Understanding Dynamics: Solving a Tricky Example

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SUMMARY

The discussion centers on a dynamics problem from a specific example in a PDF resource. The user, robcowlam, seeks clarification on the integration process leading to the equation F/k [sin(kt)+Uk(cos kt - 1)] - mgt = mv. The key point is the proper evaluation of the definite integral, which includes the evaluation at the limits, resulting in the additional term of -1. This clarification emphasizes the importance of correctly applying calculus principles in dynamics problems.

PREREQUISITES
  • Understanding of definite integration in calculus
  • Familiarity with dynamics concepts, particularly forces and motion
  • Knowledge of trigonometric functions and their properties
  • Basic proficiency in evaluating integrals
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  • Review the principles of definite integration in calculus
  • Study dynamics problems involving forces and motion
  • Learn about the application of trigonometric identities in physics
  • Practice solving similar dynamics problems from academic resources
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This discussion is beneficial for students studying dynamics, educators teaching calculus and physics, and anyone looking to enhance their understanding of integration in the context of motion and forces.

robcowlam
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HI guys,

Im having some trouble with a dynamics question, I am only looking at a worked example however I can't see where the final answer comes from.
The problem is sample 3/5 on http://www.eng.auburn.edu/~marghitu/MECH2110/Dynamics/D2_1_examples.pdf
I follow the working until the line after "which becomes" I cannot see why the integral becomes
F/k [sin(kt)+Uk(cos kt - 1)] - mgt = mv
rather than:

F/k [sin(kt)+Uk(cos kt)] - mgt = mv

Its probably just something small but I can't figure it out,
Can anybody help me out?

Thanks!
 
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robcowlam: That is normal calculus (definite integration). Thus, integral[-mu*F*sin(k*t)*dt], integrated from t = 0 to t, is equal to -mu*(F/k)*{-cos(k*t) - [-cos(k*0)]} = -mu*(F/k)*[-cos(k*t) - (-1)] = mu*(F/k)*[cos(k*t) - 1], not mu*(F/k)*cos(k*t).
 
Ah yes I see it now, I wasn't evaluating the integral, thanks for your help.
 

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