Motion under influence of a resistive force

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SUMMARY

The discussion focuses on the motion of a body influenced by a resistive force defined by F = -m*γ*vα, where m is mass, γ is a constant, v is velocity, and α is a dimensionless constant. The equation of motion v'' = -γ*vα leads to the time equation t = (1/γ)*(V0^(1-α))/(1-α), indicating that for γ > 1, the time to stop is finite. However, the sign of t becomes negative for certain values of α, particularly when α = 2, which raises questions about the physical realism of such scenarios. The discussion concludes that while t is positive for α = 0, it becomes non-physical for α values greater than 1.

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  • Understanding of classical mechanics and Newton's laws of motion
  • Familiarity with differential equations and their applications in physics
  • Knowledge of resistive forces and their mathematical representation
  • Basic grasp of dimensional analysis and physical realism in mathematical models
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  • Explore the implications of different values of α on motion dynamics
  • Study the behavior of resistive forces in various physical contexts
  • Investigate the solutions to differential equations in classical mechanics
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Students and professionals in physics, particularly those studying dynamics and resistive forces, as well as researchers interested in mathematical modeling of motion under various forces.

Ananthan9470
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Consider the 1d motion of a body under the influence of the force given by F = -m*γ*vα. m is mass, γ is a constant of appropriate dimension, v is velocity and α is dimensionless constant. The value of α for which the motion will come to a stop in finite time is to be calculated. I solved the equation of motion given by v'' = -γ*vα and got an equation for t given by t = (1/γ)*(V01-α)/(1-α); V0 is the initial velocity. According to this, for all γ>1, time is finite. But the sign of t is negative. Why is this happening? Am I doing something wrong?
 
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Your solution for t is correct. What do you get for t if ##\alpha = 0##? Is this a finite time? Is it positive?
 
Chestermiller said:
Your solution for t is correct. What do you get for t if ##\alpha = 0##? Is this a finite time? Is it positive?
For ##\alpha = 0## it is positive but what about something like α = 2? It is turning out to be negative unless I am mistaken. right?
 
Ananthan9470 said:
For ##\alpha = 0## it is positive but what about something like α = 2? It is turning out to be negative unless I am mistaken. right?
Yes. That's the region that is not physically realistic.

What do you get when ##\alpha = -1##? Is that a finite time? What do you get for v vs t in the special case when ##\alpha = 1##?
 

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