Who Can Find the Time for a Falling Object with Resistive Force?

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Discussion Overview

The discussion centers around determining the time it takes for a falling object to reach terminal velocity while considering resistive forces. Participants explore the dynamics involved in setting up the equations of motion, particularly focusing on the role of different forms of resistive forces.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how to find the time for a falling object in terms of velocity before reaching terminal velocity, factoring in resistive forces.
  • Another participant suggests choosing a specific form of resistive force, either proportional to inverse velocity or inverse square velocity, and writing down the corresponding differential equation.
  • A participant emphasizes the importance of terminology and clarifies that the discussion is about setting up the dynamic equation rather than differentiating a function.
  • There is a disagreement regarding the nature of the resistive force, with one participant stating that drag typically increases with velocity, countering the suggestion of using "inverse" formulas.
  • One participant proposes a specific equation for the forces acting on the object, stating that the resistive force can be modeled as proportional to velocity, leading to a separable equation for further analysis.
  • Another participant agrees with the approach of using a resistive force proportional to velocity but cautions about the correct signs in the equations, noting that drag acts opposite to the direction of velocity.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate form of the resistive force and its implications for the equations of motion. There is no consensus on the best approach to model the resistive force or the terminology used in the discussion.

Contextual Notes

Participants mention various forms of resistive forces, including linear and quadratic dependencies on velocity, but do not resolve which model is most appropriate for the scenario described. The discussion also highlights the importance of correctly applying signs in the equations.

Fiasal teslla
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For a falling object, who can find the time in terms of the velocity for this journey( before reaching terminal velocity) with considering the resistive force?
 
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Force equals mass times acceleration
Choose what resistive force to use, inverse velocity or inverse square velocity
Write down the differential equation
Solve it
 
I can"! But are you clear on the terminology? Your title asked "how can differentiate this one" implying you have a function to differentiate but that doesn't appear to be the case. You are just asking about setting up the dynamic equation.

As g edgar says, "Force equals mass times acceleration" so ma= m dv/dt= -g- f(v) where "f(v)" is the resistive force. That can be a very complicated function of the velocity depending on the situation. I do not agree with g edgar's "inverse" formulas. Typically, the faster something is going, the greater the drag, not the other way around. Normally, the drag is simplified to either -kv or -kv2 where k is the constant of proportionallity and v is the speed.
 
HallsofIvy said:
I can"! But are you clear on the terminology? Your title asked "how can differentiate this one" implying you have a function to differentiate but that doesn't appear to be the case. You are just asking about setting up the dynamic equation.

As g edgar says, "Force equals mass times acceleration" so ma= m dv/dt= -g- f(v) where "f(v)" is the resistive force. That can be a very complicated function of the velocity depending on the situation. I do not agree with g edgar's "inverse" formulas. Typically, the faster something is going, the greater the drag, not the other way around. Normally, the drag is simplified to either -kv or -kv2 where k is the constant of proportionallity and v is the speed.
When I tried to solve it, I started with this equation
F-R=ma, where F= mg, snd R is the risitive force which equals bv (b is a constant)
So, mg-bv=ma
mg-bv=m dv/dt then we separate the variables to get the time in terms of the velocity

Am I correct with that?
 
Yes, in that case, with resistive force proportional to v, you get a fairly simple equation:m dv/dt= mg- bv, a separable equation. But be careful about signs. If you are taking "upward" to be positive, then it is m dv/dt= -mg- bv. Since drag always acts opposite to velocity, the coefficient of v is always negative.
 

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