| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
15 Mar 2005 08:20:24 PM |
| Object: |
Rolling rigid object |
Hi,
I'm trying to study the motion of a rigid object on a rail. I have a
very basic question in statics. Suppose, a ball is rolling on a rail (
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by equating to
the weight. The horizontal components are also equal and opposite. But
is there any way to find the value for the horizontal component?
I will highly appreciate an answer to this question.
Hank you,
Priya
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| User: "David Cross" |
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| Title: Re: Rolling rigid object |
15 Mar 2005 08:31:02 PM |
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<priya2@umbc.edu> wrote in message
news:1110939624.072942.194770@o13g2000cwo.googlegroups.com...
Hi,
I'm trying to study the motion of a rigid object on a rail. I have a
very basic question in statics. Suppose, a ball is rolling on a rail (
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by equating to
the weight. The horizontal components are also equal and opposite. But
is there any way to find the value for the horizontal component?
If it's just sitting there, or moving with constant velocity, you can use the
fact that the net torque is zero.
--
David Cross
dcross1 AT shaw DOT ca
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| User: "Franz Heymann" |
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| Title: Re: Rolling rigid object |
16 Mar 2005 04:56:59 AM |
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"David Cross" <nospam@spammenot.com> wrote in message
news:GTMZd.688450$Xk.332682@pd7tw3no...
<priya2@umbc.edu> wrote in message
news:1110939624.072942.194770@o13g2000cwo.googlegroups.com...
Hi,
I'm trying to study the motion of a rigid object on a rail. I have
a
very basic question in statics. Suppose, a ball is rolling on a
rail (
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by
equating to
the weight. The horizontal components are also equal and opposite.
But
is there any way to find the value for the horizontal component?
If it's just sitting there, or moving with constant velocity, you
can use the
fact that the net torque is zero.
How does that help?
{:-((
The truth of the matter is that the system is overdetermined, and the
actual numerical value of the pair of transverse horizontal
components will depend on the elastic properties of the Vee on which
the ball sits.
--
Franz
"One Galileo in 2000 years is enough."
Pope Pius XII
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| User: "David Cross" |
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| Title: Re: Rolling rigid object |
16 Mar 2005 02:09:20 PM |
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On Wed, 16 Mar 2005 10:56:59 +0000 (UTC), "Franz Heymann"
<notfranz.heymann@btopenworld.com> wrote:
If it's just sitting there, or moving with constant velocity, you
can use the
fact that the net torque is zero.
How does that help?
{:-((
The truth of the matter is that the system is overdetermined, and the
actual numerical value of the pair of transverse horizontal
components will depend on the elastic properties of the Vee on which
the ball sits.
I agree that I supplied only one of the constraints for a system which has
several variables. Often though, when one specifies at least one constrant it
is sometimes found how to get some of the others.
At least this is what I found when solving statics problems although it's been
a long time since I had to do any. :-)
---
David Cross
dcross1 AT shaw DOT ca
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| User: "PD" |
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| Title: Re: Rolling rigid object |
16 Mar 2005 10:36:02 AM |
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wrote:
Hi,
I'm trying to study the motion of a rigid object on a rail. I have a
very basic question in statics. Suppose, a ball is rolling on a rail
(
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by equating
to
the weight. The horizontal components are also equal and opposite.
But
is there any way to find the value for the horizontal component?
I will highly appreciate an answer to this question.
Hank you,
Priya
Define a different set of coordinates for convenience: along the
inclined rail and perpendicular to the rail.
There are four forces acting on the ball.
1. Gravity acting on the center of mass of the ball.
2. The component of the rail's contact force that is perpendicular to
plane of the rails.
3. The component of the rail's contact force that is parallel to the
rail, commonly known as static friction. It is static friction because
there is no slipping between the surfaces if the ball rolls.
4. The component of the rail's contact force that is perpendicular to
the rail and points toward the other rail.
(3) is constrained to be the coefficient of friction times the vector
sum of (2) and (4).
Because it's rolling, the rotational acceleration is coupled to the
linear acceleration.
Your other constraints are Newton's 2nd law for translation, in each of
the two directions specified in my first sentence, and Newton's 2nd law
for rotation.
Now you'll see you have just enough constraints for all your unknowns.
PD
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| User: "Franz Heymann" |
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| Title: Re: Rolling rigid object |
16 Mar 2005 03:36:05 PM |
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"PD" <pdraper@yahoo.com> wrote in message
news:1110990962.487599.218020@f14g2000cwb.googlegroups.com...
priya2@umbc.edu wrote:
Hi,
I'm trying to study the motion of a rigid object on a rail. I have
a
very basic question in statics. Suppose, a ball is rolling on a
rail
(
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by
equating
to
the weight. The horizontal components are also equal and opposite.
But
is there any way to find the value for the horizontal component?
I will highly appreciate an answer to this question.
Hank you,
Priya
Define a different set of coordinates for convenience: along the
inclined rail and perpendicular to the rail.
There are four forces acting on the ball.
1. Gravity acting on the center of mass of the ball.
2. The component of the rail's contact force that is perpendicular
to
plane of the rails.
3. The component of the rail's contact force that is parallel to the
rail, commonly known as static friction. It is static friction
because
there is no slipping between the surfaces if the ball rolls.
4. The component of the rail's contact force that is perpendicular
to
the rail and points toward the other rail.
(3) is constrained to be the coefficient of friction times the
vector
sum of (2) and (4).
Because it's rolling, the rotational acceleration is coupled to the
linear acceleration.
Your other constraints are Newton's 2nd law for translation, in each
of
the two directions specified in my first sentence, and Newton's 2nd
law
for rotation.
Now you'll see you have just enough constraints for all your
unknowns.
No you don't.
Do remember that the ball exerts two equal and opposed horizontal
force components on the rail, at different points. This will put a
stress on the rail, so something will have to be known about the
elastic properties of the rail if the story is to be sorted out fully.
Try it your way for the simple case in which the ball is stationary.
--
Franz
"One Galileo in 2000 years is enough."
Pope Pius XII
.
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| User: "Franz Heymann" |
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| Title: Re: Rolling rigid object |
17 Mar 2005 02:31:12 AM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:d1a8s5$ia0$3@titan.btinternet.com...
[snip]
Whatever I have said so far in this thread is quite irrelevant.
Please regard it all as unsaid, and please accept my apologies for
possibly misleading folk.
--
Franz
"One Galileo in 2000 years is enough."
Pope Pius XII
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| User: "PD" |
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| Title: Re: Rolling rigid object |
16 Mar 2005 03:57:11 PM |
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Franz Heymann wrote:
"PD" <pdraper@yahoo.com> wrote in message
news:1110990962.487599.218020@f14g2000cwb.googlegroups.com...
priya2@umbc.edu wrote:
Hi,
I'm trying to study the motion of a rigid object on a rail. I
have
a
very basic question in statics. Suppose, a ball is rolling on a
rail
(
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by
equating
to
the weight. The horizontal components are also equal and
opposite.
But
is there any way to find the value for the horizontal component?
I will highly appreciate an answer to this question.
Hank you,
Priya
Define a different set of coordinates for convenience: along the
inclined rail and perpendicular to the rail.
There are four forces acting on the ball.
1. Gravity acting on the center of mass of the ball.
2. The component of the rail's contact force that is perpendicular
to
plane of the rails.
3. The component of the rail's contact force that is parallel to
the
rail, commonly known as static friction. It is static friction
because
there is no slipping between the surfaces if the ball rolls.
4. The component of the rail's contact force that is perpendicular
to
the rail and points toward the other rail.
(3) is constrained to be the coefficient of friction times the
vector
sum of (2) and (4).
Because it's rolling, the rotational acceleration is coupled to the
linear acceleration.
Your other constraints are Newton's 2nd law for translation, in
each
of
the two directions specified in my first sentence, and Newton's 2nd
law
for rotation.
Now you'll see you have just enough constraints for all your
unknowns.
No you don't.
Do remember that the ball exerts two equal and opposed horizontal
force components on the rail, at different points. This will put a
stress on the rail, so something will have to be known about the
elastic properties of the rail if the story is to be sorted out
fully.
Try it your way for the simple case in which the ball is stationary.
Unlike the clearly impossible task of trying to hang a weight from a
string while trying to keep the string perfectly horizontal, I don't
see any obvious reasons why assuming the rails are arbitrarily stiff is
a poor assumption in this case. Assume the rails are horizontal for a
moment. The rails push inward and upward on the ball, toward the center
of the ball, in the plane consisting of the two contact points and the
ball's center. The upward components are what cancel the weight of the
ball, and the horizontal components are determined by the angle of the
contact point of the rails.
PD
--
Franz
"One Galileo in 2000 years is enough."
Pope Pius XII
.
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| User: "Pete" |
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| Title: Re: Rolling rigid object |
16 Mar 2005 08:50:39 PM |
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"PD" <pdraper@yahoo.com> wrote in message news:<1110990962.487599.218020@f14g2000cwb.googlegroups.com>...
priya2@umbc.edu wrote:
Hi,
I'm trying to study the motion of a rigid object on a rail. I have a
very basic question in statics. Suppose, a ball is rolling on a rail
(
with two point of contacts) what are the reaction forces ?. The
vertical component can be assumed to be equal and found by equating
to
the weight. The horizontal components are also equal and opposite.
But
is there any way to find the value for the horizontal component?
I will highly appreciate an answer to this question.
Hank you,
Priya
Define a different set of coordinates for convenience: along the
inclined rail and perpendicular to the rail.
There are four forces acting on the ball.
1. Gravity acting on the center of mass of the ball.
2. The component of the rail's contact force that is perpendicular to
plane of the rails.
3. The component of the rail's contact force that is parallel to the
rail, commonly known as static friction. It is static friction because
there is no slipping between the surfaces if the ball rolls.
4. The component of the rail's contact force that is perpendicular to
the rail and points toward the other rail.
(3) is constrained to be the coefficient of friction times the vector
sum of (2) and (4).
Because it's rolling, the rotational acceleration is coupled to the
linear acceleration.
Your other constraints are Newton's 2nd law for translation, in each of
the two directions specified in my first sentence, and Newton's 2nd law
for rotation.
Now you'll see you have just enough constraints for all your unknowns.
PD
Hi Priya,
PD's explanation is basically correct, although I think that
introducing coefficients of friction muddies the waters a bit.
The main point is, I think, that this is a problem in *dynamics*, not
statics. You can't equate vertical force components or horizontal
force components (or components along the slope, as PD suggests).
Remember, the ball is *accelerating* down the slope, so Newtons 2nd
law applies, as PD says. This means that gain in KE (translational KE
- half m v squared, plus rotational KE - half I omega squared) equals
loss in potential energy - mgh where h is the vertical height dropped.
The solution is trivial, although a bit messy and trigonometrical and
I can't be bothered to work it out now for three reasons:
1) It's 2.45 a.m my (UK) time
2) I can't remember what the expression for the moment of inertia (I)
of a solid uniform sphere about its centre is.
3) I've been drinking.
Cheers,
Pete
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