# Drop test of an object, are my physics right?

Hi, I will keep this short and sweet.

I am drop testing an object of complicated geometry froma set height in 10 different orientations.
Due to the complexity I decided to make a free falling rig where it slides (via roller bearings for example) when released.

Simple freefall analysis using energy equation from dropping it at a set height h:

mgh=1/2m(v^2)

v = √2gh

Falling attached on the rig:

Mgh - Fh = 1/2M(v^2)

where M is mass including the attachment and arm. and -Fh is friction energy lost from the bearing rolling/sliding down the rig. (i want to keep it within 2% of freefall velocity so will need to find a coefficent of friction which allows this)

However we can see the mass doesn't effect the velocity, doesn't the mass change the force/energy it hits the ground with?

Hence:

mgh(freefall) = mgh(rig)

So solving for the height will give me the height at which the object will hit the floor with the same energy/force.

Is this right?

Due to terminal velocity, would I need to solve for this as well ?

Thanks.

I have attached a diagram to help.

#### Attachments

• diagram.doc
24 KB · Views: 74

pbuk
Gold Member
When you attach the rig to the object you are no longer drop-testing the object, you are drop-testing the rig-object system.

Stand on the desk. Jump off. That wasn't too bad was it? Now try it with a load strapped to your back.

I know that hence why I said where the Mass of the 2nd equation is the mass of the object AND rig and that friction must be included.

You didn't answer my question at all.

nasu
Gold Member
Even if you eliminate the friction, the velocity will be the same but energy will not. The energy depends on the mass.
So the body on the rig will have more energy at the impact, assuming zero friction.
The force of impact is harder to estimate. The rig may change the impact time so the force will change in a quite unpredictable way.

pbuk