Integral of non-constant acceleration

[Karol]

Homework Statement

A stone falls from a high tower. what is the distance, in the x direction, parallel to the ground, that the stone reaches, caused only by the x component, the tangential component, of the gravity, after t seconds of free fall.

Homework Equations

The constant acceleration formula:
$$x=\frac{1}{2}a\cdot t^2$$

The Attempt at a Solution

According to the drawing attached, the acceleration in the x direction is:
$$a=g\cdot\sin\theta=g\sin(\omega t)$$
Where ω is the angular velocity of the earth.
I insert this a into the constant acceleration formula, since i don't know anything else, and, maybe, in short intervals of time the acceleration is approximately constant.
So, i get:
$$dx=\frac{1}{2}g\sin(\omega dt)\cdot dt^2$$
And:
$$x=\frac{g}{2}\int_{0}^{4} \sin(\omega t)\cdot t^2$$
Is this approach correct? since, numerically, the results aren't.

 

[SammyS]

Homework Statement

A stone falls from a high tower. what is the distance, in the x direction, parallel to the ground, that the stone reaches, caused only by the x component, the tangential component, of the gravity, after t seconds of free fall.

Homework Equations

The constant acceleration formula:$$x=\frac{1}{2}a\cdot t^2$$

The Attempt at a Solution

According to the drawing attached, the acceleration in the x direction is:
$$a=g\cdot\sin\theta=g\sin(\omega t)$$Where ω is the angular velocity of the earth.
I insert this a into the constant acceleration formula, since i don't know anything else, and, maybe, in short intervals of time the acceleration is approximately constant.
So, i get:
$$dx=\frac{1}{2}g\sin(\omega dt)\cdot dt^2$$And:
$$x=\frac{g}{2}\int_{0}^{4} \sin(\omega t)\cdot t^2$$Is this approach correct? since, numerically, the results aren't.

Since the numerical result isn't correct, I doubt that the approach is correct either.


Kind of a strange problem.

I guess that the tower is so high that stone doesn't land at a point directly below the position it was dropped from.

It also looks like you were to to assume that the tower is short enough to consider g to be a constant value.


What class is this for?

 

[rude man]

Yes, weird. Looks like a Coriolis problem except the latitude is not given, nor the tower height.

 

[haruspex]

I agree, strange question. What is 'tangential component of gravity', unless caused by a gravitational anomaly? What does the diagram represent? Perhaps theta therein is supposed to be pi/2 - latitude, but the rest of the picture still doesn't make sense, and as Rude man says you still need the tower height. (I get ωht sin(θ), where h is the tower height and ω = 2π/24hours.)

Edit: What rubbish, haruspex . The tower height is not needed. My h above should be the vertical distance fallen in time t, so it reduces to ωgt³ sin(θ)/2 = 2.3 cm at the equator.
But I don't see that you need any of that fancy non-inertial frame calculation - at a distance h below the release point, the tangential velocity difference between the stone and the tower is ωh sin(θ), and that is constant, so the horizontal separation is ωht sin(θ).

 

[TSny]

I believe this question is related to another post: https://www.physicsforums.com/showthread.php?t=688963

 

[haruspex]

I believe this question is related to another post: https://www.physicsforums.com/showthread.php?t=688963

Ah yes, even the diagram is almost identical. And in that post it is stated that the tower is at the equator.