In an amusement park ride called The Roundup, passengers stand inside a 19.0m -d

Question

In an amusement park ride called The Roundup, passengers stand inside a 19.0m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.Suppose the ring rotates once every 4.60s .

A) If a rider's mass is 55.0kg , with how much force does the ring push on her at the top of the ride?

B) Suppose the ring rotates once every 4.60s . If a rider's mass is 55.0kg , with how much force does the ring push on her at the bottom of the ride?

C)What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Explanation / Answer

The tangential speed of the ring = (pi D)/4.6s, =12.97m/sec. when still horizontal.
Centripetal acceleration = (v^2/r) = (12.97^2)/9.5m., =17.70 m/sec^2.
It is then tilted to the vertical.
A) The acceleration at the top of the spin = (17.70 - g) =17.70 - 9.8, = 7.90 m/sec^2.
Force on 55kg. person = (ma) = 55 x 7.90, = 434.91 N.

B) The acceleration at the bottom of the spin = (17.70 + g) =17.70 + 9.8, = 27.50 m/sec^2.
Force on 55kg. person = (ma) = 55 x 27.50 = 1512.91 N.

C)   v^2(top) = minimum to be in circle = r g and T(top) =0 that is just sufficient to provide acceleration = g at top so that it crosses top point.

v (top) = sqrt [(19/2) * 9.8] = 9.648 m/s

for T (top)=0 just pass without fall
V (top) = 9.648 m/s = r w = 2 pi r / T(period)
T = 2*3.14*19/2*9.648 = 6.18 secs

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.