How Can We Calculate Rollercoaster Loop Dynamics With Friction?

[igforce]

In a lab we were supposed to produce a rollercoaster ramp while it had a loop, and then to drop a marble at a height where it would finish the entire path without falling off, with enough centripetal acceleration at the loop. we were not given the mass of the marble, and we were only given a meter stick to determine any measurements eg. (height, radius, etc..). We must account for friction.

http://session.masteringphysics.com/problemAsset/1041727/8/YF-07-32.jpg

We must find the velocity at the bottom, side and top of the loop. As well as the g force at each point in terms of mg.

so we did it using conservation of energy to figure out the velocity of the ball at many paths, but we were told to account for friction. (we were not given the mass of the marble, the coeficient of friction, or the efficiency of a ramp). is this problem possible? and how would it be done.

our teacher also said this lab can be done using kinematics as well. how would this be done using kinematics?

thanks

 

[K^2]

Force of friction is typically proportional to the normal force, which in this case, is going to be proportional to the mass of the marble through the entire path.

There is also aerodynamic drag, but you can usually neglect that at speeds you are going to be dealing with.

With the loop included, there is no simple way to account for friction exactly, but you can estimate the average kinetic energy loss per length of the path. On average, it should be fairly consistent, but it will vary through the loop.

With kinematics, you'll find that mass ultimately cancels. However, it's still difficult to account for friction correctly, because on curved sections, normal force will depend on velocity, and rate of change of velocity will depend on friction. This dependence will depend on position, and the rate of change of that depends on velocity. So you end up with a complex set of differential equations that you are unlikely to be able to solve analytically.