Vertical Motion with Linear Drag Derivation

In summary, the user is asking for an explanation of a formula in Taylor's Classical Mechanics chapter regarding the approach of a projectile's speed. The formula is m\dot{v}_{y} = -b(v_{y} - v_{ter}), where v_{ter} is the terminal velocity found by setting m\dot{v}_{y}=0 and solving for v_{y}. They also ask if this formula is only applicable in the 1D case where gravity is the counteracting force to drag, and how v_{y} was replaced with v_{y} - v_{ter} while excluding gravity. The responder clarifies that the formula is specific to derivations and drag is not normally linear.
  • #1
opprobe
17
0
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
opprobe said:
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
Haha. Alright - I got it thanks!
 

1. What is the equation for vertical motion with linear drag?

The equation for vertical motion with linear drag is m(d^2y/dt^2) = mg - kv, where m is the mass of the object, g is the acceleration due to gravity, k is the drag coefficient, and v is the velocity.

2. How is velocity affected by linear drag?

Velocity is constantly decreasing due to the drag force, which is directly proportional to the velocity. As the object falls, the drag force increases, causing the velocity to decrease at a faster rate.

3. What is the relationship between the drag coefficient and the surface area of the object?

The drag coefficient is directly proportional to the surface area of the object. This means that the larger the surface area, the greater the drag force and the slower the object will fall.

4. Can the mass of the object affect its vertical motion with linear drag?

Yes, the mass of the object does affect its vertical motion with linear drag. A heavier object will experience a greater force of gravity, but it will also have a greater inertia, making it more difficult for the drag force to slow it down. A lighter object will be more easily affected by the drag force.

5. How does the initial velocity of the object affect its motion with linear drag?

The initial velocity of the object will determine how quickly it will reach its terminal velocity. If the initial velocity is high, the object will reach its terminal velocity sooner and continue to fall at a constant speed. If the initial velocity is low, it will take longer for the object to reach its terminal velocity and it will continue to accelerate until it reaches that point.

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