vec v_{C'}=-vecomegatimesvec{OC}+vecomegatimesvec{OC'}
=vecomegatimes(vec{OC'}-vec{OC})=vecomegatimes(vec{CC'}).
$$
This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
$endgroup$
The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $vec V$ of the center $O$ and the velocity $vecomegatimesvec{OP}$ of rotation about the center, where $vecomega$ is angular velocity (perpendicular to the plane of the wheel).
A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have
$$
vec v_C=vec V+vecomegatimesvec{OC}
quadtext{and}quad
vec v_{C'}=vec V+vecomegatimesvec{OC'}
$$
If $vec v_C=0$ then $vec V=-vecomegatimesvec{OC}$ and
$$
vec v_{C'}=-vecomegatimesvec{OC}+vecomegatimesvec{OC'}
=vecomegatimes(vec{OC'}-vec{OC})=vecomegatimes(vec{CC'}).
$$
This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
answered Jun 10 '18 at 14:12
AretinoAretino
25.2k21445
25.2k21445
add a comment |
add a comment |
$begingroup$
Starting with three assumptions
Both wheels stay concentric all the way,
Both make full circle without sliding,
At the end the distance is equal to the perimeter
you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.
Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.
answered Jun 10 '18 at 13:47
RégisRégis
678310
$endgroup$
1
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
add a comment |
$begingroup$
Starting with three assumptions
Both wheels stay concentric all the way,
Both make full circle without sliding,
At the end the distance is equal to the perimeter
you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.
Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.
answered Jun 10 '18 at 13:47
RégisRégis
678310
$endgroup$
1
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
add a comment |
$begingroup$
Starting with three assumptions
Both wheels stay concentric all the way,
Both make full circle without sliding,
At the end the distance is equal to the perimeter
you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.
Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.
answered Jun 10 '18 at 13:47
RégisRégis
678310
$endgroup$
Starting with three assumptions
Both wheels stay concentric all the way,
Both make full circle without sliding,
At the end the distance is equal to the perimeter
you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.
Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.
answered Jun 10 '18 at 13:47
RégisRégis
678310
edited Jun 12 '18 at 0:55
answered Jun 10 '18 at 13:47
RégisRégis
678310
answered Jun 10 '18 at 13:47
RégisRégis
678310
answered Jun 10 '18 at 13:47
RégisRégis
678310
678310
1
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
add a comment |
1
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
1
1
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
This restates the first explanation given by OP, which was rejected.
$endgroup$
– Ross Millikan
Jun 10 '18 at 13:50
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
$begingroup$
No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction.
$endgroup$
– Régis
Jun 10 '18 at 14:06
add a comment |
$begingroup$
Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:
Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.
Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.
In conclusion, the smaller the smaller circle is, the greater it slides.
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
$endgroup$
add a comment |
$begingroup$
Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:
Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.
Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.
In conclusion, the smaller the smaller circle is, the greater it slides.
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
$endgroup$
add a comment |
$begingroup$
Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:
Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.
Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.
In conclusion, the smaller the smaller circle is, the greater it slides.
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
$endgroup$
Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:
Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.
Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.
In conclusion, the smaller the smaller circle is, the greater it slides.
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
edited Jun 10 '18 at 15:13
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
answered Jun 10 '18 at 14:27
farruhotafarruhota
21.1k2841
21.1k2841
add a comment |
add a comment |
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$begingroup$
It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance.
$endgroup$
– Aretino
Jun 10 '18 at 13:55