Re: [math-fun] spherical pendulums?
Surely there must be conditions more than simply "smooth". For example, if the cross-section is smooth, but not convex, then the particle will leave the surface when the curvature becomes negative. At 11:08 AM 12/6/2011, Eugene Salamin wrote:
The curved cylinder imposes a normal force on the point mass, and this does not affect the motion parallel to the surface. So indeed the particle moves in a parabola wrapped around the cylinder, no matter what the cross section, at least for smooth surfaces.
-- Gene
________________________________ From: Henry Baker <hbaker1@pipeline.com> To: Adam P Goucher <apgoucher@gmx.com> Cc: math-fun@mailman.xmission.com Sent: Tuesday, December 6, 2011 8:01 AM Subject: Re: [math-fun] spherical pendulums?
Hmmmm....
Does it matter if we wrap the transparent paper around a cylinder with circular cross section, or could the cross section be another shape -- e.g., an ellipse or something else?
Obviously, if the shape included some sort of cusp, it wouldn't stay on the shape.
Aren't we assuming that any accelerations are perpendicular to the shape of the cross section? But if the moving object is accelerating/decelerating along other dimensions, wouldn't that change things?
At 10:43 AM 12/5/2011, Adam P. Goucher wrote:
Here's a related problem:
A point particle of mass M (i.e., zero moment of inertia) is sliding frictionlessly around inside a vertical cylinder; gravity is downwards of strength G. What are the paths?
Imagine drawing a parabolic arc on a sheet of transparent paper, and wrapping it around the inside of the cylinder. This is parametrised by:
x = r cos(wt) y = r sin(wt) z = -½gt²
(w, r and g are constants, namely the angular speed, radius and gravitational field strength, respectively.)
Sincerely,
Adam P. Goucher
If the particle is constrained to lie in the surface, then the normal force can be inward or outward. On the other hand, if you are thinking of a ball touching the surface, you are correct. But in this case, the physics is much more complicated unless you impose the condition of zero frictional force, so that the ball has constant angular momentum. This means that the ball will slip rather than roll. -- Gene
________________________________ From: Henry Baker <hbaker1@pipeline.com> To: Eugene Salamin <gene_salamin@yahoo.com>; math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, December 6, 2011 11:11 AM Subject: Re: [math-fun] spherical pendulums?
Surely there must be conditions more than simply "smooth".
For example, if the cross-section is smooth, but not convex, then the particle will leave the surface when the curvature becomes negative.
At 11:08 AM 12/6/2011, Eugene Salamin wrote:
The curved cylinder imposes a normal force on the point mass, and this does not affect the motion parallel to the surface. So indeed the particle moves in a parabola wrapped around the cylinder, no matter what the cross section, at least for smooth surfaces.
-- Gene
________________________________ From: Henry Baker <hbaker1@pipeline.com> To: Adam P Goucher <apgoucher@gmx.com> Cc: math-fun@mailman.xmission.com Sent: Tuesday, December 6, 2011 8:01 AM Subject: Re: [math-fun] spherical pendulums?
Hmmmm....
Does it matter if we wrap the transparent paper around a cylinder with circular cross section, or could the cross section be another shape -- e.g., an ellipse or something else?
Obviously, if the shape included some sort of cusp, it wouldn't stay on the shape.
Aren't we assuming that any accelerations are perpendicular to the shape of the cross section? But if the moving object is accelerating/decelerating along other dimensions, wouldn't that change things?
At 10:43 AM 12/5/2011, Adam P. Goucher wrote:
Here's a related problem:
A point particle of mass M (i.e., zero moment of inertia) is sliding frictionlessly around inside a vertical cylinder; gravity is downwards of strength G. What are the paths?
Imagine drawing a parabolic arc on a sheet of transparent paper, and wrapping it around the inside of the cylinder. This is parametrised by:
x = r cos(wt) y = r sin(wt) z = -½gt²
(w, r and g are constants, namely the angular speed, radius and gravitational field strength, respectively.)
Sincerely,
Adam P. Goucher
participants (2)
-
Eugene Salamin -
Henry Baker