| Topic: |
Science > Physics |
| User: |
"mjb" |
| Date: |
09 Jun 2005 01:15:04 AM |
| Object: |
Rigid Body Mechanics Question |
As I understand it, if I apply an impulse (J) to a mass (M) then I will get
a velocity change (dv) given by:
dv = J/M
But what if the impulse is not applied in line with the centre of mass?
It seems to me that the movement, at the point that the impulse is applied,
will be:
dv = J/M + linear velocity change due to rotation change
I worked out that this additional velocity change is given by:
(r x J) = [I](dv x r)
where:
r = position of impulse relative to cm
J = impulse vector
x = cross product
[I] = inertia tensor
dv = linear velocity change due to rotation change
I derived this by integrating:
torque = r x F = [I] alpha
So my question is, is this equation (r x J) = [I](dv x r) correct?
If so how can I solve for dv ?
What I am trying to get to is an equation for the change in linear velocity
of a point on a solid body due to an impulse at that point.
Thanks,
Martin
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| User: "PD" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 02:47:40 AM |
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mjb wrote:
As I understand it, if I apply an impulse (J) to a mass (M) then I will get
a velocity change (dv) given by:
dv = J/M
But what if the impulse is not applied in line with the centre of mass?
It seems to me that the movement, at the point that the impulse is applied,
will be:
dv = J/M + linear velocity change due to rotation change
No, it's not additive. Linear momemtum and angular momentum are
conserved *separately*
so in your notation
dv = J/M
AND
dw = (r x J)/I (you have a math error somewhere. check dimensions.)
where w = angular velocity.
It *might* be that there is a coupling between v and w, if for example
it's rolling, in which case this becomes
dv/r = (r x J)/I
And if this is the case, you can solve the simultaneous equations.
I worked out that this additional velocity change is given by:
(r x J) = [I](dv x r)
where:
r = position of impulse relative to cm
J = impulse vector
x = cross product
[I] = inertia tensor
dv = linear velocity change due to rotation change
I derived this by integrating:
torque = r x F = [I] alpha
So my question is, is this equation (r x J) = [I](dv x r) correct?
If so how can I solve for dv ?
What I am trying to get to is an equation for the change in linear velocity
of a point on a solid body due to an impulse at that point.
Thanks,
Martin
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| User: "mjb" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 04:08:38 AM |
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I realise that linear momentum and angular momentum are conserved separately
but I was speculating that the linear velocities of a point due to the
linear and rotation could be added.
I was not proposing any constrains, if the body is free floating and if the
impulse is applied off-centre then I imagine that there will be an equal
and opposite impulse acting on the centre-of-mass due to inertia and since
these impulses are not inline then this would produce an impulse equivalent
of torque?
It just seems to me that this situation is so simple it must be possible to
derive an equation for the total velocity change due to an impulse but I
still cant think a way to proceed?
Thanks,
Martin
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| User: "" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 02:22:51 PM |
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In article <11ag1fchaq5s5b6@corp.supernews.com>, mjb <mjb4567@yahoo.com> writes:
I realise that linear momentum and angular momentum are conserved separately
but I was speculating that the linear velocities of a point due to the
linear and rotation could be added.
Yes, certainly. For an extended body, the local velocity at a
specific point is given by
V = V_cm + w x r
where V_cm is the velocity of the center of mass, w is the angular
velocity of rotation around the CM and r is the location (relative
to the CM). the "x" signifies a cross-product.
I was not proposing any constrains, if the body is free floating and if the
impulse is applied off-centre then I imagine that there will be an equal
and opposite impulse acting on the centre-of-mass due to inertia and since
these impulses are not inline then this would produce an impulse equivalent
of torque?
Sure, but the easiest is not to think in terms of torque but in terms
of linear and angular momenta.
It just seems to me that this situation is so simple it must be possible to
derive an equation for the total velocity change due to an impulse but I
still cant think a way to proceed?
The impulse gives you a momentum change. (r x impulse) gives you
angular momentum change. Note that for the second you need to know
where on the body is the impulse applied.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "PD" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 04:37:40 AM |
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mjb wrote:
I realise that linear momentum and angular momentum are conserved separately
but I was speculating that the linear velocities of a point due to the
linear and rotation could be added.
No, you can't. Classic problem is to find the height of a billiard rail
cushion in terms of the ball radius R, so that when the ball rolls into
the rail, it recoils without slipping.
I was not proposing any constrains, if the body is free floating and if the
impulse is applied off-centre then I imagine that there will be an equal
and opposite impulse acting on the centre-of-mass due to inertia and since
these impulses are not inline then this would produce an impulse equivalent
of torque?
It just seems to me that this situation is so simple it must be possible to
derive an equation for the total velocity change due to an impulse but I
still cant think a way to proceed?
Thanks,
Martin
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| User: "mjb" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 09:31:49 AM |
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PD wrote:
I realise that linear momentum and angular momentum are conserved
separately but I was speculating that the linear velocities of a point
due to the linear and rotation could be added.
No, you can't. Classic problem is to find the height of a billiard rail
cushion in terms of the ball radius R, so that when the ball rolls into
the rail, it recoils without slipping.
I think perhaps I was not explaining properly? All I was saying is that:
vp = vcm + w x r
where:
vp = velocity of particle on solid body
vcm = velocty of centre of mass
w = angular velocity
r = position of particle relative to centre of mass
In other words there is a part due the linear velocity of the centre of mass
and a part due to the particles rotation around the centre of mass.
Martin
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| User: "PD" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 09:43:30 AM |
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mjb wrote:
PD wrote:
I realise that linear momentum and angular momentum are conserved
separately but I was speculating that the linear velocities of a point
due to the linear and rotation could be added.
No, you can't. Classic problem is to find the height of a billiard rail
cushion in terms of the ball radius R, so that when the ball rolls into
the rail, it recoils without slipping.
I think perhaps I was not explaining properly? All I was saying is that:
vp = vcm + w x r
where:
vp = velocity of particle on solid body
vcm = velocty of centre of mass
w = angular velocity
r = position of particle relative to centre of mass
In other words there is a part due the linear velocity of the centre of mass
and a part due to the particles rotation around the centre of mass.
Martin
That's fine, because you don't have a connection between vcm and w yet.
vcm is determined by the linear impulse
w is determined by the angular impulse
Yeah, it's fine, I think.
PD
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| User: "mjb" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 10:19:01 AM |
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That's fine, because you don't have a connection between vcm and w yet.
vcm is determined by the linear impulse
w is determined by the angular impulse
Yeah, it's fine, I think.
What I am saying is that the linear impulse also generates an angular
impulse because the equal and opposite reaction (due to the mass) is at the
centre of mass which is offset from the point that the impulse is applied
so we have a turning impulse.
So if the linear impulse and angular impulse are caused by the same impulse
we must be able to work out an equation that combines them?
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| User: "Ron Verrall" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 01:31:59 PM |
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"mjb" <mjb4567@yahoo.com> wrote in message
news:11agn5p4tgp201@corp.supernews.com...
That's fine, because you don't have a connection between vcm and w yet.
vcm is determined by the linear impulse
w is determined by the angular impulse
Yeah, it's fine, I think.
What I am saying is that the linear impulse also generates an angular
impulse because the equal and opposite reaction (due to the mass) is at
the
centre of mass which is offset from the point that the impulse is applied
so we have a turning impulse.
So if the linear impulse and angular impulse are caused by the same
impulse
we must be able to work out an equation that combines them?
Martin:
You are quite right that the total motion of the solid body can be split
into the linear motion (of the centre of mass) and the angular rotation
(about the centre of mass).
The motion of the centre of mass is given simply by:
F = ma, where m is the total mass and F is the net external force.
Equivalently, the change in momentum of the centre of mass is given by the
impulse delivered by the force.
The angular velocity (about the centre of mass) is given by
Torque = rate of change of angular momentum
In simple problems, where you can think of the moment of inertia as a
number, rather than a tensor, you can express this as:
Torque (about the centre of mass) = moment of inertia (about the center of
mass) times the rate of change of angular velocity. Or, the change in
angular velocity is given by the angular impulse divided by the moment of
inertia.
The velocity of any particle in the body can be determined by a vector
addition of the two velocities - one due to the linear motion of the centre
of mass and the other due to the rotation of the body about the centre of
mass.
Simple problems associated with this include finding the "sweet spot" of a
billiard ball. This the height that a cue would strike the cue ball so that
the ball rolls without sliding on the table. It is the same height as the
cushion - as mentioned by PD.
A similar problem is finding the sweet spot of a baseball bat, and this is
much easier if you assume the bat is of uniform cross-section, where the
moment of inertia I = mL^2/12
Let's assume the ball hits the bat at a distance 'a' away from the centre of
the bat (i.e., from the centre of mass).
Assume the impulse is 'p'.
The velocity given to the centre of mass is
v1 = p/m
The angular velocity around the centre of mass is the angular impulse
divided by the moment of inertia.
w = pa/I = 12 pa/mL^2
The velocity at the end where your hand is holding the bat (due to this
rotation) is
v2 = -w L/2 = -6pa/mL
For the total velocity to be zero (i.e., no stinging hands) we must have
v1 + v2 = 0 = p/m - 6pa/mL
Which leads to
a = L/6
Since a is the distance from the center, the distance from your hands is L/2
+ L/6 = L/3. This is the location of the sweet spot for a very simple bat.
Ron Verrall
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| User: "PD" |
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| Title: Re: Rigid Body Mechanics Question |
09 Jun 2005 10:53:46 AM |
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mjb wrote:
That's fine, because you don't have a connection between vcm and w yet.
vcm is determined by the linear impulse
w is determined by the angular impulse
Yeah, it's fine, I think.
What I am saying is that the linear impulse also generates an angular
impulse because the equal and opposite reaction (due to the mass) is at the
centre of mass which is offset from the point that the impulse is applied
so we have a turning impulse.
So if the linear impulse and angular impulse are caused by the same impulse
we must be able to work out an equation that combines them?
Yes. The connection, as you noted, is that the angular impulse is r x
J, where r is the application point of the impulse.
It is still true that momentum conservation laws are conserved
separately, but you do have the connection between the two impulses.
PD
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| User: "mjb" |
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| Title: Re: Rigid Body Mechanics Question |
10 Jun 2005 03:01:40 AM |
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OK, thanks, I think I'm getting there.
Taking Rons suggestion to assume object is symmetrical so 'I' can be a
scalar, so,
Vp = Vcm + w x r
deltaVp = deltaVcm + deltaw x r
deltaVp = J/M + ((J x r) x r)/I
using triple product identity gives:
deltaVp = J/M + (-J(r.r) + r(J.r))/I
deltaVp = J( 1/M -(r.r)/I) + r(J.r)/I
This looks about right in terms of units although I think a sign has gone
wrong somewhere?
However, I would still like to get a more general solution for 3x3 inertia
tensor, would this be right?
deltaVp = J/M + [I^-1](-J(r.r) + r(J.r))
Martin
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