Yes, normal force = m v^2/r
max friction force up = mu m v^2/r
weight down = mg
slip when m g = m v^2/r
v^2 = r g
The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out.
A-What force keeps the people from falling out the bottowm of the ride?
I think that it's the Normal force?
B-If the coefficient of friction is 0.40 and the cylinder has a radius of 2.5m, what is the minimum angular speed the of the cylinder so the people don't fall out?
Do I use NL #2 Sum F=mv2/r
which=ma(sub)c ??
Also, how do I set up the math to solve for the angular speed?
Thank you
2 answers
Yes, normal force = m v^2/r
max friction force up = mu m v^2/r
weight down = mg
slip when m g = mu m v^2/r
v^2 = r g / mu
max friction force up = mu m v^2/r
weight down = mg
slip when m g = mu m v^2/r
v^2 = r g / mu
1 answer