A 30 ft chain lies straight out on a a smooth table 10ft. above the floor. The c

Question

A 30 ft chain lies straight out on a a smooth table 10ft. above the floor. The chain is allowed to slide off the table and then held in place so that one end of the chain just touches the floor. The chain is then released. Find the velocity of the chain at the instant the chain leaves the table. Assume the chain has a linear density of rho pounds per foot. Also assume acceleration due to gravity is g=32 feet per second squared. Assume that once a part of the chain touches the floor it is not moving anymore. When we say the table is smooth, we assume that it is frictionless. Leave your answers in terms of rho

Explanation / Answer

total initial energy=initial potential energy+initial kinetic energy

as initial speed=0, initial kinetic energy=0

20 ft of chain lies on the table i.e at a height of 10 ft.

mass of this length=20*pho

then potential energy of this portion=mass*g*height

=20*pho*32*10=6400*pho Joules

for the 10 ft length which is vertical,

center of mass will be at a height of 5 ft from ground.

mass=10*pho

so potential energy=10*pho*32*5=1600*pho Joules

so total initial energy=6400*pho+1600*pho=8000*pho Joules

at the instant when the chain leaves the table, only 10 ft of the length is vertical and rest of the length is at rest o ground.

so if speed is v,

then total energy=potential eergy+kinetic energy

=10*pho*32*5+0.5*(10*pho)*v^2

=1600*pho+5*pho*v^2

as energy is conserved,

equating initial and final energy values:

8000*pho=1600*pho+5*pho*v^2

==>6400*pho=5*pho*v^2

==>v=sqrt(6400/5)=35.777 ft/sec

speed of the chain when it leaves the table is 35.777 ft/sec

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