- #1
Bill M
- 11
- 0
Hey guys,
I'm looking to calculate the initial velocity of a large object sliding to a stop over a surface. This is easy without the drag force of the air, but I'd like to include that to prove a point in my research. As a quick background, I haven't done a differential equation in over a decade since I was an undergrad, so I'm trying to relearn on the fly.
Treating the vehicle as beginning at x=0 and moving in the postitive x direction a known distance D, the equation I came up with is
-fMg - kV2(x) = MV'(x)V(x)
The left side are the forces and the right side is mass times acceleration. f is the adjusted coeff. of friction, M is the object's mass, g is gravitational acceleration, k is a constant encompassing the other constants in front of v2 in the drag force.
I want to find the velocity V with respect to position x. The solution I came up with is:
[itex]V(x) = \sqrt{-\frac{Mfg}{k}(1-e^{\frac{-2k(x-D)}{M}})}[/itex]
This uses the fact that when x = D, V = 0. Does this look correct? My calculations make sense, but I want to be sure. I can include the intermediate steps if necessary, but that's a lot of typing!
Thanks from an old out of practice former physics nerd who now only uses this stuff on occasion.
I'm looking to calculate the initial velocity of a large object sliding to a stop over a surface. This is easy without the drag force of the air, but I'd like to include that to prove a point in my research. As a quick background, I haven't done a differential equation in over a decade since I was an undergrad, so I'm trying to relearn on the fly.
Treating the vehicle as beginning at x=0 and moving in the postitive x direction a known distance D, the equation I came up with is
-fMg - kV2(x) = MV'(x)V(x)
The left side are the forces and the right side is mass times acceleration. f is the adjusted coeff. of friction, M is the object's mass, g is gravitational acceleration, k is a constant encompassing the other constants in front of v2 in the drag force.
I want to find the velocity V with respect to position x. The solution I came up with is:
[itex]V(x) = \sqrt{-\frac{Mfg}{k}(1-e^{\frac{-2k(x-D)}{M}})}[/itex]
This uses the fact that when x = D, V = 0. Does this look correct? My calculations make sense, but I want to be sure. I can include the intermediate steps if necessary, but that's a lot of typing!
Thanks from an old out of practice former physics nerd who now only uses this stuff on occasion.