The easy way to do this is with conservation of energy, since you have to deal with an increasing acceleration rate, which would make the equation of motion rather messy. You can use energy conservation because it is frictionless.
When it starts out, the center of mass of the chain is (1/2)*(L/4) = L/8 below the table surface.
When 1/4 of the chain's length remains ion the table, the center of mass is
(1/2)(3L/4) = 3L/8 below the table surface. The change in CM height is L/4. At that time, the velocity V is given by
M g L/4 = (1/2) M V^2
g L = 2 V^2
V = sqrt(gL/2) = 2.96 m/s
A uniform chain of total mass m is laid out straight on a frictionless table and held stationary so that one-quarter of its length, L = 1.79 m, is hanging vertically over the edge of the table. The chain is then released. Determine the speed of the chain at the instant when only one-quarter of its length remains on the table.
3 answers
LEt total mass be m
force pulling: masshanging over *g
force puling: mg*x/1.79 where x is the amount hanging over.
well, then force pulling is linear with the length hanging over, so the average force between .25x and .75 x is at .5x
or avg force= mg/2
Vf^2=Vi^2+2ad
Vf^2=0+2(avgforce/m)(1/2 1.79)
Vf= sqrt (g*1.79/2)
check my thinking.
Vf= sqrt (2*
force pulling: masshanging over *g
force puling: mg*x/1.79 where x is the amount hanging over.
well, then force pulling is linear with the length hanging over, so the average force between .25x and .75 x is at .5x
or avg force= mg/2
Vf^2=Vi^2+2ad
Vf^2=0+2(avgforce/m)(1/2 1.79)
Vf= sqrt (g*1.79/2)
check my thinking.
Vf= sqrt (2*
ignore the last line,old typo. I agree with drwls, it is the same argument.