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Free fall time in an accelerating frame of reference

  • Thread starter archaic
  • Start date
128
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Homework Statement
An elevator car whose floor-to-ceiling distance is equal to ##2.7m## starts ascending with constant acceleration ##1.2 m/s^2##. ##2.0 s## after the start a bolt begins falling from the ceiling of the car. Find the bolt's free fall time.
Homework Equations
##y(t)=\frac{1}{2}at^2+v_0t+y_0##
##l=2.7m## and ##w=1.2m/s^2##
This is my wrong attempt, the bolt's equation for ##t\geq 2## would be ##y_b(t)=-\frac{1}{2}g(t-2)^2+e(2)+l## where ##e(t)=\frac{1}{2}wt^2## the position of the elevator's floor in the absolute frame of reference.
##d(t)=-\frac{1}{2}(w+g)t^2+2gt+2(w-g)+l## the distance between the bolt and the elevator.
I get ##\Delta=4g^2+(2g+2w)(2w-2g+l)## and so ##t=\frac{-2g\pm\sqrt{4g^2+(2g+2w)(2w-2g+l)}}{-(g+w)}## which yields a wrong answer.
My mistake is probably in the bolt's position equation, I don't see how it's wrong though.
The correct answer is ##0.7s##.
Thank you!
 
Last edited:

Doc Al

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That expression for the bolt's position is a bit confusing. How about redoing it, measuring the time from the moment the bolt drops and the position from the car floor at that moment. Then do something similar for the position of the car floor and solve for their intersection.
 
Problem Statement: An elevator car whose floor-to-ceiling distance is equal to ##2.7m## starts ascending with constant acceleration ##1.2 m/s^2##. ##2.0 s## after the start a bolt begins falling from the ceiling of the car. Find the bolt's free fall time.
Relevant Equations: ##y(t)=\frac{1}{2}at^2+v_0t+y_0##
##l=2.7m## and ##w=1.2m/s^2##

This is my wrong attempt, the bolt's equation for ##t\geq 2## would be ##y_b(t)=-\frac{1}{2}g(t-2)^2+e(2)+l## where ##e(t)=\frac{1}{2}wt^2## the position of the elevator's floor in the absolute frame of reference.
##d(t)=-\frac{1}{2}(w+g)t^2+2gt+2(w-g)+l## the distance between the bolt and the elevator.
I get ##\Delta=4g^2+(2g+2w)(2w-2g+l)## and so ##t=\frac{-2g\pm\sqrt{4g^2+(2g+2w)(2w-2g+l)}}{-(g+w)}## which yields a wrong answer.
My mistake is probably in the bolt's position equation, I don't see how it's wrong though.
The correct answer is ##0.7s##.
Thank you!
Simply add the two accelerations, 1.2 + g, and use the standard SUVAT equations.
I get t = 0.70033 s.
 

TSny

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My mistake is probably in the bolt's position equation, I don't see how it's wrong though.
Did you take into account that the bolt has an upward velocity at the moment the bolt begins to fall?
 
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That expression for the bolt's position is a bit confusing. How about redoing it, measuring the time from the moment the bolt drops and the position from the car floor at that moment. Then do something similar for the position of the car floor and solve for their intersection.
Well, at first the bolt is moving with the elevator so its position relative to the ground would be ##y_b(t)=e(t)+l## but then it starts falling at ##t=2s##, hence the ##-\frac{1}{2}g(t-2)^2##, and ##e(t=2s)+l## is its distance relative to the ground when it goes into the free fall. So ##y_b(t)=-\frac{1}{2}g(t-2)^2+e(t=2s)+l##.
Did you take into account that the bolt has an upward velocity at the moment the bolt begins to fall?
Oh, so when the ball starts to free fall, we interpret it as not being accelerated anymore and so it has a velocity by virtue of its previous state of being accelerated? Makes sense.
 

TSny

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Oh, so when the ball starts to free fall, we interpret it as not being accelerated anymore and so it has a velocity by virtue of its previous state of being accelerated? Makes sense.
Yes. Of course the bolt still has acceleration g after starting free fall.
 
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I didn't take into account the fact that the bolt would have a velocity when it's set free. After editing my equations, however, I still get a wrong result (two seconds later/earlier than it should be).
Here's what I did :
$$y_b(t)=-\frac{1}{2}g(t-2)^2+2w(t-2)+(l+2w)$$
$$e(t)=\frac{1}{2}wt^2$$
$$\begin{align*}
d(t)&=y_b(t)-e(t)\\
&=-\frac{1}{2}g(t^2-4t+4)-\frac{1}{2}wt^2+2wt+(-4w+l+2w)\\
&=-\frac{1}{2}gt^2+2gt-2g-\frac{1}{2}wt^2+2wt+(l-2w)\\
&=-\frac{1}{2}(g+w)t^2+2(g+w)t+(l-2w-2g)
\end{align*}$$
We have then :
$$\begin{align*}
\Delta &=4(g+w)^2+2(g+w)(l-2w-2g)\\
&=2(g+w)[2(g+w)+l-2w-2g]\\
&=2l(g+w)
\end{align*}$$
Finally :
$$\begin{align*}
t&=\frac{-2(g+w)\pm\sqrt{2l(g+w)}}{-(g+w)}\\
&=2\mp\sqrt{\frac{2l}{w+g}}
\end{align*}$$
Does someone know where I might have gone wrong?
 

TSny

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Finally :
$$\begin{align*}
t&=\frac{-2(g+w)\pm\sqrt{2l(g+w)}}{-(g+w)}\\
&=2\mp\sqrt{\frac{2l}{w+g}}
\end{align*}$$
Does someone know where I might have gone wrong?
I don't seen anything wrong. You will need to choose the appropriate sign in front of the square root. Also, keep in mind that the bolt didn't start its free fall until t = 2 s.
 

TSny

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Note that @Doc Al and @Michael Price have suggested ways to get to the answer with less symbolic manipulations. But it's nice to see that your way will get there.
 
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I don't seen anything wrong. You will need to choose the appropriate sign in front of the square root. Also, keep in mind that the bolt didn't start its free fall until t = 2 s.
That's the problem, neither plus nor minus gives the correct answer.
Yes, that's why I used ##(t-2)## in ##y_b(t)##.
Note that @Doc Al and @Michael Price have suggested ways to get to the answer with less symbolic manipulations. But it's nice to see that your way will get there.
I, in fact, know how to solve this. In the elevator's frame of reference :
$$-\frac{1}{2}(g+w)t^2+l=0\Leftrightarrow t=\pm\sqrt{\frac{2l}{g+w}}$$
I want to go about it using the ground's frame of reference. I want to see the error in my reasoning <-<
 

TSny

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If ##t_1## is the value of ##t## when the bolt starts free fall, and if ##t_2## is the value of ##t## when the bolt hits the floor of the elevator, how would you calculate the time that the bolt is in free fall?
 
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If ##t_1## is the value of ##t## when the bolt starts free fall, and if ##t_2## is the value of ##t## when the bolt hits the floor of the elevator, how would you calculate the time that the bolt is in free fall?
I don't see why I'd have to use ##\Delta t##, isn't the equation set up to give the result directly?
 

TSny

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I don't see why I'd have to use ##\Delta t##, isn't the equation set up to give the result directly?
No. From the way you set up the equations, using##(t-2)##, you are taking ##t## to represent the time the elevator has been ascending, not the time that the bolt has been in free fall.
 
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@TSny my bad actually, I was thinking that I am trying to find the time at which the bolt makes contact with the floor.
 
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No. From the way you set up the equations, using##(t-2)##, you are taking ##t## to represent the time the elevator has been ascending, not the time that the bolt has been in free fall.
Indeed, I should've chosen ##d(t)=y_b(t)-e(t+2)## instead. (without the ##t-2## in ##y_b(t)##)
Thank you for your help!
 

TSny

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Indeed, I should've chosen ##d(t)=y_b(t)-e(t+2)## instead.
OK. But then you wouldn't have used ##(t-2)## in your expression for ##y_b(t)##. If you did it this way, then I think you are following @Doc Al 's suggestion.
 
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OK. But then you wouldn't have used ##(t-2)## in your expression for ##y_b(t)##. If you did it this way, then I think you are following @Doc Al 's suggestion.
Yes of course, I have mentioned that in my last comment!
 

TSny

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OK. Hopefully, now everything is good.
 
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I didn't take into account the fact that the bolt would have a velocity when it's set free. After editing my equations, however, I still get a wrong result (two seconds later/earlier than it should be).
Here's what I did :

yb(t)=−12g(t−2)2+2w(t−2)+(l+2w)​
This equation is dimensionally inconsistent!!!!!
 
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This equation is dimensionally inconsistent!!!!!
If you're talking about the ##2w## then no, it is consistent, ##e(2)=\frac{1}{2}w(2)^2##
 

TSny

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This equation is dimensionally inconsistent!!!!!
In the equation ##y_b(t)=−\frac{1}{2}g(t−2)^2+2w(t−2)+(l+2w)## the terms ##2w(t-2)## and ##2w## appear to have the wrong dimensions. @archaic has explained the term ##2w## where the ##2## carries units of s2 . The term ##2w(t-2)## is OK if you realize that both ##2##'s carry units of time.

This is a good illustration of confusion coming from not carrying all the units along.
 

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