4. A chain of length l is held stretched out on a
frictionless horizontal table, with a length x hanging
down through a hole in the table. The chain is re-
leased. As a function of time, find the length that
hangs down through the hole (don't bother with t
after the chain loses contact with the table). Also,
find the speed of the chain right when it loses contact
with the table.
1Let σ = mass density of rope
Now, momentum of rope= σL dx/dt
F=dp/dt = σL d2x/dt2 = (σx)g, as gravitational force is responsible for change in momentum.
d2x/dt2 = gx/L
On solving the differential,
x(t)= Ae^(t √(g/L) ) + Be^(-t √(g/L) )
1Finding the velocity at the end is pretty simple.
KE at end=mv2/2 = (σL) (v2) / 2
KE gained=loss in PE
PE loss= mgh = (σL) (g) (L/2)
Equating KE and PE expressions,
v= √(Lg)
1concept: force generated due to variable mass
F_r=v_r\frac{dm}{dt} \\ m=\sigma x\\ \frac{dm}{dt}=\sigma\frac{dx}{dt}\\ F_r=\sigma v_r^2\\ F_r=\sigma (2gx)
F_{net}=\sigma g (x+X) -F_r \\ \sigma x \frac{dv}{dx}=\sigma g(X-x)
now integrate with proper limits to get
v(x)
and then write
v(x)=dxdt
and again integrate to get
x as a function of time