Ask Question
23 February, 23:21

At an amusement park, a Physics 114 takes a ride on a fast-moving Ferris wheel. The apparent weight of the student is different at the top than at the bottom. What is the ratio of the student's apparent weight at the top to the student's apparent weight at the bottom, given that the radius of the loop-the-loop is 27.0 m and the student completes three revolutions every 150 seconds? The mass of the student is 55 kg.

+3
Answers (1)
  1. 24 February, 01:03
    0
    N1 / N2 = 1.0016

    Explanation:

    The apparent weight of the student is the value that a balance would have this corresponds in this case to the normal of the student. Let's write Newton's second law in the lower and upper part of the loop

    For lower rotating wheel

    N1 - W = m a

    a = v² / r

    N1 = W + v² / r

    The relationship between linear and angular velocity is

    v = w r

    As the wheel rotates at a constant speed, we can use angular kinematics

    w = θ / t

    θ = 3 rev (2π rad / 1rev) = 6π rad

    w = 6π / 150

    w = 0.04π rad / s = 0.1257 rad / sec

    calculate

    N1 = mg + w² r

    N1 = 55 9.8 + 0.1257² 27

    N1 = 539 + 0.4267

    N1 = 539.426 N

    Now we perform the same calculation for the top

    -N2 - W = - m a

    N2 = - W + ma

    N2 = 539 - 0.4267

    N2 = 538.5733 N

    The relationship between the weight at the bottom and top is

    N1 / N2 = 539.4267 / 538.5733

    N1 / N2 = 1.0016
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “At an amusement park, a Physics 114 takes a ride on a fast-moving Ferris wheel. The apparent weight of the student is different at the top ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers