i've already done it in an earlier post...
√[L/g]*ln(2x+√[4x²-L²]) (+C)=t
C=-√[L/g]*ln(L) (by using x=L/2 and t=0)
then solving for x yields x=(L²+L²*e^(2t√[g/L]))/(4L*e^(t√[g/L])
x=L*(1+e^(2t√[g/L]))/(4e^(t√[g/L])), which equals (L/2)*cosh(t√[g/L])
a=xg/L
substitute...
my logic is that if i solve that integrated equation for x, then i can plug that in for a=xg/L, getting an equation that gives acceleration as a function of time
is that what you're trying to tell me to do? x gives the amount of chain hanging vertically, not the acceleration.
I'm plugging in the equation into http://integrals.wolfram.com/index.jsp
using what you have, Sqrt[L/g]*Sqrt[1/(x*x-L*L/4)]
integrates into √[L/g]*ln(2x+√[4x²-L²]) (+C)
i checked by taking the derivative of that, and it works.
am I making a mathematical mistake?
So if I take the square root of (dx/dt)^2=(4*x*x*g-g*L*L)/(4*L)
I get dx/dt=√[(4x²g-gL²)/(4L)]
dx*√[4L/(4x²g-gL²)]=dt , integrate
ln(2gx+g√[4x²-L²])/√[g] + C = t
test x=L/2, t=0 yields C=-ln(gL)/√[g]
solve for x
x=L(1+e^(2t√[g]))/(4*e^(t√[g]))
plug into a=xg/L...
then I'd get: .5[(dx/dt)^2]=x*x*g/(2*L)-g*L/8
(dx/dt)^2=x*x*g/L-g*L/4
(dx/dt)^2=(4*x*x*g-g*L*L)/(4*L)
move x so it's with dx
[(dx)^2]/(4*x*x*g-g*L*L)=4*L*[(dt)^2]
integrate
(dx)(-[tanh^-1[2*x/L]]/(2*g*L)=(dt)4*L*t
integrate again
-(x*[tanh^-1[2*x/L]]+(L/4)ln(L*L-4*x*x))/(2*g*L)=L*t*t/2...
sorry, I don't think I'm making any progress. I can't figure out when t comes into the answer.
I tried using F=m*dv/dx
m*a=m*dv/dx
a=dv/dx
x*g/L*dx=dv
integrate
x*x*g/(2*L)+C=v
C=-L*g/8
I can't use v=distance/time since that's avg velocity.
With energy, I tried
K+Ug=constant...
oops, my energy equation is wrong, the first term, to the left of the = should be negative. And the first term to the right of the = should also have *g.
dv/dt=dv/dx*dx/dt
So a=dv/dx*dx/dt.
Taking the differential of my energy equation, simplified, yields dv/dx=x/(L*v). Should I substitute...
Homework Statement
A chain lies on a frictionless table at rest, half off the edge, and half on. As soon as it is let go, it begins accelerating due to gravity only. Determine the acceleration of the chain as a function of time. The mass is m, gravity is g, and the length of the chain is L...