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harina [27]
3 years ago
7

On a popular amusement park ride, the riders stand along the inside wall of a large cylinder with a diameter of 8.27 m. The cyli

nder rotates and when it reaches a rotational frequency of 0.66 revolutions per second the floor drops out. What is the minimum coefficient of static friction needed between a rider and the cylinder wall to prevent the rider from moving down?
Physics
1 answer:
____ [38]3 years ago
3 0

Answer:

Explanation:

D = 8.27 m   ⇒  R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W  ⇒  μ*N = m*g   <em>(I)</em>

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation <em>I</em>

<em />

μ*N = m*g    ⇒    μ*m*ω²*R = m*g   ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

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   P₁ = 2.3506 10⁵ Pa

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          A₁ v₁ = A₂ v₂

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          A₂ = π 0.1²

          A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

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          P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

           P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

           P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

           P₁ = 2.3506 10⁵ Pa

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