Calculate G Force Impact & Curve Speed

[jchively]

I hope this is the correct place... I need help with a project... We are dropping 300lbs from 100 - 150 feet.. I need to know what the speed on impact would be.. Also, I need to know if this was running on a track and we start curving the rail 20 feet above impact and in a distance of 20 feet curve the rail 180 degrees, basically a 1 to 1 ratio turn, how many g's would the object pull in the bottom of the turn prior to going vertical..

 

[cepheid]

Welcome to PF jchively,

There are a couple of methods of arriving at the answer to the free-fall problem:

Method 1: Conservation of Energy

Before the drop, the object has gravitational potential energy (PE) (which we'll measure relative to the ground), but no kinetic energy (KE) (since it is not moving). Once it is released, as it descends it loses gravitational potential energy. But any PE that it loses turns into KE that is gained (since energy is conserved). Therefore, at the bottom of the drop, where the gravitational PE is 0, all of the energy must have been converted to kinetic, meaning that the final KE is equal to the initial PE. The expression for gravitational PE is:

PE = mgh

where

- m is the mass of the object
- g is the acceleration due to gravity (9.81 m/s² in SI units)
- h is the height above the surface.

The expression for kinetic energy is:

(1/2)mv²

where

- v is the speed

So conservation of energy says:

E_final = E_initial

KE_final = PE_initial

$$\frac{1}{2}mv^2 = mgh$$
$$v^2 = 2gh$$
$$v = \sqrt{2gh}$$

Plug in the value for h (whatever it is from 100 ft to 150 ft), and the value for g in imperial units (ft/s²), and you're good to go.

 

[cepheid]

Method 2: Basic Kinematics

We know from kinematics that if y is the vertical position of the object above the ground, and it moves under constant vertical acceleration, then the most general formula for y vs. time is:

$$y = y_0 + v_0t + \frac{1}{2}at^2$$

where:

- y₀ is the initial vertical position (the position at t = 0)
- v₀ is the initial vertical speed (the speed at t = 0)
- a is the vertical acceleration
- t is elapsed time

You just have to accept this for now -- I'm not going to get into a derivation. We also know that for constant acceleration, the expression for the vertical speed v vs. time is:

$$v = v_0 + at$$

This just comes from the definition of acceleration -- it is the rate at which speed changes with time. So if you multiply acceleration by time, you get the change in speed. Now, in our case, we know that the acceleration is due to gravity: a = -g, with g = +9.81 m/s². We also know that v₀ = 0 since the object is dropped from rest. We also know that the initial position y₀ is just the height h from which the object is dropped. With that in mind, we get for the first equation:

$$y = h - \frac{1}{2}gt^2$$

and for the second equation

$$v = - gt$$

Using the second equation, solve for t:

$$t = -\frac{v}{g}$$

Plug that result for t into the first equation:

$$y = h - \frac{1}{2}g\left(-\frac{v}{g}\right)^2$$

Now, we want to know what the speed is when the object reaches the ground, so set the vertical position y = 0 and solve:

$$h = \frac{1}{2}g\frac{v^2}{g^2}$$

$$2h = \frac{v^2}{g}$$

$$2gh = v^2$$

This is exactly the same equation as we derived before using the conservation of energy. You could also have derived it using the work-energy theorem, but that's really not much different from the conservation of energy method.

 

[jchively]

Thank you.. I am building a small scale version of what is anticipated and needed that information in order to move forward prior to involving the expensive engineers.. Much appreciated!

 

[cepheid]

Thank you.. I am building a small scale version of what is anticipated and needed that information in order to move forward prior to involving the expensive engineers.. Much appreciated!

If this is for some serious application, you should know that the above discussion was for free-fall in a vacuum. It neglects the effects of air resistance (drag). Depending on the object geometry, it may reach a terminal velocity (maximum speed beyond which it does not accelerate). Figuring out what that might be is more complicated.

If you are having the object slide (or roll) down a nearly vertical ramp until it enters a circular loop at the bottom, then the speed entering the loop will be close to what we computed above,, except it will point in the "along the rail" direction (again, neglecting the effects of air drag and friction).

To figure out the acceleration of the object while in the loop, you need to figure out what then centripetal acceleration is.

 

[jchively]

There will be very little drag. the object is 4 feet by feet square... I was hoping to keep it around 60 to 80 mph in the free fall and no more than 3.5 g's at the circular loop... Based on your calculations what do you think they would be if we use height of 100 feet and object weight of 300 lbs.