Physics question

Question

An amusement park ride consists of a large vertical cylinder that spins around its axis fast enough such that any person inside is held up against the wall when the floor drops away. The coefficient of static friction between person and wall is µ_s, and the radius of the cylinder is R.

Determine the maximum period of revolution necessary to keep the person from falling assuming that R = 3.50 m and µ_s = 0.600.

Answer 1

Hi Edwin,

Let us consider the balanced forces acting on a person that is inside this spinning cylinder:

1. vertically: force of gravity (F_g, acting downward) is balanced by the force of static friction (F_s, acting upward)

(1) F_g = F_s

(2) mg = µ_s*N where N is the normal force (the force of the wall on his back -- see next step)

1. horizontally: force of the wall on his back (N, acting towards center of cylinder) is balanced by the centrifugal force that he feels is pushing him into the wall (F_c, acting toward the wall)

(3) N = F_c

1. now, let us plug equation (3) into equation (2) (because both of them contain the N vector). Also, remember that centripetal acceleration, F_c = mv²/R

(4) mg = µ_s*F_c

(5) mg = µ_s*(mv²/R)

(6) g = µ_s*(v²/R) (the m's cancel from both sides)

1. we can now express equation (6) in terms of the velocity, v

(7) v² = gR/µ_s

(8) v = square root (gR/µ_s) (we now have the tangential velocity needed to keep this person still)

1. to solve for the period of revolution, we use the simple formula and plug in our calculated tangential velocity v

(9) T = 2πR/v (where 2πR is simply the circumference of this cylinder)

And solving for T here gives you the answer (T represents the rotational period of the spinning cylinder; the amount of time it takes to complete one full turn)