Circular Motion of a motorcycle

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SUMMARY

The discussion focuses on calculating the necessary speed for a motorcycle to maintain a horizontal circular path inside a vertical cylinder with a radius of 8 meters and a coefficient of static friction of 0.9. The centripetal force equation, Fc = mv²/R, is utilized alongside the frictional force equation, Ff = μN, leading to the conclusion that the motorcycle must travel at a speed of approximately 9.333 m/s. This speed is derived from the simplified formula v = √(gr/μ), confirming that the motorcyclist can travel in either direction with the same speed.

PREREQUISITES
  • Understanding of centripetal force and its equation (Fc = mv²/R)
  • Knowledge of static friction and its coefficient (μ)
  • Familiarity with Newton's laws of motion
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of the centripetal force equation in detail
  • Learn about the effects of different coefficients of friction on circular motion
  • Explore real-world applications of circular motion in vehicles
  • Investigate the dynamics of motorcycle handling in various conditions
USEFUL FOR

Physics students, motorcycle enthusiasts, and engineers interested in vehicle dynamics and circular motion principles.

clipperdude21
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Circular Motion! :)

1. 12. A motorcycle moves in a horizontal circular path on the inside surface of a vertical
cylinder of a radius 8 m. Assuming that the coefficient of static friction between the
wheels of the motorcycle and the wall is 0.9, how fast must the motorcycle move so that
it stays in the horizontal path?




2. Fc=mv^2/R, Force friction= uN, Fg=mg



3. The way I did it was I did the sum of forces in the y direction as sum of forces in the y direction=uN-mg=0. I solved for N and got N=mg/u.

I then plugged it into Force centripetal=N. I got mv^2?r= mg/u. I solved for v and got +/- 9.333 m/s. Did I do this right? :) Thanks!
 
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Yes, the answer is correct. In this case the motorcyclist could travel in either direction with the same magnitude of tangential velocity.

A more straightforward approach to the problem would be to realize that

\mu \frac{mv^2}{r}\,=\,mg and the m's divide out so

\mu \frac{v^2}{r}\,=\,g or

v\,=\,\sqrt{\frac{gr}{\mu}}
 

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