In an amusement park ride called The Roundup, passengers stand inside a 16.0 m -

Question

In an amusement park ride called The Roundup, passengers stand inside a 16.0 m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure

A) Suppose the ring rotates once every 4.90 s . If a rider's mass is 57.0 kg , with how much force does the ring push on her at the top of the ride?

B)Suppose the ring rotates once every 4.90 s . If a rider's mass is 57.0 kg , 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

velcoity V = W/t = 2piR/t

V = (2*3.14 * 8/4.9) = 10.25 m/s

now for net force

-mv^2/R = - N - mg

N + mg = mv^2/R

N = mv^2 / r - mg

N = m (v^2 / r - g)

N = 57 * (10.25^2/8 - 9.8)

N = 189.81 N is the forcr at top

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B. force at Bottom

N = mg + mv^2 / r

N = m (g + v^2 / r)

N = 57*(9.8 + 10.57^2/8)

N = 1354.63 N
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C. or that matter , N = 0 at the top ,

- mg = -mv^2 / r

v^2 = gr

v at top = sqrt (9.8 * 8 )

V = 8.85 m/s

T = 2piR/V

T = (2*3.14 * 8/8.85)

T = 5.67 secs

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