Elevator Question (Rotation of rigid body and Energy)

[zewei1988]

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

 

[LiorE]

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.

 

[zewei1988]

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.

 

[jd120457]

You forgot to subtract the potential energy of the counterweight due to its lowering of "d"

 

[rwisz]

I would suggest approaching this problem by breaking it down into smaller parts and using the relevant equations for each part. Let's start with the elevator car and passengers.

We know that the car and passengers have a combined mass of 710 kg + 80n kg. They are moving upward at 3.23 m/s, so their initial kinetic energy is given by KEi = 0.5(710+80n)(3.23)^2 = 3632.8 + 131.2n^2.

Next, let's consider the counterweight. It has a mass of 810 kg and is moving at the same speed as the car and passengers, so its initial kinetic energy is KEi = 0.5(810)(3.23)^2 = 4163.3.

Now, let's look at the pulley. It has a mass of 280 kg and is rotating at a certain speed, which we can calculate using the equation w = v/R, where R is the radius of the pulley (0.700 m). The speed of the pulley is therefore given by w = (3.23)/0.700 = 4.614 rad/s. Using the equation KE = 0.5Iw^2, we can calculate the initial kinetic energy of the pulley as KEi = 0.5(0.5*280*0.700^2)(4.614)^2 = 1163.4.

Now, let's consider the potential energies of the system. The car and passengers are moving upward, so their initial gravitational potential energy is Ug = (710+80n)gh. The counterweight is moving downward, so its initial gravitational potential energy is Ug = (810)(-9.8)d, where d is the distance the car coasts upward. The pulley does not have any gravitational potential energy.

Now, we can set up the equation KEi + Ugi = KEf + Ugf and solve for d. This gives us:

3632.8 + 131.2n^2 + 4163.3 + 1163.4 + (710+80n)(9.8)d = 0 + 0 + 0 + (810)(9.8)(d)

Simplifying, we get:

(4956.1 + 131.2n^2 + 710d +