| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1320.1 | PE+KE= constant | SMAUG::ABBASI | | Sun Oct 28 1990 22:51 | 36 |
| 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
|
| 1320.2 | | TRACE::GILBERT | Ownership Obligates | Mon Oct 29 1990 20:11 | 2 |
| In this problem, you should take care regarding whether the dancer's feet
slide, or stay in the same spot.
|
| 1320.3 | bumping into reality | CSSE::NEILSEN | I used to be PULSAR::WALLY | Tue Oct 30 1990 14:48 | 3 |
| .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.
|
| 1320.4 | depending on directions | SMAUG::ABBASI | | Wed Oct 31 1990 01:49 | 15 |
| 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
|
| 1320.5 | typo | SMAUG::ABBASI | | Wed Oct 31 1990 01:52 | 8 |
| 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"
|
| 1320.6 | a slight clarification | CSSE::NEILSEN | I used to be PULSAR::WALLY | Wed Oct 31 1990 14:54 | 7 |
| > 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!
|
| 1320.7 | solution | HERON::BUCHANAN | combinatorial bomb squad | Fri Nov 02 1990 10:35 | 83 |
| 1320.8 | direct answer to the questions in .0 | HERON::BUCHANAN | combinatorial bomb squad | Fri Nov 02 1990 13:14 | 71 |
| 1320.9 | yet more | EAGLE1::BEST | R D Best, sys arch, I/O | Fri Nov 02 1990 13:24 | 30 |
| 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
|
| 1320.10 | is this right? | HERON::BUCHANAN | combinatorial bomb squad | Wed Nov 07 1990 09:39 | 59
|