The maximum frictional force that the wall can exert must equal the person's weight at the minimum required rotational speed. Call that angular speed, in radians per second), w
The centripetal force is M R w^2 and the maximum frictional force is
M R w^2*0.49
Set that equal to M g and solve for w. The mass M cancels out.
w = sqrt[g/(0.49 R)]= ?
Multiply w in rad/s by
60 s/min/(2 pi rad/rev) to get r.p.m.
The Wall of Death in an amusement park is comprised of a vertical cylinder that can spin around the vertical axis. Riders stand against the wall of the spinning cylinder and the floor falls away leaving the riders held up by friction. The radius of the cylinder is 3.8 m and the coefficient of static friction between the rider and the wall is 0.49. Find the minimum number of revolutions per minute necessary so that the riders do not slip down the wall.
I have no clue which equation/steps I need to do for this. Can I please get some help?
1 answer
1 answer