Low Jerk Elevator Ride: max speed & acceleration

[PFuser1232]

Homework Statement

For a smooth (“low jerk”) ride, an elevator is programmed to start from rest and accelerate according to

$$a(t) = \frac{a_m}{2}[1 − \cos{\frac{2\pi t}{T}}] \:\:\:\:0 ≤ t ≤ T$$

$$a(t) = -\frac{a_m}{2}[1 − \cos{\frac{2\pi t}{T}}] \:\:\:\:T ≤ t ≤ 2T$$

Where $a_m$ is the maximum acceleration and $2T$ is the total time for the trip.

(a) Draw sketches of $a(t)$ and the jerk as functions of time.
(b) What is the elevator’s maximum speed?
(c) Find an approximate expression for the speed at short times near the start of the ride, $t ≪ T$.

2. Homework Equations

$j = \dot{a} = \ddot{v}$

The Attempt at a Solution

[/B]
The sketch required for part (a) is trivial, so I won't go over the details of the two sketches.
Now, for part (b), I found the value of $t$ for which the acceleration is zero, and used integration to find the velocity (or rather, maximum velocity) at that value of $t$. The problem is, there's more than one value of $t$ for which the acceleration of the elevator is zero. In fact, the acceleration is zero at $t = 0$ inspite of the fact that $v(0) = 0$. I feel like I have made some kind of error somewhere.

 

[DEvens]

Maybe if you plot the speed as a function of time? That might help with your intuition and your confusion about maximums and so on.

Remember that a full plot of a function includes things like maximums, minimums, zeroes, and inflection points.

 

[PFuser1232]

Maybe if you plot the speed as a function of time? That might help with your intuition and your confusion about maximums and so on.

Remember that a full plot of a function includes things like maximums, minimums, zeroes, and inflection points.

$a(t) = 0$ implies maximum speed (maximum/minimum velocity), right? Why am I getting a contradiction for $t = 0$?

 

[gneill]

$a(t) = 0$ implies maximum speed (maximum/minimum velocity), right? Why am I getting a contradiction for $t = 0$?

At time t=0 the elevator starts from rest. You don't expect it to suddenly jump to maximum speed instantaneously, right?

Did you sketch the acceleration profile (say for the first acceleration formula)? Does the acceleration ever change sign through the acceleration phase?

Do you expect the elevator to attain maximum speed before or after all the acceleration has taken place?

 

[PFuser1232]

At time t=0 the elevator starts from rest. You don't expect it to suddenly jump to maximum speed instantaneously, right?

Did you sketch the acceleration profile (say for the first acceleration formula)? Does the acceleration ever change sign through the acceleration phase?

Do you expect the elevator to attain maximum speed before or after all the acceleration has taken place?

I did plot acceleration as a function of time, it didn't change signs through the first phase. I guess my question is more of a calculus question; doesn't $f'(a) = 0$ imply that $f$ has a maximum/minimum value when evaluated at $a$?

 

[gneill]

I did plot acceleration as a function of time, it didn't change signs through the first phase. I guess my question is more of a calculus question; doesn't $f'(a) = 0$ imply that $f$ has a maximum/minimum value when evaluated at $a$?

That's the case unless it's a saddle point. Use the second derivative test to confirm max or min. Here it should be clear from the physical scenario whether you have max or min when t = 0 or t = T.

 

[PFuser1232]

That's the case unless it's a saddle point. Use the second derivative test to confirm max or min. Here it should be clear from the physical scenario whether you have max or min when t = 0 or t = T.

What about $t = 2T$?

 

[gneill]

What about $t = 2T$?

T = 2T corresponds to the time when the elevator has been brought to rest.

Your first a(t) expression describes taking the elevator from rest to its "cruising" speed, while the second a(t) expression describes slowing the elevator from cruising speed back to rest. That's why the second expression is negative: it's a deceleration.

 

[PFuser1232]

T = 2T corresponds to the time when the elevator has been brought to rest.

Your first a(t) expression describes taking the elevator from rest to its "cruising" speed, while the second a(t) expression describes slowing the elevator from cruising speed back to rest. That's why the second expression is negative: it's a deceleration.

So I should find the velocity at time $T$ by integration as follows:

$$v(T) = v(0) + \int_0^T a(t) dt = \int_0^T a(t) dt$$

For part (c), I could use integration again to find velocity as a function of time, which would yield the sum of a linear term and a sinusoidal term. Would the small angle approximation be appropriate in that case?

 

[haruspex]

For part (c), I could use integration again to find velocity as a function of time, which would yield the sum of a linear term and a sinusoidal term. Would the small angle approximation be appropriate in that case?

Since they indicate the conditions for that would be met, I would say so.

 

[PFuser1232]

Since they indicate the conditions for that would be met, I would say so.

$$v(t) = \int_0^t a(t') dt' = \int_0^t [\frac{a_m}{2} - \frac{a_m}{2}\cos{\frac{2\pi t'}{T}}] dt' = \frac{a_m}{2} t - \frac{a_m}{2} \frac{T}{2\pi} \sin{\frac{2\pi t}{T}} = \frac{a_m}{2} t - \frac{a_m T}{4\pi} \sin{\frac{2\pi t}{T}}$$

For very small $t$, the second term becomes approximately equal to the first term (by the small angle approximation), so $v(t) ≈ 0$

Is there another way to find an approximation to $v(t)$ for very small $t$?

 

[haruspex]

For very small $t$, the second term becomes approximately equal to the first term (by the small angle approximation), so $v(t) ≈ 0$

So include the next term in the expansion of sin().

 

[PFuser1232]

So include the next term in the expansion of sin().

How do we write the most general expression for $v(t)$? I'm confused because acceleration is piecewise defined. Should we integrate the first part of the acceleration function from $0$ to $T$ (giving us $v_{max}$), then add the integral of the second part of the acceleration function from $T$ to $t$ where $t$ lies in the interval $[0,2T]$?

 

[haruspex]

How do we write the most general expression for $v(t)$? I'm confused because acceleration is piecewise defined. Should we integrate the first part of the acceleration function from $0$ to $T$ (giving us $v_{max}$), then add the integral of the second part of the acceleration function from $T$ to $t$ where $t$ lies in the interval $[0,2T]$?

Yes. The full description of v(t) is likely to involve separation into the two time ranges, just as it did for the acceleration.

 

[PFuser1232]

Yes. The full description of v(t) is likely to involve separation into the two time ranges, just as it did for the acceleration.

So to find the distance $D$ traveled by the elevator in time $2T$ I would have to integrate the velocity function (which itself is split into two integrals; one from $0$ to $T$ and the other from $T$ to $t$) from $0$ to $2T$.

 

[haruspex]

So to find the distance $D$ traveled by the elevator in time $2T$ I would have to integrate the velocity function (which itself is split into two integrals; one from $0$ to $T$ and the other from $T$ to $t$) from $0$ to $2T$.

Yes.