course mth174

challenge physics problemA chain of uniform mass per unit length, with total length l, is loosely coiled on a table at height H above the floor. It starts to fall through a hole in the table.a. How long does it take before the chain first touches the floor?b. How long does it take before the whole chain has reached the floor?

Initially, the chain has no kinetic energy, but it has gravitational potential energy mgh=u1

once it falls through the hole its potential energy is converted into kinetic energy; it has both kinetic energy of the falling chain and uncoiling chain

K2= 1/2(m/L)v^2+ 1/2(mr^2)(omega^2)

we can use the relationship omega= v/r then

1/2(m/l)v^2+1/2(mr^2)(v^2/r^2)

1/2 (m/L)v^2 can't be a KE term; it has dimensions of energy / length.

Having fallen through the hole, assuming none of it is on the floor yet (i.e., L < H), its PE will be 1/2 m g * L/2, where m is its mass.

However it's what happens between the initial and the final state that determines how long it takes to fall.

I think the appropriate assumption is that a link of the chain on the table remains stationary until it falls through the hole. So there would be no angular KE involved. In any case the chain can't be falling and rotating at the same time; and in any case no radius is given or implied.

giving the relationship

mgh= 1/2(m/l)v^2+mv^2

we can solve for v= sqrt( gh/ (1+1/2L)

(apparently, the radius has no significance here because it cancels out, I just assumed that since the chain was in coil, which is circular.

the velocity function is v(t)= g(t)+ sqrt(gh/(1+1/2L)

The PE loss depends on the height y of the end of the chain above the floor. Thus your velocity function, and hence the right-hand side of your differential equation, will have some y dependence.

dy/dt= gt+sqrt(gh/(1+1/2L)

then set dy=(dt)*gt+sqrt(gh/(1+1/2L)

y(t)= 1/2gt^2+2/3*t*(gh/(1+1/2L)^3/2

hopefully this makes sense, it seems like the is plausible. If we assume that this is correct, then t can be found by using the quadratic equation if we had any actual numbers). However, I am a little confused on finding how long it will be before the whole entire chain hits the ground.

The PE loss depends on the height y of the end of the chain above the floor. Thus your velocity function, and hence the right-hand side of your differential equation, will have some y dependence.

Not bad, but not quite on the right track yet. See my notes and give it another shot.

Not bad, but not quite on the right track yet.

See my notes and give it another shot.