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 5.0s. If a rider's mass is 58.0kg, with how much force does the ring push on her at the top of the ride?
b)How much force does it 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?
Thank you :)
b)How much force does it 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?
Thank you :)
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Hello
The centrifugal force is
Fc = mω^2*R (with R = radius, ω = 2pi*f , and f = frequency = 1/5 Hz)
Fc = 58*(2pi*1/5)^2*8.5 = 778.5 N.
The forces at top and bottom, respectively, are
778.5 - 9.81*58 N (top)
778.5 + 9.81*58 N (bottom)
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For the lowest possible velocity is the centrifugal force = weight.
mω^2R = mg
w^2 = g/R = 9.81/8.5 = 1.154
ω = 1.074 rad/s
ω = 2pi/T
T = 2pi/ω = 2pi/1.074
T = 5.85 seconds = longest rotation period
-------------------
Regards
The centrifugal force is
Fc = mω^2*R (with R = radius, ω = 2pi*f , and f = frequency = 1/5 Hz)
Fc = 58*(2pi*1/5)^2*8.5 = 778.5 N.
The forces at top and bottom, respectively, are
778.5 - 9.81*58 N (top)
778.5 + 9.81*58 N (bottom)
-------------------------------
For the lowest possible velocity is the centrifugal force = weight.
mω^2R = mg
w^2 = g/R = 9.81/8.5 = 1.154
ω = 1.074 rad/s
ω = 2pi/T
T = 2pi/ω = 2pi/1.074
T = 5.85 seconds = longest rotation period
-------------------
Regards