In an amusement park ride called The Roundup, passengers stand inside a 19.0 m -
ID: 2143721 • Letter: I
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, as shown in the figure (Figure 1) .
A) Suppose the ring rotates once every 4.60
. If a rider's mass is 59.0kg , with how much force does the ring push on her at the bottom of the ride?
C) Part C
Explanation / Answer
m = 59 kg
r = d/2 = 9.5 m
T = 4.6s
T = 2*pi*r/v
==> v = 2*pi*r/T = 12.97 m/s
A) at top point
F = m*g - m*v^2/r
F = -466.54 N(down ward)
B) F = m*g + m*v^2/r
F = 1622.94 N
C) vmin = sqrt(g*r)
vmin = sqrt(9.8*9.5) = 9.65 m/s
Tmax = 2*pi*r/Vmin = 6.183 s
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