# Problem with Centripetal velocity on a banked corner

1. Feb 28, 2015

### Astraithious

first of all thanks for reading and especially thank you to people who helped with my previous question here https://www.physicsforums.com/threa...on-change-with-the-angle.798572/#post-5019349

Unfortunately i am having more problems with this instructor insisting she is right and refusing to elaborate so i am here again to ask how her solution here is correct

Example #4:
A bobsled goes into a turn (r=25m) with an banked angle of 40 degrees. Assuming no friction (pretty good assumption), a) what is the velocity of the bobslead b) what happens if the bobsled comes into the same turn above this speed? c) what happens if the bobsled comes into the same turn below this speed?

2. Relevant equations
F=ma
centripetal acceleration = v2/R
Force of static friction = µs * N

3. The attempt at a solution
I have found through another diagram

however i am unfamiliar with how to get
that vmax from the x and y and would love some help there deriving it.

either way though using that vmax for frictionless i get 14.3ms. I think the answer she gets is just not using the right order of operations

2. Feb 28, 2015

### haruspex

It's not clear to me which is your solution and which your teacher's. The handwritten one is fine except for the numerical calculation at the end. It should produce 14.33, or thereabouts.

3. Feb 28, 2015

### Astraithious

the hand written is hers, and yeah i thought so as well. very annoying taking a course with so many errors