Question

In an amusement park ride called The Roundup, passengers stand inside a 17.0 m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.50 s . If a rider's mass is 56.0 kg , with how much force does the ring push on her at the top of the ride?

Suppose the ring rotates once every 4.50 s . If a rider's mass is 56.0 kg , with how much force does the ring push on her at the bottom of the ride?

Answer 1

we are given T = 4.5 ,

so,

w = 2 / T = 2 / 4.5 = 1.396 rad/sec

so,

v = rw

v = (17/2) * 1.396

v = 11.866 m/s

so,

At the top, both normal force and weight act in same direction

N + mg = mv² / r

N = mv² / r - mg

put m = 56 kg, r = 8.5 m, v = 11.866 , g = 9.8 m/s²

N = 379 Newtons

At the bottom, the normal force acts upward and weight is downward

N - mg = mv² / r

N = mv² / r + mg

N = 1477 Newtons