Minimum Time for Elevator to Reach Level B?

springwave
Messages
18
Reaction score
0

Homework Statement



An elevator of mass (m), is pulled upwards by a massless rope, which can bear a maximum tension of nmg (n-any natural number > 1). If the elevator starts from rest at point A, and it rises to level B where it comes to rest again.(momentary rest is sufficient).
Find the minimum time taken by the elevator in doing so. (tension may vary in different manners, but the initial conditions of rest must be satisfied)
Level b is at height (h) above levelA.

Homework Equations



ƩF = ma

KE1 + PE1 = KE2 + PE2

The Attempt at a Solution



Well, in my attempt, and also in the given solution, it is assumed that the minimum time case is when, till a certain height (y)<h, tension = nmg acts, and then suddenly falls to zero,
such that the elevator retards to rest by the time it reaches level B
.
If this assumption is made the solution the question is fairly easy, and can be found out using conservation of energy.

The solution turns out to be = √(2hn/((n-1)g))

But, I don't see any proper logical/ mathematical reason why the above case will yield minimum time. I'm guessing there should be other well defined functions of tension which varies with time and distance, which can yield even lesser time, though I'm unable to find
any.

I think it can be mathematically solved by minimizing the integral, using calculus of variations.
( which I unfortunately do not know)

It would be nice to know everyones views about this, and I would be grateful if anyone can find a case where the time is lesser, proving the given solution wrong, or give a proper logical reason to why the above case yields the minimum time.

I hope I will find some direction on this !
 
Physics news on Phys.org
springwave said:
I think it can be mathematically solved by minimizing the integral, using calculus of variations.
( which I unfortunately do not know)
There's no need to resort to calculus of variations here. Elementary calculus should be sufficient, I think. Essentially you have four constraints. You wish to find the smallest t_0 such that y(t_0) = h whilst y^{\prime\prime}(t) \leq (n-1)g for n\in\mathbb{N} with y(0) = 0.

The first step is to solve the boundary value problem. Can you do that?
 
Last edited:
Thanks for the quick reply!
Nope I'm unable to do that. There seem to be more constraints.

y'(0) = 0 & y'(t0)= 0

And additionally y''(t) can never be less than -g (since tension can't be -ve)

Not very familiar with boundary value problems, though I can solve differential equations.
Hope I can get some help with this.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top