In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-d

Question

In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed. It this into a vertical plane, as shown in the figure (Figure 1). Suppose the ring rotates once every 5.10 s. If a rider's mass is 58.0 kg, with how much force does the ring push on her at the top of the ride? Express your answer with the appropriate units. Suppose the ring rotates once every 5.10 s. If a riders mass is 58.0 kg, with how much force does the ring push on her at the bottom of the ride? Express your answer with the appropriate units. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top? Express your answer with the appropriate units.

Explanation / Answer

a)

From the Free Body diagram of the man:

N + mg = mv^2/R

So, N = mv^2/R - mg = m*(v^2/R - g)

= m*((2*pi*R/T)^2/R - g )

= 58*((2*pi*9/5.1)^2/9 - 9.8)

= 223.9 N <--------answer

b)

At the bottom :

N - mg = mv^2/R

So, N = m*(v^2/R + g)

So, N = 58*((2*pi*9/5.1)^2/9 + 9.8) = 1360.7 N <--------answer

c)

At the top: for N = m*((2*pi*R/T)^2/R - g )

For falling off, N = 0

So, m*((2*pi*R/T)^2/R - g ) = 0

So, (2*pi*R/T)^2/R - g = 0

So, 2*pi*R/T = sqrt(Rg)

So, T = (2*pi*R)/sqrt(Rg)

So, T = (2*pi*9)/sqrt(9*9.8) = 6.02 s <--------answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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