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| User: "James Stokes" |
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| Title: Re: Rigid rotations about multiple axes |
25 Sep 2003 10:53:17 PM |
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"Mel Lep" <mel_lep@yahoo.se> wrote:
"James Stokes" wrote:
Consider a sphere rolling down an inclined surface while being pressed
by
some arbitrary force against a side surface. The shear force generated
at
the interface of the sphere and each surface produces an angular
acceleration about two axes.
Looks similar to the case where a ball freely rolls down a \/-groove
(with a 90-deg apex angle).
Yes, except the \/ groove is turned 45 deg to produce an |_ shaped groove.
Show that the linear acceleration of the
sphere is given by (5*g*sin[theta])/(9*sqrt[2]). I have attempted this
a
number of times and continue to yield (5*g*sin[theta])/9. My answer is
in
agreement with energy conservation assuming the sphere rotates about two
axes with identical angular speed. What am I missing?
Thanks in advance.
James
Did you reproduce the problem text literally? (From what physics/mechanics
book is it?)
This is not a textbook question, it's actually part of a more complex
problem I'm attempting to tackle which involves a sphere rolling down a
helix-shaped incline with a side wall. If you consider an infinitesimal
section of the curve, it is clear where the |_ is derived from. The
equation (5*g*sin[theta])/(9*sqrt[2]) is quoted from a website whose address
I no longer possess. Although no derivation was provided, I am quite
certain this is the correct equation.
In the '\/-groove' case the expression for the CM acceleration coincides
with your answer.
That's correct, so why is the answer different for the case I'm describing?
Thanks in advance.
James
M.L.
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| User: "Mel Lep" |
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| Title: Re: Rigid rotations about multiple axes |
27 Sep 2003 12:20:33 PM |
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"James Stokes" wrote:
"Mel Lep" <mel_lep@yahoo.se> wrote:
"James Stokes" wrote:
Consider a sphere rolling down an inclined surface while being
pressed by some arbitrary force against a side surface. The shear
force generated at the interface of the sphere and each surface
produces an angularacceleration about two axes.
Looks similar to the case where a ball freely rolls down a \/-groove
(with a 90-deg apex angle).
Yes, except the \/ groove is turned 45 deg to produce an |_ shaped
groove.
Show that the linear acceleration of the sphere is given by
(5*g*sin[theta])/(9*sqrt[2]). I have attempted this a
number of times and continue to yield (5*g*sin[theta])/9.
My answer is in agreement with energy conservation assuming the
sphere rotates about two axes with identical angular speed.
Yes, the angular velocity (omega) is parallel to the instantaneous
axis of rotation, which passes through the two contact points (if
the sphere rolls without slipping):
.. \|..r.\ CM of sphere
.. |\ '\
.. | \ 'r \
.. |____\'___ \
.. \ \ omega
inst axis
Also, if the track is straight, the angular acceleration (alpha)
points in the omega-direction. Hence,
a = alpha * r/sqrt(2).
Let F1 and F2 be the contact forces. Using the torque eqn about
the CM axis we get
F1 * r/sqrt(2) + F2 * r/sqrt(2) = (2/5)mr^2 * alpha,
and, combining the eqns above,
F1 + F2 = (4/5)ma;
i.e. the sqrt(2) is gone. The force eqn gives
mg sin(theta) - F1 - F2 = ma,
so
mg sin(theta) = ma + (4/5)ma,
a = (5/9)g sin(theta).
What am I missing?
Nothing, I think.
[...]
This is not a textbook question, it's actually part of a more complex
problem I'm attempting to tackle which involves a sphere rolling down a
helix-shaped incline with a side wall. If you consider an infinitesimal
section of the curve, it is clear where the |_ is derived from. The
equation (5*g*sin[theta])/(9*sqrt[2]) is quoted from a website whose address
I no longer possess. Although no derivation was provided, I am quite
certain this is the correct equation.
In the '\/-groove' case the expression for the CM acceleration coincides
with your answer.
That's correct, so why is the answer different for the case I'm describing?
Maybe they fiddled?
M.L.
Thanks in advance.
James
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| User: "James Stokes" |
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| Title: Re: Rigid rotations about multiple axes |
27 Sep 2003 07:56:49 PM |
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"Mel Lep" <mel_lep@yahoo.se> wrote
[snip]
Yes, the angular velocity (omega) is parallel to the instantaneous
axis of rotation, which passes through the two contact points (if
the sphere rolls without slipping):
. \|..r.\ CM of sphere
. |\ '\
. | \ 'r \
. |____\'___ \
. \ \ omega
inst axis
Also, if the track is straight, the angular acceleration (alpha)
points in the omega-direction. Hence,
a = alpha * r/sqrt(2).
Let F1 and F2 be the contact forces. Using the torque eqn about
the CM axis we get
F1 * r/sqrt(2) + F2 * r/sqrt(2) = (2/5)mr^2 * alpha,
and, combining the eqns above,
F1 + F2 = (4/5)ma;
i.e. the sqrt(2) is gone. The force eqn gives
mg sin(theta) - F1 - F2 = ma,
so
mg sin(theta) = ma + (4/5)ma,
a = (5/9)g sin(theta).
What am I missing?
Nothing, I think.
I agree with this derivation. However, check out
http://www.xeq.se/labyrinth/design_algorithms.html#BounceCalc
This website considers a similar problem which suggests the presence of the
factor 1/sqrt[2]. Presumably one must consider the component of gravity
perpendicular to the rotation axis, however I'm not sure if that applies to
this situation.
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| User: "Mel Lep" |
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| Title: Re: Rigid rotations about multiple axes |
28 Sep 2003 05:14:36 AM |
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"James Stokes" wrote:
"Mel Lep" <mel_lep@yahoo.se> wrote
[snip]
Yes, the angular velocity (omega) is parallel to the instantaneous
axis of rotation, which passes through the two contact points (if
the sphere rolls without slipping):
. \|..r.\ CM of sphere
. |\ '\
. | \ 'r \
. |____\'___ \
. \ \ omega
inst axis
Also, if the track is straight, the angular acceleration (alpha)
points in the omega-direction. Hence,
a = alpha * r/sqrt(2).
Let F1 and F2 be the contact forces.
Better: Take the x-axis along the center of mass path, and let
F1 and F2 be the x-components of the contact forces.
Using the torque eqn about the CM axis we get
F1 * r/sqrt(2) + F2 * r/sqrt(2) = (2/5)mr^2 * alpha,
and, combining the eqns above,
F1 + F2 = (4/5)ma;
i.e. the sqrt(2) is gone. The force eqn gives
mg sin(theta) - F1 - F2 = ma,
so
mg sin(theta) = ma + (4/5)ma,
a = (5/9)g sin(theta).
What am I missing?
Nothing, I think.
I agree with this derivation. However, check out
http://www.xeq.se/labyrinth/design_algorithms.html#BounceCalc
This website considers a similar problem which suggests the presence of the
factor 1/sqrt[2]. Presumably one must consider the component of gravity
perpendicular to the rotation axis, however I'm not sure if that applies to
this situation.
Their "accelerating force" must still point in the x-direction,
i.e. that force should be the x-component of mg.
Click your link and go to the text around this picture:
http://www.xeq.se/labyrinth/images/algo_inclined2.gif
If the right wall serves as a constraint, the ball can't
penetrate it.
M.L.
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| User: "James Stokes" |
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| Title: Re: Rigid rotations about multiple axes |
28 Sep 2003 08:09:50 AM |
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"Mel Lep" <mel_lep@yahoo.se> wrote:
"James Stokes" wrote:
"Mel Lep" <mel_lep@yahoo.se> wrote
[snip]
Yes, the angular velocity (omega) is parallel to the instantaneous
axis of rotation, which passes through the two contact points (if
the sphere rolls without slipping):
. \|..r.\ CM of sphere
. |\ '\
. | \ 'r \
. |____\'___ \
. \ \ omega
inst axis
Also, if the track is straight, the angular acceleration (alpha)
points in the omega-direction. Hence,
a = alpha * r/sqrt(2).
Let F1 and F2 be the contact forces.
Better: Take the x-axis along the center of mass path, and let
F1 and F2 be the x-components of the contact forces.
Okay.
Using the torque eqn about the CM axis we get
F1 * r/sqrt(2) + F2 * r/sqrt(2) = (2/5)mr^2 * alpha,
and, combining the eqns above,
F1 + F2 = (4/5)ma;
i.e. the sqrt(2) is gone. The force eqn gives
mg sin(theta) - F1 - F2 = ma,
so
mg sin(theta) = ma + (4/5)ma,
a = (5/9)g sin(theta).
What am I missing?
Nothing, I think.
I agree with this derivation. However, check out
http://www.xeq.se/labyrinth/design_algorithms.html#BounceCalc
This website considers a similar problem which suggests the presence of
the
factor 1/sqrt[2]. Presumably one must consider the component of gravity
perpendicular to the rotation axis, however I'm not sure if that applies
to
this situation.
Their "accelerating force" must still point in the x-direction,
i.e. that force should be the x-component of mg.
Yes, but is the accelerating force the x-component of mg or the x-component
of mg/sqrt[2]? This website seems to suggest the latter. Why is it so?
Click your link and go to the text around this picture:
http://www.xeq.se/labyrinth/images/algo_inclined2.gif
If the right wall serves as a constraint, the ball can't
penetrate it.
M.L.
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| User: "James Stokes" |
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| Title: Re: Rigid rotations about multiple axes |
27 Sep 2003 09:17:31 AM |
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Stumped are we? Well, here's what I came up with.
The frictional force f at the interface of the sphere and the two surfaces
gives rise to an angular acceleration alpha about two axes.
If b is the radius of the sphere,
b*f = 2*I*alpha
the frictional force is thus
f = (2*I*alpha)/b
The total linear acceleration is now
a = g*Sin[theta] - (2*I*alpha)/b
= g*Sin[theta] - (2*I*(a/r))/b
which when suitably rearranged yields
a = (5*g*Sin[theta])/9
However this is not correct. Any ideas?
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