Calculating a bolt's free fall time in an accelerating elevator

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

A) W.r.t. Earth's frame,
After 2s , velocity of bolt = at = 1.2*2 = 2.4 m/s
distance traveled = (1/2) 1.2*2² = 2.4 m
Let's say that the bolt reaches the floor at time t.
Using s = ut + (1/2)at²
2.4 = 2.4t +(1/2)(-9.8) t²
Solving this equation doesn't give proper answer.

After leaving the car, the bolt goes up till its speed becomes 0 m/s. Let's say that this process takes the time t₁ and by this time the bolt travels a distance s₁.
0 = 2.4 - 9.8t ₁
t₁ = 2.4/9.8 = 0.245
s₁ = 2.4* 0.245 + (1/2)(-9.8)0.245² =0.294 m
After the speed of the bolt becomes 0, the bolt starts falling down.
Let's say that the bolt takes time t₂ to reach the floor from now.
So,
s + s₁ = (1/2)9.8 t₂
Solving this gives t₂ = 0.741s

What is meant by free fall time ?
t₂ = 0.741 s or t₂ + t₁ = 0.986 s ?

b) W.r.t. elevator,Taking t₂ as the free fall,
The displacement is
d = s + s₁ +[ {1.2*(2 + t₁ )} t₂ + (1/2) 1.2 * t₂² = 5.019 m where the bold part is distance traveled by the car in time t₂

The distance is D = d+ s₁ = 5.019+0.294= 5.313 m,

 

[Pushoam]

After leaving the car, the bolt goes up till its speed becomes 0 m/s. Let's say that this process takes the time t1 and by this time the bolt travels a distance s1.

Bolt doesn't go up. Here, it is given in the question that the bolt directly comes down.So, the bolt's speed is 0 at the moment it leaves the ceiling.

So, s = (1/2)9.8 t₂²
t₂ = 0.7 s , free fall time

 

[Pushoam]

The bolt falls inside the elevator.
So, after 2s, the ceiling is at a distance 2.7 m + s =2.70+ 2.4 =5.1 m
The bolt falls from the ceiling and touches the floor of the elevator under free fall and the free fall takes time t.
distance traveled by the bolt in downward direction + distance traveled by the floor in upward direction = 5.1 m
⇒ (1/2)g t² + 2.4t + (1/2)1.2 t² + 2.4 m = 5.1 m
t = 0.52 s
which one is correct?

 

[Doc Al]

Bolt doesn't go up. Here, it is given in the question that the bolt directly comes down.So, the bolt's speed is 0 at the moment it leaves the ceiling.

Not sure what you mean. At the moment it leaves the ceiling, the bolt's moving at the same speed as the ceiling (and rest of the elevator).

 

[Pushoam]

Not sure what you mean. At the moment it leaves the ceiling, the bolt's moving at the same speed as the ceiling (and rest of the elevator).

o.k.
Now I understood.
First of all, the bolt falls inside the elevator.
W.r.t. Earth's frame,

At the moment of leaving, both the bolt and ceiling moves upward with same speed i.e. 2.4 m/s.

Due to gravity , the speed of the bolt goes on decreasing to 0 and then the bolt falls down while the ceiling goes up with acceleration 1.2 m/s².

After leaving the car, the bolt goes up till its speed becomes 0 m/s. Let's say that this process takes the time t₁ and by this time the bolt travels a distance s₁.
0 = 2.4 - 9.8t₁
t₁ = 2.4/9.8 = 0.245
s₁= 2.4* 0.245 + (1/2)(-9.8)0.2452 =0.294 m

Let's say that the bolt takes time t₂ to reach the floor from now.
So,

Distance traveled by the bolt after its speed becomes 0 is d = ( 1/2) g t₂²

Now,
2.7+ 0.294 = ( 1/2) g t₂² + 2.4( t₁ + t₂) + ( 1/2) 1.2( t₁ + t<sub>2[/SUB)² , the bold part is distance traveled by the floor during free fall
2.37 = 4.9 t2² + 2.4 t2 + 0.6 t2² + 0.294 t2
2.37 = 5.5t2²² +2.694 t2
t2 = 0.456 s

Time of free fall =t1+ t2 = 0.7 s</sub>

 

[Pushoam]

I didn't know how to solve part (B).
Frame attached to the shaft is equivalent to a frame attached to the floor for our purpose.
Now, a person on the floor will see the bolt traveling from ceiling to the floor. So, the displacement is the distance between the ceilling and the floor i.e. 2.7m.

Is this correct?

 

[Doc Al]

I didn't know how to solve part (B).
Frame attached to the shaft is equivalent to a frame attached to the floor for our purpose.
Now, a person on the floor will see the bolt traveling from ceiling to the floor. So, the displacement is the distance between the ceilling and the floor i.e. 2.7m.

Is this correct?

That's what I would say. But that seems a strange question.

Another (equivalent) way to solve part (a) from the frame of the Earth is to write two equations: one for the position of the floor as a function of time and one for the position of the bolt as a function of time. Solve for when bolt and floor are at the same position.

 

[TSny]

I didn't know how to solve part (B).
Frame attached to the shaft is equivalent to a frame attached to the floor for our purpose.
Now, a person on the floor will see the bolt traveling from ceiling to the floor. So, the displacement is the distance between the ceilling and the floor i.e. 2.7m.

Is this correct?

I believe the frame of the elevator shaft is the same as the Earth frame.
https://www.vocabulary.com/dictionary/elevator shaft

 

[Pushoam]

From this dictionary, what I understood is elevator shaft is something very long tube like and it is attached to the roof of the elevator.

Hence, the shaft,too, is moving.
The shaft frame is equivalent to the elevator's floor frame.

 

[Hiero]

No no, I agree with TSny... An "elevator shaft" is the shaft in which the elevator moves, not something that moves with the elevator.

I think Irodov also meant it this way, or else the two parts of (b) are redundant (the displacement would equal the distance).

 

[Doc Al]

I believe the frame of the elevator shaft is the same as the Earth frame.

You're right! I misread that as saying the "frame of the elevator". (Which is why I thought it a strange -- trivial -- question!) D'oh!

 

[Pushoam]

Taking elevator frame as the Earth frame,

distance traveled by the floor during free fall = 2.4( t₁ + t₂) + ( 1/2) 1.2( t₁ + t₂)² = 1.977 m =2 m

So, the displacement traveled by the bolt =[( 2.7 -2.0) m = 0.7 m ]toward the floor

the distance traveled by the bolt = s₁ + ( 1/2) g t₂² = 0.294 + 1.019 = 1.312 =1.3 m

So, the problem is solved.

Thanks to all.

 

[Ray Vickson]

Taking elevator frame as the Earth frame,

distance traveled by the floor during free fall = 2.4( t₁ + t₂) + ( 1/2) 1.2( t₁ + t₂)² = 1.977 m =2 m

So, the displacement traveled by the bolt =[( 2.7 -2.0) m = 0.7 m ]toward the floor

the distance traveled by the bolt = s₁ + ( 1/2) g t₂² = 0.294 + 1.019 = 1.312 =1.3 m

So, the problem is solved.

Thanks to all.

An easier way is to note that for somebody making measurements inside the elevator, it looks exactly like a free-fall problem on a new planet with a stronger force of gravity than earth's: on this new planet, the acceleration of "gravity" is $g_0 = 9.8 + 1.2 = 11.0\: m/s^2$. The bolt starts with 0 "velocity" from a height of 2.7 m.

 

[Pushoam]

An easier way is to note that for somebody making measurements inside the elevator, it looks exactly like a free-fall problem on a new planet with a stronger force of gravity than earth's: on this new planet, the acceleration of "gravity" is g0=9.8+1.2=11.0m/s2g0=9.8+1.2=11.0m/s2g_0 = 9.8 + 1.2 = 11.0\: m/s^2. The bolt starts with 0 "velocity" from a height of 2.7 m.

Thanks for this insight.This helps in solving part a easily as

w.r.t. floor frame,
2.7 = (1/2) *11*t_fall²
t_fall= 0.7 s

Now, t_fall for both the floor and the Earth frame is same i.e. t_fall = 0.7 s. I am assuming it so. But how to prove this? It is assumed so under Galilean transformation . But, Galilean transformation deals with only inertial frames. And here, floor frame is non - inertial.
But, how can it help in solving part (b)?
W.r.t. the Earth frame,

Knowing that t_fall= 0.7 s

the displacement can be given as
D = 2.4 t_fall - (1/2) g t_fall² = 0.7 m
But, for calculating distance, I will have to do what I did in the earlier post i.e. post 12.

 

[Ray Vickson]

Thanks for this insight.This helps in solving part a easily as

w.r.t. floor frame,
2.7 = (1/2) *11*t_fall²
t_fall= 0.7 s

Now, t_fall for both the floor and the Earth frame is same i.e. t_fall = 0.7 s. I am assuming it so. But how to prove this? It is assumed so under Galilean transformation . But, Galilean transformation deals with only inertial frames. And here, floor frame is non - inertial.
But, how can it help in solving part (b)?
W.r.t. the Earth frame,

Knowing that t_fall= 0.7 s

the displacement can be given as
D = 2.4 t_fall - (1/2) g t_fall² = 0.7 m
But, for calculating distance, I will have to do what I did in the earlier post i.e. post 12.

You can transform to non-Galilean frames. In this case the transformation would be something like $x' = x - v_0 t - \frac{1}{2} a t^2, \; t' = t$. From the behavior of $x(t)$ you can get the behavior of $x'(t') = x'(t).$

A more familiar example of transformation to a non-Galilean frame is to transfer to a rotating coordinate system, where such effects as "centrifugal force" and "Coriolis force" appear naturally.

 

[Pushoam]

You can transform to non-Galilean frames.

Are non-Galilean frames nothing but non - inertial frames without considering special relativity?
What are Galilean and non - Galilean frames?

 

[Ray Vickson]

Are non-Galilean frames nothing but non - inertial frames without considering special relativity?
What are Galilean and non - Galilean frames?

I probably should have said non-inertial frames and non-Galilean transformations.

 

[Pushoam]

You can transform to non-Galilean frames. In this case the transformation would be something like

Floor frame ⇒ (x, t)
Earth frame⇒(x', t')t' =t, assumption
x' = x + v₀t + (1/2) at² where v₀ is the initial speed of the floor and a is the acceleration of the floor measured from the Earth's framew.r.t. floor frame,
2.7 = (1/2) *11*t<sub>fall[/SUB²
tfall= 0.7 s

x = 2.7 - 2.4*0.7 - 0.5*1.2*0.7² = 0.7 m

This transformation helps in calculating displacement , not distance.</sub>