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Yanka [14]
2 years ago
10

A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670

N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (that is, the normal force exerted on her by the bench) of the person as she passes through the highest pointof her motion?
Physics
1 answer:
Nikolay [14]2 years ago
4 0

Answer:  459.14 N

Explanation:

from the question, we have

diameter = 10 m

radius (r)  = 5 m

weight (Fw) = 670 N

time (t) = 8 seconds

Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below

∑ F = F c =  F w − Fn ..............equation 1

Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2

substituting the value of v as (2πr / T) we now have

Fn = mg − (m(2πr / T )^2) / r

Fn= mg − (4(π^2)mr / T^2)   ..........equation 3

Fw (mass of the person) = mg

therefore m = Fw / g

                m = 670 / 9.8 = 68.367 kg

now substituting  our values into equation 3

Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)

Fn = 670 - 210.86

Fn = 459.14 N

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Answer:

X-Positions:                                         Y-Positions

x(0) = 0                                                   y(0) = 0

x(2) = 120 m                                           y(2) = 19.6 m

x(4) = 240 m                                          y(4) = 78.4 m

x(6) = 360 m                                          y(6) = 176.4 m

x(8) = 480 m                                          y(8) = 313 m

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Explanation:

X-Positions

  • First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
  • After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
  • So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

       x = v_{ox} * t = 60.0 m/s * t(s)

  • It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

  • We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
  • As both axes are  perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
  • At any moment, it is subject to the acceleration of gravity, g.
  • As the acceleration is constant, we can find the vertical displacement (taking the  height of the cliff as the initial reference level), using the following kinematic equation:

       y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}

  • Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
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  • y(6) = 18*9.8 m/s2 = 176.4 m
  • y(8) = 32*9.8 m/s2 = 313.6 m
  • y(10)= 50 * 9.8 m/s2 = 490.0 m
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