Conference rusure::math

I'll give this a try:

Potential energy + kinetic energy = K (constant)
    
     i=Number of particles in body
Let ---
    \   M(i) L(i)^2   = K1   (represents Moment of inertia about rotaion axis)
    /    
    ---
    i=1
where M(i)= mass of particle i, L(i)= distance of particle i from axis of
rotation.

     i=Number of particles in body
Let ---
    \   M(i) L(i)   = K2   (to use with potional energy calcualtions)
    /    
    ---
    i=1
    
                                                  .
so  g(K2 cos(phi)) + (1/2) Moment of inertia * ( w )^2
       . 
where  w = angular velocity = d(phi)/dt

but Moment of inertia about the axis of rotation= K1

so: g k2 cos(phi) +1/2 K1 (d(phi)/dt)^2 = K

 d(phi)/dt = SQRT (   (k- g k2 cos(phi)) /  (1/2  K1) )

integrate and solve for phi, the integration constants found from
initial conditions, phi=0 at t=0.

/naser   
    

    In this problem, you should take care regarding whether the dancer's feet
    slide, or stay in the same spot.

.0 is not considering another common case, where the dancer has a significant
linear momentum.  This can translate to a large angular momentum around the
fixed point, whether it is the floor or some part of another dancer.

    ref .-1
    if you have a rod of length L moving at constant speed V from left to
    right, at one point it starts to tilt to the right at the top only
    while the lower point still in contact with the floor and moving
    at the same speed V. so in this case it is the same as if the rod
    was not moving, since the speed is uniform, it is equivelant to
    stationary rod (right ?).
    now if the rod had an accelaration ACC from left to right also, then
    and the rod started to tilt from the top to the left, Then the linear
    accelaration will have an effect of reducing the angular velosity
    of the top point of the rod, since both are movin from left to right ,
    but if the accelaration was from right to left, and the top of
    the rod tilted to the right, then the fall will be in less time,
    (right?)
    /naser

    ref .-1 
    for whateve its is worth, I meant to say
    
>    now if the rod had an accelaration ACC from left to right also, then
>    and the rod started to tilt from the top to the left, Then the linear
                                                     ^^^^
                                               should be "right"
    

>    if you have a rod of length L moving at constant speed V from left to
>    right, at one point it starts to tilt to the right at the top only
>    while the lower point still in contact with the floor and moving
>    at the same speed V. so in this case it is the same as if the rod

Right, for your case.  But the case I was thinking of is where the lower point,
in contact with the floor, becomes stationary.  Ouch!

	Let's simplify the problem slightly, to show the underlying
structure better.   I'll imagine that I have a uniform pencil, so the
centre of gravity is in the centre and the moment of inertia about one
end will be 4ml�/3 (m being the weight and 2l the length).   Then we can
track the pencil until the moment when it slips.   The key parameter is
then the coefficient of friction, �.

	Result: let c be the cosine of the angle that the pencil makes with the
vertical.   c begins as 1, and decreases as the pencil falls.   

	If � < 15*sqrt(10)/128 then the pencil will slip backwards, at some
point with 9/11 < c < 1.

	If � > 15*sqrt(10)/128 then the pencil will slip forwards, at some
point with 1/3 < c < (48*sqrt(14) - 35)/231.

	To me, the surprising feature of this puzzle is that such numbers pop
up from a setting in which they don't seem to be inherent.

	To solve the more general setting, you have only to change the centre
of gravity and the moment of inertia.   I think that the same qualitative
behaviour will hold.

-------------------------------------------------------------------------------

Proof:

	Let W be the weight, N the normal reactive force where the pencil
touches the surface, and F the friction force.   Let a be the angle with the
vertical, and a' & a" the first and second derivatives of a with respect to
time.

radial force:		(W-N)cos(a) - Fsin(a) = ml(a')�
tangential force:	(W-N)sin(a) + Fcos(a) = mla"
moments about 1 end:	Wlsin(a) = (4/3)ml�a"

	This summarizes what we know about the system.   Now take the moment
equation, and integrate with respect to a:

			W(1-cos(a)) = (2/3)ml(a')�

	We can now get rid of a' & a" in the force equations, to give:

			W((5/2)cos(a) - 3/2) = Ncos(a) + Fsin(a)
			W(1/4)sin(a)	     = Nsin(a) - Fcos(a)

	Combining to remove W, and substituting kN for F, we get:

			|k| = 3 * sqrt(1-c�) * (3c-2) / (3c-1)�

	The pencil will slip when |k| exceeds �, and graphing the above
equation will show you why there are two regions where the pencil might slip,
and a safe zone in between.

-------------------------------------------------------------------------------

	The original question was looking for the behaviour in time.   To get
this, we return to the equation:

			W(1-cos(a)) = (2/3)ml(a')�

whence:			dt = b/sqrt((1-c)/2).da
	[b = sqrt(ml/3W)]

cos(a) = 2cos�(a/2) - 1

sqrt((1-c/2) = sqrt(1-cos�(a/2)) = sin(a/2)

			dt = b/sin(a/2).da

			A*exp(t/(2b)) = csc(a/2) - cot(a/2)

Rearranging:		sin(a/2) = 2A�exp(t/b)/(1+A�exp(t/b))

At t = 0, sin(a/2) = epsilon, which allows us to get a value for the constant
of integration A.

	To reiterate, the general problem won't be any more complicated than
this, it will just have different values for the location of the centre of
gravity, and for the moment of inertia.

Regards,
Andrew.

	First, one clarification.   In my reply, I was explicitly assuming that
the foot of the pencil was stationary.

	Next, I want to return to the original note, and look at the balletic
mechanic's workings.   I translate his notation to mine, and restrict myself
to the uniform pencil.
    
>    Think of a pencil perfectly balanced on end on your perfectly
>    motionless desk (with absolutely no air currents).  It won't fall
>    over.  Now carefully move the top of the pencil such that the pencil's 
>    long axis forms an angle theta[0] with vertical.  Given t (time), what
>    is theta?
 
        sin(Theta[0]/2) = epsilon
   
>    Imagine the body as an idealized stick of length L, somewhat heavier at
>    the upper end than the lower.  This stick can be lanaced vertically on
>    the floor.  If it displaced from the vertical by a small initial angle
>    theta[0] it will start toppling, and the angle theta will increase at
>    an accelerating rate. 

        I haven't come across the word "lanacing" before.   Is it balletic?

        L = 2l.
    
>    He gives the equation
>    
>        mgR[c]sin(theta) = I(alpha) = m(R[g])**2(alpha)
>    
>    Where
>      
>       m = mass of of the body
>       g = acceleration due to gravity
>       R[c] = the distance from the point of support (I assume the floor)
>          to the center of gravity.
>       I = moment of inertia of the body toppling around an axis thru
>         the point of support (I assume the point is where the idealized body
>         - or stick - touches the floor)
>       R[g] = radius of gyration  = sqrt(I/m)
>       alpha = toppling  angular accelaration
 
	This is just taking a moment around the foot of the pencil.

		Wlsin(a) = (4/3)ml�a"
   
>    Since theta is small, he substitutes theta for sin(theta).  

	This seems a bit feeble.   theta[0] is small, but my calculus
of -.1 did not need to assume that theta remains small.   My sums will remain
valid until the pencil slips, which could be as late as c = 1/3.

>	Ok - so far I follow.  But then he waves his hands and says
>    
>       theta = theta[0] cosh(Kt)
>    
>    where 
>    
>        theta[0] is the initial angular displacement
>        K = sqrt(gR[c])/R[g]
>    
>    What is the differential equation he uses and solves to get this?
    
	Basically, he's solving the equation:

	x = k�x", where k is a real number.   This has the solution
x = Acosh(kt) + Bsinh(kt), and boundary conditions when t = 0 give values for
A & B.   He assumes that at t=0, theta is theta[0], and the pencil is
initially stationary.

Regards,
Andrew.

re .0:

>    What is the differential equation he uses and solves to get this?

:	=	2nd derivative wrt time

                        :
Cancel m and substitute theta for alpha:
         :
R[g]^2 * theta = g*R[c] * sin( theta )


Substitute theta for sin( theta ) and divide through by R[g]^2:
:
theta = ( g*R[c]/R[g]^2 ) * theta


Bring theta term to left:
:
theta - ( g*R[c]/R[g]^2 ) * theta  =  0


Call the coefficient of theta 'k'.

Goto PHYSICS notesfile # 159.1 and compare this diff eqn with the bead
equation making the substitutions:

	theta		<->	r
	theta[0]	<->	R
	k		<->	W1

	Hmmm, I want to revisit the time equation more carefully.

	Stripping away all the camouflage, we have:

		x" = k�.sin(x)

where x = x(t), x(0) = p << 1, x'(0) = 0.

	I integrate both sides with respect to x.   Note that int(x".dx) is
�(x')�.   Thus:

		�(x')� = k�.(cos(p)-cos(x))

	Now since p is arbitrarily tiny, it's reasonable that we take cos(p)
to be 1.   However, we couldn't do the same for x.   [Note the painful 
spectacle of a pure mathematician agonizing over each approximation that he
makes.]

=>		cos(x) = cos�(x/2) - sin�(x/2)

=>		(x')� = 4k�.sin�(x/2)

	Now, we can take the sqrt of both sides, keeping the *positive* roots.

=>		x' = 2k.sin(x/2)

=>		int(csc(x/2).dx) = int(2k.dt)

	Now, the lhs = 2ln(csc(x/2) - cot(x/2))

=>		csc(x/2) - cot(x/2) = A.e^(kt)

where A is a constant of integration.   

		A = csc(p/2) - cot(p/2)

		  =ish (1-1+(p/2)�/2)/(p/2)

		  = p/8

	Now, to get it explicit in x:

	Let c = cos(x/2)

		(1-c)�/(1-c�) = p�.e^(2kt)/64 = K

		(1-K) - 2c + (1+K)c� = 0

		c = (1 � K)/(1 + K)

	c = 1 corresponds to the solution where the pencil is in unstable
equilibrium (p = 0), but the real solution is:

		x = 2.arccos( (1-K)/(1+K) )

where K = p�.e^(2kt)/64.

Regards,
Andrew.