marks@tekigm2.UUCP (Mark D. Salzman) (04/29/86)
Hello Out There,
Here's a little thought problem that might stir things up a bit.
Picture yourself on a space station similar to the one used in 2001
(i.e. a spinning ring or toroid). You are standing in the middle of
one of the decks near the outside edge of the ring and the spin of
the station is providing a "gravity" about equal to that found on
the surface of the Earth.
If you were to drop a ball (a simple release with no additional
forces applied), would it fall straight down (along a line through
the center of the ring and the point of release) or would it follow
another path (relative to the aforementioned line)?
Since my physics is a bit rusty, I can't say which is true. So I
invite comments from more knowledgeable people on which they feel
is right. Please include your reasons to support your views.
Forgive me if this has been covered before. Otherwise, Have A Nice Day!
Mark Salzman, Tektronix Inc. | The more complex the mind,
P.O. Box 3500, Mail Stop C1-937 | the greater the need for
Vancouver, Washington. 98668 | the simplicity of play.
Phone (206) 253-5542. |
{world}!tektronix!tekigm2!marks | James T. Kirkgwyn@brl-smoke.ARPA (Doug Gwyn ) (05/03/86)
In article <632@tekigm2.UUCP> marks@tekigm2.UUCP (Mark D. Salzman) writes: >Picture yourself on a space station similar to the one used in 2001 >(i.e. a spinning ring or toroid). You are standing in the middle of >one of the decks near the outside edge of the ring and the spin of >the station is providing a "gravity" about equal to that found on >the surface of the Earth. > >If you were to drop a ball (a simple release with no additional >forces applied), would it fall straight down (along a line through >the center of the ring and the point of release) or would it follow >another path (relative to the aforementioned line)? Once you remove constraining forces from the ball (i.e., let go), you can apply Newton's law of inertia: the ball will continue in a straight line with the velocity it had at the moment of release. (Actually, things are very slightly complicated by the fact that this experiment is being done in orbit rather than in empty space, but that can be ignored for purposes of a first-order discussion.) An observer attached to the space station at the point of release would see the ball follow a curved path toward the outer wall. It would not be a radial path, since the observer is subject to an acceleration whose direction changes as he moves (always centrally directed). You can see some of the effects by playing with chalky marbles on a kid's phonograph turntable.
desj@brahms.BERKELEY.EDU (David desJardins) (05/03/86)
In article <632@tekigm2.UUCP> marks@tekigm2.UUCP (Mark D. Salzman) writes: >Picture yourself on a space station similar to the one used in 2001 >(i.e. a spinning ring or toroid). You are standing in the middle of >one of the decks near the outside edge of the ring and the spin of >the station is providing a "gravity" about equal to that found on >the surface of the Earth. > >If you were to drop a ball (a simple release with no additional >forces applied), would it fall straight down (along a line through >the center of the ring and the point of release) or would it follow >another path (relative to the aforementioned line)? Assuming a vacuum, or that air resistance effects are negligible, there would be no significant forces on the ball, and so it would follow a straight line in space. Of course since you are moving in a great circle this line would not appear straight to you. The actual path would appear to move downward with the acceleration you would expect, and would curve antispinward (i.e. in the direction against the one the ring is spinning in. The curvature would depend, of course, on the diameter of the ring. If the ring were small enough, or the ball was allowed to fall for long enough, it would also be noted that it would start to fall slightly more slowly than expected. -- David desJardins
ins_akaa@jhunix.UUCP (Ken Arromdee) (05/03/86)
>Picture yourself on a space station similar to the one used in 2001 >(i.e. a spinning ring or toroid). You are standing in the middle of >one of the decks near the outside edge of the ring and the spin of >the station is providing a "gravity" about equal to that found on >the surface of the Earth. > >If you were to drop a ball (a simple release with no additional >forces applied), would it fall straight down (along a line through >the center of the ring and the point of release) or would it follow >another path (relative to the aforementioned line)? From the point of view of someone not spinning and standing outside the station, the ball would move, without accelerating, in a straight line tangent to the circle whose center is at the space station's center and with a radius of the distance from the center to where the ball was released. This straight line would eventually intersect the rim of the space station. An observer on the space station would see the ball appear to drop and then move in an antispinward direction, eventually hitting the floor. -- Kenneth Arromdee | | BITNET: G46I4701 at JHUVM, INS_AKAA at JHUVMS -|------|- CSNET: ins_akaa@jhunix.CSNET -|------|- ARPA: ins_akaa%jhunix@hopkins.ARPA -|------|- UUCP: {allegra!hopkins, seismo!umcp-cs, ihnp4!whuxcc} -|------|- !jhunix!ins_akaa | |
bl@hplabsb.UUCP (Bruce T. Lowerre) (05/05/86)
> Hello Out There, > > Here's a little thought problem that might stir things up a bit. > > Picture yourself on a space station similar to the one used in 2001 > (i.e. a spinning ring or toroid). You are standing in the middle of > one of the decks near the outside edge of the ring and the spin of > the station is providing a "gravity" about equal to that found on > the surface of the Earth. > > If you were to drop a ball (a simple release with no additional > forces applied), would it fall straight down (along a line through > the center of the ring and the point of release) or would it follow > another path (relative to the aforementioned line)? Due to angular momentum (or lack there of) it will fall in a curved path toward the opposite direction of the rotation of the toroid.
rdp@teddy.UUCP (Richard D. Pierce) (05/05/86)
In article <3461@hplabsb.UUCP> bl@hplabsb.UUCP (Bruce T. Lowerre) writes: >> Hello Out There, >> >> Here's a little thought problem that might stir things up a bit. >> >> Picture yourself on a space station similar to the one used in 2001 >> (i.e. a spinning ring or toroid). You are standing in the middle of >> one of the decks near the outside edge of the ring and the spin of >> the station is providing a "gravity" about equal to that found on >> the surface of the Earth. >> >> If you were to drop a ball (a simple release with no additional >> forces applied), would it fall straight down (along a line through >> the center of the ring and the point of release) or would it follow >> another path (relative to the aforementioned line)? > >Due to angular momentum (or lack there of) it will fall in a curved >path toward the opposite direction of the rotation of the toroid. Well, I don't know exactly what is meant by this explanation. So here is my own: Upon releasing the ball, it is no longer experiencing the centripetal force that keeps it in it's circular path (that force is supplied by your hand, which is supplied by your body, which is supplied by the fact that you are standing on the "floor" which is connected to the center of rotation, etc.) The ball is now free to move under the influence of its own inertia, meaning that it will attempt to persue a straight path, tangent to the circle it was describing at the point of its release, at a speed equal to the circumfrential velocity of the ring. It will continue on this path until interrupted, by the floor, by you, or by the great outfielder at the edge of the universe. Dick Pierce
marks@tekigm2.UUCP (Mark D. Salzman) (05/06/86)
<Nobody knows where the Line Eater goes.> Thank's to all those that responded to my original posting. The consensus is that the falling ball will curve back against the direction of spin of the space station. Holding the angular velocity of the station constant, the instantaneous velocity increases as you go farther away from the center of spin. Since the ball, once released, maintains the same velocity that it had at the release point, it will appear to slow down as it nears the outer rim (floor) of the station. Now that we are all aware of these facts, playing catch on such a space station will be simple, right? :-) Enjoy!
rb@ccird1.UUCP (Rex Ballard) (05/11/86)
In article <3461@hplabsb.UUCP> bl@hplabsb.UUCP (Bruce T. Lowerre) writes: >> Hello Out There, >> >> Here's a little thought problem that might stir things up a bit. >> >> Picture yourself on a space station similar to the one used in 2001 >> (i.e. a spinning ring or toroid). You are standing in the middle of >> one of the decks near the outside edge of the ring and the spin of >> the station is providing a "gravity" about equal to that found on >> the surface of the Earth. >> >> If you were to drop a ball (a simple release with no additional >> forces applied), would it fall straight down (along a line through >> the center of the ring and the point of release) or would it follow >> another path (relative to the aforementioned line)? > >Due to angular momentum (or lack there of) it will fall in a curved >path toward the opposite direction of the rotation of the toroid. Theoretically, there should be the same curved path for a ball dropped on Earth. It is not noticable because the earths rotation is so slow, and gravity so strong, that it would appear to fall straight. As to whether you would see an observable curve, depends on the, diameter of the toroid. Since you have specified pseudogravitation equal to earths, a small toroid would have to spin very quickly, and curvature would be quite pronounced. If the toroid were sufficiently large, the arc might only be a few seconds. Does anybody have a formula for the relationship?
desj@brahms.BERKELEY.EDU (David desJardins) (05/14/86)
In article <415@ccird1.UUCP> rb@ccird1.UUCP (Rex Ballard) writes: >>> Picture yourself on a space station similar to the one used in 2001 >>> (i.e. a spinning ring or toroid) .... spin of the station is providing >>> a "gravity" about equal to that found on the surface of the Earth. >>> >>> If you were to drop a ball ... would it fall straight down ... or >>> would it follow another path? >> >>Due to angular momentum (or lack there of) it will fall in a curved >>path toward the opposite direction of the rotation of the toroid. > >As to whether you would see an observable curve, depends on the, >diameter of the toroid. Since you have specified pseudogravitation >equal to earths, a small toroid would have to spin very quickly, and >curvature would be quite pronounced. If the toroid were sufficiently >large, the arc might only be a few seconds. > >Does anybody have a formula for the relationship? The exact equation of motion in the coordinate frame of the observer moving with the rotation (+x = spinward, +y = down, observer at (0,0), R = radius of rotation, w = rate of rotation = sqrt (g/R)) is: (x, y) = (wRt cos wt - R sin wt, wRt sin wt + R (cos wt - 1)). To third order in time, and substituting sqrt (g/R) for w, this is: (x, y) ~= (-1/2 g sqrt (g/R) t^3, 1/2 g t^2). The y term at least is not surprising! Now, eliminating t and solving for x in terms of y, x ~= - sqrt(2) y^3/2 R^-1/2. So, if y = 1 meter, we have the following results: R x (distance of impact point from "straight down") ------ ----- 10 m .44 m 100 m .14 m 1000 m .04 m -- David desJardins
bl@hplabsb.UUCP (Bruce T. Lowerre) (05/14/86)
> In article <3461@hplabsb.UUCP> bl@hplabsb.UUCP (Bruce T. Lowerre) writes: > >> Hello Out There, > >> > >> Here's a little thought problem that might stir things up a bit. > >> > >> Picture yourself on a space station similar to the one used in 2001 > >> (i.e. a spinning ring or toroid). You are standing in the middle of > >> one of the decks near the outside edge of the ring and the spin of > >> the station is providing a "gravity" about equal to that found on > >> the surface of the Earth. > >> > >> If you were to drop a ball (a simple release with no additional > >> forces applied), would it fall straight down (along a line through > >> the center of the ring and the point of release) or would it follow > >> another path (relative to the aforementioned line)? > > > >Due to angular momentum (or lack there of) it will fall in a curved > >path toward the opposite direction of the rotation of the toroid. > > Theoretically, there should be the same curved path for a ball dropped > on Earth. It is not noticable because the earths rotation is so slow, > and gravity so strong, that it would appear to fall straight. > > As to whether you would see an observable curve, depends on the, > diameter of the toroid. Since you have specified pseudogravitation > equal to earths, a small toroid would have to spin very quickly, and > curvature would be quite pronounced. If the toroid were sufficiently > large, the arc might only be a few seconds. > > Does anybody have a formula for the relationship? An interesting observation is that the falling ball inside of the toroid would NOT accelerate as it "fell" while on Earth it would. Actually, the ball would mearly proceed in a straight line (seen from an unaccelarted frame of reference). If the ball were to be released on a point along the axis of rotation, then it would not "fall".
msnthrp@fluke.UUCP (Jeffrey Gueble) (05/20/86)
> In article <3461@hplabsb.UUCP> bl@hplabsb.UUCP (Bruce T. Lowerre) writes: > >> Hello Out There, > >> > >> Here's a little thought problem that might stir things up a bit. > >> > >> Picture yourself on a space station similar to the one used in 2001 > >> (i.e. a spinning ring or toroid). You are standing in the middle of > >> one of the decks near the outside edge of the ring and the spin of > >> the station is providing a "gravity" about equal to that found on > >> the surface of the Earth. > >> > >> If you were to drop a ball (a simple release with no additional > >> forces applied), would it fall straight down (along a line through > >> the center of the ring and the point of release) or would it follow > >> another path (relative to the aforementioned line)? > > > >Due to angular momentum (or lack there of) it will fall in a curved > >path toward the opposite direction of the rotation of the toroid. > > Theoretically, there should be the same curved path for a ball dropped > on Earth. It is not noticable because the earths rotation is so slow, > and gravity so strong, that it would appear to fall straight. > > As to whether you would see an observable curve, depends on the, > diameter of the toroid. Since you have specified pseudogravitation > equal to earths, a small toroid would have to spin very quickly, and > curvature would be quite pronounced. If the toroid were sufficiently > large, the arc might only be a few seconds. > > Does anybody have a formula for the relationship? The ball will not follow a curved trajectory; it will depart from the toroid on a linear path. This path may not appear linear to the observer located on the moving toriod, however, it will still be linear. After the release, the ball will travel on a linear path, rotating about an axis through its center of mass. The problem is analogous to a ball attached to a string which is swung in a circular arc. If the string is released, the ball sails off in a linear trajectory (actually, the effects of gravity make this a parabolic trajectory). The important thing to realize here is that there are no forces acting on the ball after release, and the momentum of the ball will be determined by the initial angular and linear velocities. The angular velocity will be the same as the angular velocity of the toroid. The linear velocity will depend on the distance from the release point to the center of the toroid, and will have a magnitude equal to the radial distance times the angular velocity. The only way that the angular velocity could affect the trajectory would be if the act of spinning induced a force which acted on the ball. In free space, I dont believe that this would occur.