Using KE in elevator-system to design brakes?

[MartinLoland]

Homework Statement

Hello! I am doing some rough design on an elevator-system and I have now come to the brake. Here is my system:

The text is norwegian but "r_driv" in the first image is "R_drivskive" in the second image and it's this wheel that is dragging the rope.

The "Brems" is where I am going to put the brake. And I am trying to figure out how much torque it must deliver to stop the elevator in time.

Required stopping time: 2 seconds
Velocity of elevator: 5 m/s
Angular velocity of drive-shaft ("Drivaksling") before braking: 43,63 rad/s

Homework Equations

(1) T = I*α
(2) Work = T*θ
(3) KE_θ = 1/2*I*ω²
(4) KE_x = 1/2*m*v²

The Attempt at a Solution

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This means that the driveshaft must have an angular acceleration of -23,3 rad/s^2 and travels about 40 rad in the two seconds it uses to brake.

If I only consider the driveshaft I could have used T=I⋅α where I is the moment of inertia of the driveshaft to calculate the required torque.

But there is a lot of energy in the system like vertical motion of the elevator and rotational energy in the gears that I guess that I have to take into consideration.

So here is my thought
If I calculate and add all the energy in the system (KE_θ of gears and shafts and KE_x of elevator and counterweight. Can I use eqn. (2) to calulate the torque since I have the θ it travels and all the energy I have to remove from the system? Or is this only for constant velocity?

Thanks :)

 

[Andrew Mason]

Required stopping time: 2 seconds
Velocity of elevator: 5 m/s
Angular velocity of drive-shaft ("Drivaksling") before braking: 43,63 rad/s

Homework Equations

(1) T = I*α
(2) Work = T*θ
(3) KE_θ = 1/2*I*ω²
(4) KE_x = 1/2*m*v²
So here is my thought
If I calculate and add all the energy in the system (KE_θ of gears and shafts and KE_x of elevator and counterweight. Can I use eqn. (2) to calulate the torque since I have the θ it travels and all the energy I have to remove from the system? Or is this only for constant velocity?

Thanks :)

BTW, welcome to PF Martin!

You have to take into account the change in gravitational potential energy as well. Since the counterweight is going in the opposite direction to the elevator, the change in potential energy depends on the difference in mass between the counterweight and the elevator. I assume you will design it for a load that results in the maximum difference in mass. Depending on which direction it is going, that difference in mass may assist the braking or may add to the braking requirement.

So, in order to stop the elevator a brake must apply a torque such that :

$\tau\theta = \Delta E = \Delta PE + KE$

where KE is the kinetic energy of the entire system including cables, pulleys, counterweight and elevator and $\Delta PE$ is the change in potential energy during the braking which is a function of the stopping distance.

You want the elevator to slow to a stop with a constant loss of speed per unit time - constant torque and constant deceleration.

In order to do that in 2 seconds you would be able to calculate $\Delta\theta$ ie. $\Delta\theta = \int \omega dt = \int (\omega_0- \alpha t)dt = \omega_0 t - \frac{1}{2}\alpha t^2$

where $\alpha = \Delta \omega/\Delta t = \omega_0/t$ which means that $\Delta\theta = \frac{1}{2}\omega_0 t$

$\tau\Delta\theta = \tau\frac{1}{2}\omega_0 t = \Delta PE + \Delta KE = \Delta E$

so: $\tau = 2\Delta E/\omega_0 t$

[Note: my previous attempt was incorrect. I was confusing $\dot\theta \text{ with } \dot\omega$. Sorry about that. I think the above is correct.]

AM

 

[MartinLoland]

Thank you for your thorugh answer, it really helps! I tried to solve it in two ways; using forces and energy and tried to compare the results. I am posting them here as they may be to help for others.

Some assumptions:

1. Constant deceleration and torque
   
   
2. The elevator and counterweight is held by 6 ropes each. All the ropes that hold the elevator has the same tension S₁. S₂ for counterweight.
3. My calculations is for when the elevator is going down at maximum load and v₀ = 5 m/s.
4. I am neglecting all mass in ropes and gears, only thinking about the mass in elevator, counterweight and flywheel.

Result
As you can see from my calculations I get somewhat the same answer, but that is only if I use positive sign on the ΔPE for the system, though it actually is negative (the elevator with the most mass is lower than before). But I thought that since this loss in energy has to be converted to heat in the brakess (or is it?) it contribute to the torque the brakes has to make. OR maybye I made a mistake somewhere else?