Conceptual problem involving centripetal accel

[physics noob]

amusement park ride...basically a large cylinder rotating about a vertical axis. when it is spinning fast enough the floor is dropped out and the riders stick to the wall well above the floor. find the minimum Vtangential needed. express answer in terms of R of circle, coefficent of static friction (mew)s and gravitational acell (g)... seems easy right, and it probably is, but i don't know why my answer is not right...btw the correct answer is

V= (gR/mews)^.5

here's what i got so far...

x direction forces) V^2/ (r) * m - static coeff *n = 0

y direction forces) static coeff *n - mg = 0


so mg = sf *n so v^2/ R = g so



V = (g/R)^.5 where does the sfriction coeff come from?!?

i worked backwards and saw that is must have been multiplied by
v^2/R...now why would you do that... thanks in advanced

 

[abercrombiems02]

Well this is actually a 2-D problem, the FBD for the rider should be something as follows.

We have the frictional force upwards, a weight force downwards, and a normal force radially inwards.

Now use sumF = ma

In the vertical direction we have
Friction - Weight = 0
u*N = mg

In the radially inward direction we have

Normal = Centripetal acceleration
N = mv^2/r

thus

u*m*v^2/r = m*g

u*v^2/r = g

v = sqrt(r*g/u)

 

[sniffer]

vertically, mg=(mew)*N i.e. equilibrium

horizontally, N= mv^2/r.

i.e. not equilibrium, but rather the normal force provides the centripetal force.

substitute N into above, get V=(g*r/(mew))^0.5.

typically, when someone learns centripetal force for the 1st time, they get confused with the whole forces producing the centripetal force.

centripetal force is the net force acting on the object, it is the sum of all external forces. If this net force is not directed perpendicular to the velocity of the object, then the path is not going to be circular.

the object is not in equilibrium on horizontal plane, but is accelerated.

you need to sit back and think deeply about this.

 

[physics noob]

thanks a lot guys,,,, what through me off was i thought there was friction on the horizontal, but it seems it only exists on the vertical forces, i really appreciate the help, this site is amazing... thanks for your time