Question

the cosmoclock 21 ferris wheel in yokohama city, japan, has a diameter of 100 m. its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). a passenger weighs 882 n at the weight - guessing booth on the ground. what is his apparent weight at the lowest point on the ferris wheel? express your answer in newtons. lowest = n

Explanation:

Step1: Find the mass of the passenger

The weight on the ground \( W = mg \), where \( g = 9.8\ m/s^2 \). So \( m=\frac{W}{g}=\frac{882\ N}{9.8\ m/s^2}=90\ kg \).

Step2: Find the radius of the wheel

The diameter \( d = 100\ m \), so radius \( r=\frac{d}{2}=50\ m \).

Step3: Find the angular velocity \( \omega \)

The wheel makes one revolution (\( 2\pi \) radians) in \( T = 60.0\ s \). So \( \omega=\frac{2\pi}{T}=\frac{2\pi}{60.0}\ rad/s=\frac{\pi}{30}\ rad/s \).

Step4: Find the centripetal acceleration \( a_c \)

The formula for centripetal acceleration is \( a_c=\omega^2r \). Substituting values: \( a_c = (\frac{\pi}{30})^2\times50\ m/s^2\approx\frac{9.87}{900}\times50\approx0.548\ m/s^2 \).

Step5: Find the apparent weight at the lowest point

At the lowest point, the apparent weight \( W_{apparent}=mg + ma_c=m(g + a_c) \). Substituting \( m = 90\ kg \), \( g = 9.8\ m/s^2 \), \( a_c\approx0.548\ m/s^2 \): \( W_{apparent}=90\times(9.8 + 0.548)=90\times10.348 = 931.32\ N\approx931\ N \).

Answer:

\( 931\) (or more precisely around \( 931\ N \))