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 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.

Answer 1

N1 / N2 = 1.0016

Step-by-step 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