Sciencemadness Discussion Board - Measuring height variation with a chronometer and a stone

Measuring height variation with a chronometer and a stone

math - 26-7-2006 at 04:41

hi,

I would like to know how is possible to measure, for example,
a pit's depth, using a stone and a chronometer.

When stone is released, the chronometer is started, and when the stone hit the bottom, the chronometer is stopped.
The result would be not much accurate, specially for short measures, but I'd be interested in how to calculate a depth starting from these.

I think gravity acceleration costant (g=9.81 m/s^2) is something needed in this.

thank you

[Edited on 26-7-2006 by math]

woelen - 26-7-2006 at 04:55

This is quite simple. The covered distance by the stone is 0.5*g*t*t, with g the acceleration of gravity (appr. equal to 10), and t the elapsed time in seconds. Of course, I assume that the initial velocity of the stone equals 0. You need to drop it from your hand, while keeping that motionless before dropping.

So, per example, if the sound of the stone hitting the bottom takes 2 seconds, then the depth of the pit is appr. 0.5*10*2*2, hence 20 meters. For 1 second the depth is appr. 5 meters and for 3 seconds the depth is appr. 45 meters.

This indeed is quite rough. Clocking the chronometer is a large source of inaccuracy, especially if the measured interval of time is short. Another source of error is the friction in air. At speeds of several tens of meters per second, this is a serious issue. This, however, only is significant for very deep pits (e.g. 50 meters or more) and if you use a large, heavy stone you do not need to worry about this. Finally, another source of error is that the sound also needs time to come back to you. It travels at appr. 300 m/s. So, with a pit of 30 meters depth, you have to subtract 0.1 second from your measured time. Of course, this source of errors also only is significant for very deep pits (e.g. 50 meters and deeper).

math - 26-7-2006 at 05:33

thank you for replying

just wondering...where does that [0.5*g*(t^2)] formula come from?
what are the thoughts behind it? just to understand it better and not just do "it is so!".

I think that an example more to measure more accurately depth/height variation/distance would be a video of a coconut falling from a palm...one would have (depending on camera's) even 1/100th of second as time measured; the time starts when coconut fall away from the tree (initial speed = 0) and finishing when it does hit (by hearing sound or better by watching in the video) the ground.

So, if D = Distance... D = 0.5*g*(t^2)

t = sqrt[D / (0.5*g)] ,right?

[Edited on 26-7-2006 by math]

DrP - 26-7-2006 at 05:48

Derive it from the equations of motion (newton).

s = ut + 1/2 at^2

u=0, so s = 1/2 at^2

math - 26-7-2006 at 05:59

I'd like if everyone post variable's meaning as well

...

do you mean that formula?

s = 1/2 * a * t^2 + v0

where:
s = space
a = acceleration
t = time
v0 = beginning speed

Using 0.5*g*(t^2) formula, I came up that, excluding much things (friction etc...), an object would travel 4.905 meters (490.5cm) every second, or 0.049 meters every 0.1 seconds?

woelen - 26-7-2006 at 06:33

The formula of DrP also is coming from "nowhere" without the underlying physics. It is just my formula, but now for non-zero initial velocity.

The real explanation is the following:

accelaration a is the time derivative of velocity v: a = dv/dt.
velocity v is the time derivative of covered distance s: v = ds/dt

Now, for constant acceleration a, one can write v as function of time: v(t) = a*t + v(0), where v(0) is the initial velocity, and t is the elapsed time. In the initial example we may assume constant acceleration, which equals g (appr. 9.8 m/s/s).

Now, if we want the covered distance, we integrate another time and we find

s(t) = 0.5*a*t*t + t*v(0) + s(0)

Here we have the acceleration a (which is constant), the initial velocity v(0) and the initial covered distance s(0). For the original example we have a = g, v(0) = 0 and s(0) = 0, and then this formula reduces to s(t) = 0.5*g*t*t.

When the acceleration is not constant, then things become much more complex. If e.g. we let an object fall towards earth, not from a distance of a few meters, but from e.g. 20000 km height, then we cannot neglect anymore that the acceleration a depends on the covered distance (and hence on elapsed time). At an heaight of 20000 km the acceleration only is about 1/7th of the acceleration at ground level. So, an object falling down from that height will "see" an increasing accelration during its falling. Solving that problem is much harder and requires you to solve a non-linear second order differential equation.

[Edited on 26-7-06 by woelen]

chromium - 26-7-2006 at 06:37

Quote:

Originally posted by math
Using 0.5*g*(t^2) formula, I came up that, excluding much things (friction etc...), an object would travel 4.905 meters (490.5cm) every second, or 0.049 meters every 0.1 seconds?

Not that easy. During first second object travels 4.905m as you stated but speed increases all the time. During 2nd second it travels distance 19.62 - 4.905 = 14.715 m. Speed can reach many 100m 's per second when ther is enough time.

woelen - 26-7-2006 at 06:39

Speed indeed is increasing every second. At t=0, speed is 0, at t=1, speed is 9.8 m/s, at t=2, speed already is 19.6 m/s, etc. So, speed is increasing linearly with time, covered distance is increasing quadratically with time.

math - 26-7-2006 at 17:32

how can one calculate, say, taking the same example, the stone or coconut force when it does hit the ground?
I was thinking about using kinetic energy formula or work’s formula…
I found a similar thread around here, but it seemed it says “force goes to infinitum then comes back to zero”…how is that possible? So, a coconut and a 100kg rock falling from same height do express same force? :o

I thought this:
m= 2kg [mass]
h= 5m [height]

if S = 0.5 * g * (t^2)

tx = sqrt [S / (0.5*g)] , so tx = sqrt [5 / (4.905)] = 1.009637 sec [object’s time to fall from 5m]

Vf = a*t = g*t = 9.81 * tx = 9.9045 m/s [object’s speed just before it hits ground]

Ek = 0.5 * m * (Vf)^2 = 0.5 * 2 * Vf^2 = 98.1 Joules [kinetic energy of object hitting the ground]
or, using work’s formula:
L = F * S * cos(ά) = (m * a) * S * cos(ά) = (2 * 9.81) * 5 * cos 0° = 98.1 Joules

Now, say that elastic/hitting time before the object lose kinetic force is the same time to fall from 5cm (0.05 meters)…think as object would embed 5cm in ground because of dirt’s elasticity then regain (the ground) its shape

ty = sqrt [0.05 / 4.905] = 0.100963 sec [object’s time to perform 0.05 meters of fall]

from these one can try to calculate power (P):

Ek = L
P = L / ty = (F * S * cos 0°) / ty = 971.64 watt [power generated by object’s kinetic energy used in ty time]

Then we would calculate what force (F) is needed to generate the same amount of power in 0.4s, performing a 30cm (0.3m) movement…obviously results would be infinite, but we fix some rules to restrict them:
T = 0.4s
S = 0.3m
P = 971.64 watt
In this example work is just L = F * S , because cos(0°) = 1

P = (F*S) / T
F = (P * T) / S

F = (971.64 * 0.4) / 0.3 = 1295.52 N [1 kg = 9.8 N]
converting…1295.52 N = 132.19 Kg

So, lifting 132.19 Kg for a distance of 0.3 meters in 0.4 seconds, would express the same power of the 2kg coconut falling from 5 meters?

[Edited on 27-7-2006 by math]

woelen - 27-7-2006 at 02:28

Computing the force is not that difficult, provided you know how the collision is. An ideal elastic collision, which reverses the direction of motion instantaneously provides infinite force, but for zero time. The force is a so-called dirac impulse at the moment of collision. In reality, we have elastic collision (or even more realistically, elastic damped collision).

Let's assume a purely undamped elastic condition.

We have F = dp/dt, where p is the momentum of the object, and F is the force, applied to the object. For constant mass, dp/dt can be written as m*dv/dt, where m is the constant mass.

Now suppose that the duration of collision is Δt seconds. In this time Δt the momentum goes from -mv (downwards) to mv, upwards. So, we can compute the average force F over the time Δt. This is 2mv/Δt.

So, for your mass of 2 kg, falling from 5 m, the speed will be appr. 10 m/s on collision (at g = 10). The momentum just before impact is 20 kgm/s. After impact it has reversed sign (mass is moving upwards with 10 m/s).

The great unknown is the duration of the collision, Δt. On a hard rubber elastic floor it will be less than 0.01 s, but in general one cannot tell too much about it. Suppose it is 0.01 s, then the average force, applied to the object is 2*20/0.01 = 4000 N. In reality, the peak force will be higher (maybe 8000 N), assuming a triangular shaped force profile over time Δt.

[Edited on 27-7-06 by woelen]

math - 27-7-2006 at 05:28

Quote:

Originally posted by math

So, lifting 132.19 Kg to a height of 0.3 meters in 0.4 seconds, would express the same power of the 2kg coconut falling from 5 meters (supposing the coconut has a hitting time of 0.100963s) ?

woelen - 27-7-2006 at 06:29

Forces do not generate power, forces generate momentum (F = dp/dt). The momentaneous power, input into a system, equals F*p/m, for constant m, this can be written as F*v.

Your question cannot be answered without additional information. Lifting a certain mass to a certain height takes energy (m*g*h), but an additional energy component may be present as kinetic energy. You did not specify that. Total energy equals m*g*h + 0.5*m*v*v (again for constant mass m, and constant acceleration g).

The hitting time does not play any role in determination of the amount of energy stored in a certain mass. It does play a role in the amount of power, put into the mass, but the total energy remains the same. It is as with momentum. Short hitting times give high force, but for short time, it also results in high power input, but for short time.

Suppose you would have a mass of 2 kg, falling from a height of 5 m, and you made a nice mechanical construction, such that all energy in this mass is used to lift a mass of 132 kg (the mass of 2 kg laying still on the ground), then that mass of 132 kg would be lifted to a highest point of 5*2/132 meters, which is appr. 7.6 cm.

math - 27-7-2006 at 07:32

thank you woelen

just

...

Quote:

Originally posted by woelen
Suppose you would have a mass of 2 kg, falling from a height of 5 m, and you made a nice mechanical construction, such that all energy in this mass is used to lift a mass of 132 kg (the mass of 2 kg laying still on the ground), then that mass of 132 kg would be lifted to a highest point of 5*2/132 meters, which is appr. 7.6 cm.

that is because application time is the same of coconut hitting time (0.1s)... power is P = L / T , where L is work and T is time...
so, if work is the same, but time is longer, you'll need less power to lift the weight than using a shorter time!

if you lift a 10kg weight in 0.2s, you'll use a power A, if you lift same weight in 0.5s, you'll use a power B.
Power A > Power B.

the power of falling 2kg coconut assuming that T = 0.1s :
P = F*S / T
P = (m * a) * S / T
P = (2 * 9.81) * 5 / 0.1 = 981 watt

so, using same power formula, but inserting our comparative example:
T = 0.4s
S = 0.3m

P = L / T = F * S / T
F = (P * T) / S

F = (981 * 0.4) / 0.3 = 1308 N
so, the force to generate ( F*S/T = P!!) 981 watt (same power of coconut falling), but in a longer time (0.4s instead of 0.1s) and in a different space (0.3m instead of 5m) is 1308 N.
1308 N = 133 Kg

if this still doesn't clear up...we will need a mind gathering

woelen - 27-7-2006 at 11:34

Quote:

the power of falling 2kg coconut assuming that T = 0.1s :
P = F*S / T
P = (m * a) * S / T
P = (2 * 9.81) * 5 / 0.1 = 981 watt

Here you make a mistake. The power of conversion of potential to kinetic energy indeed is 981 W for a coconut, falling towards earth in 0.1 second at constant velocity. But.... the coconut does not fall in 0.1 second, but it takes appr. 1 second to fall 5 meters. The average power is 98.1 Watt, but in reality, the power starts off at 0 and ends with 196.2 Watt.

What happens with a falling coconut is that potential energy m*g*h (where h is the distance from ground) is converted to kinetic energy 0.5*m*v*v. This conversion goes faster and faster, because h is decreasing at an ever increasing rate, due to acceleration of the coconut.

When the coconut (2 kg mass) leaves the tree, it has a potential energy equal to 2*9.81*5 is 98.1 J. When it hits the ground, it has zero potential energy, but the same kinetic energy 0.5*m*v*v. Solving for this, you see that v is almost 10 m/s.

In order to have your formula, you need a strong catapult, which shoots the coconut from the tree, straight to the ground at an initial speed just below 50 m/s.

math - 27-7-2006 at 17:04

Quote:

Originally posted by woelen

Quote:

The average power is 98.1 Watt, but in reality, the power starts off at 0 and ends with 196.2 Watt.

where did you get this from?

how can I express that kinetic energy generated in the falling is used in 0.1s ?
calculating ek or work (in the falling coconut example ek=L) is ok, but then about power...the time that / the work is meant work's time, I hadn't thought to this...
so, how can one say the result of coconut hitting, say 30cm^2 of ground in 0.1s, in kg/s or something like that?

[Edited on 28-7-2006 by math]