How Fast Should a Motorcyclist Go to Stay on a Loop?

[ptrainerjoe]

Heres the problem:

A stunt motorcyclist rides with uniform speed on the inside rim of a vertical circular ramp of radius 8 m. How fast must the motor cyclist travel to aviod leaving the surface at the top of the loop.

the answer in the back of the book says 11m/s but I keep getting 8.9m/s. After working out the formula isn't the final formula v=sqrt{RG}?

Thanks ahead of time for the help!

 

[BLaH!]

What do you mean by "A stunt motorcyclist rides with uniform speed". If the cyclist is traveling at UNIFORM speed as he travels up the ramp, there is a nonconservative force acting on the cyclist and ENERGY IS NOT CONSERVED. Does the problem give you any information about the coefficient of static friction? If so, you can use the following:

W_{nc} = \mu_{s}\int_{0}^{<br /> \pi/2} N(\theta) R d\theta = E_{f} - E_{i} = \frac{1}{2}mv_{f}^2 + mg(2R) - \frac{1}{2}mv_{i}^2

where W_{nc}, N(\theta) is the work done by the nonconservative force and the normal force as a function of angle, respectively. The subscripts i, f denote final and inital points. I took the initial point to be when the cyclist enters the ramp and the final point to be where the cyclist is at the top of the loop. You can simplify the above equation because v_{f} = v_{i}

However, read the question again carefully. You may have confused the wording. It sounds like this is a typical freshman level physics problem and usually they aren't this complicated.

 

[ptrainerjoe]

yes it is a freshman level physics 1 course. We have gone through forces and are just starting work and kinetic energy. The problem is written down exactly as it is in the book. I had a hard time with the way it was written as well that's why i needed the help.

Heres an example of how someone else worked it out. I did it the sameway.

http://physics.mercer.edu/balduz/GenPhys/phy161images/B/b19.pdf

I think my answer is right but the answer in the back of the book says otherwise. I just want to make sure that it is right before i turn it in today at 1pm