Circular Motion of a motorcycle

[clipperdude21]

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!

 

[Astronuc]

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

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

[tex]\mu \frac{v^2}{r}\,=\,g[/tex] or

[tex]v\,=\,\sqrt{\frac{gr}{\mu}}[/tex]

 

[ogb p]

Yes, you have correctly applied the equations for circular motion and forces to solve for the speed of the motorcycle. Your solution of +/- 9.333 m/s is correct. However, it is important to note that since the motorcycle is moving in a circular path, the velocity must always be positive and in the direction of the motion. So the correct answer would be 9.333 m/s. Additionally, it is important to consider the direction of the forces when solving for circular motion problems. In this case, the force of friction is acting towards the center of the circle, while the force of gravity is acting downwards. Therefore, the equations should be written as Fc = mv^2/R, Ffriction = uN, and Fg = mg. Overall, you have correctly applied the concepts of circular motion and forces to solve for the speed of the motorcycle in this scenario.