An amusement park ride consists of a rotating vertical cylinder with rough canvas walls. The floor is initial about halfway up the cylinder wall. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down. The rider has a mass of 50 kg, the radius of the cylinder is 5m, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficent of static friction between the rider and the wall is .6

Question

3psilon · Answer

@Algebraic! What would be the Centripetal force on the rider when the floor drops down. I know there is Friction and FG but what is the 3rd force pulling him in? Thanks

anonymous · Answer

the normal force of the cylinder wall...

anonymous · Answer

N = mV^2/r = (50*(2*5)^2 )/5

3psilon · Answer

Ah thanks! I forgot the normal force was perpendicular to the surface of contact ! THANKS

anonymous · Answer

you need it to find the force from friction.