- #1
dt00073
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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.
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.