Vertical Motion with Linear Drag Derivation

  • Thread starter opprobe
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  • #1
opprobe
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I was looking through a Taylor's Classical Mechanics chapter and I have a question about the derivation.

So basically, it starts out with m[itex]\dot{v}_{y}[/itex]=mg - bv[itex]_{y}[/itex] and you find v[itex]_{ter}[/itex] by letting m[itex]\dot{v}_{y}[/itex]=0 where when you solve for v[itex]_{y}[/itex] you will get v[itex]_{ter}[/itex]. Now the book states how we must now discuss how the projectile approaches that speed and the formula it writes is m[itex]\dot{v}_{y}[/itex] = -b(v[itex]_{y}[/itex] - v[itex]_{ter}[/itex]). Although I get the general idea behind it, can someone explain to me exactly what's going on?

Is this function specific to the 1D case where gravity is the counteracting force to drag?
How was he able to replace v[itex]_{y}[/itex] with v[itex]_{y}[/itex] - v[itex]_{ter}[/itex] and exclude gravity?

Thanks!
 

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  • #2
Simon Bridge
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I was looking through a Taylor's Classical Mechanics chapter and I have a question about the derivation.

So basically, it starts out with m[itex]\dot{v}_{y}[/itex]=mg - bv[itex]_{y}[/itex] and you find v[itex]_{ter}[/itex] by letting m[itex]\dot{v}_{y}[/itex]=0 where when you solve for v[itex]_{y}[/itex] you will get v[itex]_{ter}[/itex]. Now the book states how we must now discuss how the projectile approaches that speed and the formula it writes is m[itex]\dot{v}_{y}[/itex] = -b(v[itex]_{y}[/itex] - v[itex]_{ter}[/itex]). Although I get the general idea behind it, can someone explain to me exactly what's going on?

Is this function specific to the 1D case where gravity is the counteracting force to drag?
How was he able to replace v[itex]_{y}[/itex] with v[itex]_{y}[/itex] - v[itex]_{ter}[/itex] and exclude gravity?

Thanks!
substitute: mg = bvter ... hint: what is the terminal velocity equal to?

This equation is very specific yes, the idea is to give you an idea about derivations. Drag is not normally linear.
 
  • #3
opprobe
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Haha. Alright - I got it thanks!
 

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