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Elevator Question (Rotation of rigid body and Energy)

  • Thread starter zewei1988
  • Start date
  • #1
22
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Homework Statement


An elevator system in a tall building consists of a 710 kg car and a 810 kg counterweight, joined by a cable that passes over a pulley of mass 280 kg. The pulley, called a sheave, is a solid cylinder of radius 0.700 m turning on a horizontal axle. The cable has comparatively small mass and constant length. It does not slip on the sheave. The car and the counterweight move vertically, next to each other inside the same shaft. A number n of people, each of mass 80.0 kg, are riding in the elevator car, moving upward at 3.23 m/s and approaching the floor where the car should stop. As an energy-conservation measure, a computer disconnects the electric elevator motor at just the right moment so that the sheave-car-counterweight system then coasts freely without friction and comes to rest at the floor desired. There it is caught by a simple latch rather than by a massive brake.
(a) Determine the distance d the car coasts upward as a function of n.
d =


Homework Equations


I feel that the relevant equations are:
KEi + Ugi = KEf + Ugf
I = 0.5MR2
Ug = mgh
KE = 0.5Mv2
KE = 0.5Iw2
w = v/R

The Attempt at a Solution


I tried using KEi + Ugi = KEf + Ugf, and make d the subject of formula in terms of n.
KE (car & passengers) + KE (counter weigh) + KE (pulley) + Ug (counter weight) = Ug (car & passengers)
I ended up having (8459 + 417n)/ (784n - 980), which is wrong.
I'm not sure how does the counter weight moves with respect to the car, and I wasn't given any diagram.

Hope you guys can help me out. Thanks :D
 

Answers and Replies

  • #2
38
0
just before the motor is turned off, the energy in the system is the sum of: elevator K, counterweight K, elevator U, counterweight U, and the rotational kinetic energy of the pulley. After the elevator comes to rest, the energy in the system is the sum of elevator U and counterweight U after the elevator has risen a distance d and the counterweight dropped a distance d. What matters is only the change in potential energy, so you are free to take some reference point and call it zero height. Now there is 1 eqn with 1 variable, which is d.
 
  • #3
22
0
just before the motor is turned off, the energy in the system is the sum of: elevator K, counterweight K, elevator U, counterweight U, and the rotational kinetic energy of the pulley. After the elevator comes to rest, the energy in the system is the sum of elevator U and counterweight U after the elevator has risen a distance d and the counterweight dropped a distance d. What matters is only the change in potential energy, so you are free to take some reference point and call it zero height. Now there is 1 eqn with 1 variable, which is d.
I'm sorry, but isn't what you are trying to do the same as what I did? I tried doing it using your method but I arrive at the same answer as I had gotten earlier.
 
  • #4
1
0
You forgot to subtract the potential energy of the counterweight due to its lowering of "d"
 

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