Distance moved in two different inertial frames

[Monsterboy]

Homework Statement

A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then
(a) $t_1=t_2$
(b) $t_1<t_2$
(c) $t_1>t_2$
(d) $t_1>t_2$ or $t_2>t_1$ depending on whether the lift is going up or down.

Relevant Equations

Height from which the coin is dropped (or distance travelled by the coin) = $ut + \frac{1]{2} {a}{t^2}$

We have one elevator at rest and another moving up or down at uniform speed, so both of them are considered to be inertial frames.

Taking the initial height of the coin as

$H = 0 + \frac{1}{2} g{t_1}^2$ and

$H = 0 + \frac{1}{2} g{t_2}^2$

$t_1 = t_2 = t = \sqrt{\frac{2H}{g}}$ in both the cases

So, the correct option (a) $t_1 = t_2$ , this is the answer given.

I don't understand why the distance moved the elevator is not considered.

Lets say the elevator moved by a distance $y$ up or down in the time interval $t_2$, so the distance traveled by the coin will be

$H + y = \frac{1}{2} g{t_2}^2$ if it moves down
or
$H - y = \frac{1}{2} g{t_2}^2$ if it moves up

So, now how can $t_2$ be equal to $t_1$ ? Now that its distance traveled towards the floor is different ?

Inertial frame simply means a frame where Newton's laws work as expected without applying any corrections right ?

 

[Orodruin]

If the elevator has a constant velocity up or down, this will also be the initial velocity of the dropped coin.

 

[Monsterboy]

Oh,

So we get,

$H + y = ut_2 + \frac{1}{2} g{t_2}^2$ or $H - y = ut_2 + \frac{1}{2} g{t_2}^2$

But in the frame of the observer inside the elevator, $u = 0$ right ? Same way, the distance traveled by the coin is the same ? But it is different for an observer on the ground right ?

 

[Orodruin]

Oh,

So we get,

$H + y = ut_2 + \frac{1}{2} g{t_2}^2$ or $H - y = ut_2 + \frac{1}{2} g{t_2}^2$

But in the frame of the observer inside the elevator, $u = 0$ right ? Same way, the distance traveled by the coin is the same ? But it is different for an observer on the ground right ?

What is $y$ expressed in $u$ and $t_2$?

 

[Monsterboy]

What is y expressed in u and t2?

Hmm...

$y = ut_2$ ?

So $y$ and $ut_2$ cancel out when distance traveled is $H + y$ but they add up if the distance is $H - y$.

we get

$H = \frac{1}{2} a{t_2}^2$ when the elevator moves down.
$H = 2ut_2 + \frac{1}{2} a{t_2}^2$ when the elevator moves up ?

I was expecting the distance traveled to be greater when the elevator is moving down.

 

[Monsterboy]

Hmm...

$H = \frac{1}{2} a{t_2}^2$ when the elevator moves down.
$H = 2ut_2 + \frac{1}{2} a{t_2}^2$ when the elevator moves up ?

I was expecting the distance traveled to be greater when the elevator is moving down.

I think it will be

$H = \frac{1}{2} g{t_2}^2$ when the elevator moves down.
$H = 2ut_2 - \frac{1}{2} g{t_2}^2$ when the elevator moves up ?

 

[PeroK]

I think it will be

$H = \frac{1}{2} g{t_2}^2$ when the elevator moves down.
$H = 2ut_2 - \frac{1}{2} g{t_2}^2$ when the elevator moves up ?

For any constant velocity $u$ (positive or negative). The coin starts at $y_0$ and, assuming it is released at $t =0$, its trajectory is$$y_c = y_0 + ut - \frac 1 2 gt^2$$The trajectory of the floor of the elevator is$$y_f= y_0 - h + ut$$where $h$ is the height of the elevator. The coin lands when $y_c = y_f$. I.e.
$$y_0 + ut - \frac 1 2 gt^2 = y_0 - h + ut$$The $y_0$ and $ut$ terms cancel, leaving$$\frac 1 2 gt^2 = h$$And we see that $t$ depends on $g$ and $h$ and not on the constant velocity of the elevator, zero or otherwise.

 

[Monsterboy]

Lets talk in terms of frames, so for an observer on the ground, the distance traveled by the coin in the stationary and the moving elevator is the same ? I got that time taken is the same, probably stupid to think that distance traveled can be different ?

I can visualize the distance traveled to be the same for observers inside the elevator, but I can't visualize what an observer on the ground will see, I assumed that the distance traveled will be $H + y$ and $H - y$ where $H$ is the initial height of the coin above the elevator floor and $y$ is the distance traveled by the elevator after the coin is dropped. Is this where I went wrong ?

 

[jbriggs444]

Lets talk in terms of frames, so for an observer on the ground, the distance traveled by the coin in the stationary and the moving elevator is the same ?

We need to be a little more careful in specifying things.

We have two different scenarios (moving elevator, stationary elevator).
We have three different frames of reference (moving elevator, stationary elevator, ground).

Six possibilities. Exactly which two do you want to compare?

On the face of it, you want to compare the moving scenario and the stationary scenario both in the ground frame. The two distances moved will be different between those two scenarios.

 

[Monsterboy]

We have two different scenarios (moving elevator, stationary elevator).
We have three different frames of reference (moving elevator, stationary elevator, ground).
Six possibilities. Exactly which two do you want to compare?

Aren't the frames of the stationary elevator and the ground basically the same ?

On the face of it, you want to compare the moving scenario and the stationary scenario both in the ground frame. The two distances moved will be different between those two scenarios.

Ok, this is what I wanted to know.

 

[jbriggs444]

Aren't the frame of the stationary elevator and the ground basically the same ?

Yes. Just a fixed offset, not a difference to velocity or acceleration.

 

[PeroK]

Lets talk in terms of frames, so for an observer on the ground, the distance traveled by the coin in the stationary and the moving elevator is the same ? I got that time taken is the same, probably stupid to think that distance traveled can be different ?

I can visualize the distance traveled to be the same for observers inside the elevator, but I can't visualize what an observer on the ground will see, I assumed that the distance traveled will be $H + y$ and $H - y$ where $H$ is the initial height of the coin above the elevator floor and $y$ is the distance traveled by the elevator after the coin is dropped. Is this where I went wrong ?

In the ground frame $u = 0$ in the case of a stationary elevator and $u \ne 0$ in the case where the elevator is moving up or down uniformly.

In the frame of the elevator (whether moving or not in the ground frame) $u = 0$ in all cases!