Question

the cosmodock 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). part c what is his apparent weight at the highest point on the ferris wheel? express your answer in newtons.

Explanation:

To solve the problem of finding the apparent weight at the highest point of the Ferris wheel, we need to consider the following:

Step 1: Identify the given values

• Diameter of the Ferris wheel, \( d = 100 \, \text{m} \), so the radius \( r = \frac{d}{2} = 50 \, \text{m} \)
• Time for one revolution, \( T = 60.0 \, \text{s} \)
• Let's assume the mass of the person is \( m \) (we need to know the mass to calculate the weight. Since it's not provided, we'll assume a typical mass, say \( m = 70 \, \text{kg} \) for illustration purposes. If the mass is different, the answer will change accordingly.)

Step 2: Calculate the angular velocity \( \omega \)

The angular velocity \( \omega \) is given by:
\[
\omega = \frac{2\pi}{T}
\]
Substituting \( T = 60.0 \, \text{s} \):
\[
\omega = \frac{2\pi}{60.0} \approx 0.1047 \, \text{rad/s}
\]

Step 3: Calculate the centripetal acceleration \( a_c \)

The centripetal acceleration \( a_c \) is given by:
\[
a_c = \omega^2 r
\]
Substituting \( \omega \approx 0.1047 \, \text{rad/s} \) and \( r = 50 \, \text{m} \):
\[
a_c = (0.1047)^2 \times 50 \approx 0.548 \, \text{m/s}^2
\]

Step 4: Calculate the apparent weight at the highest point

At the highest point of the Ferris wheel, the apparent weight \( w_{\text{highest}} \) is given by:
\[
w_{\text{highest}} = m(g - a_c)
\]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Substituting \( m = 70 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a_c \approx 0.548 \, \text{m/s}^2 \):
\[
w_{\text{highest}} = 70 \times (9.8 - 0.548) = 70 \times 9.252 \approx 647.64 \, \text{N}
\]

Final Answer

If the mass of the person is \( 70 \, \text{kg} \), the apparent weight at the highest point is approximately \(\boxed{648}\) Newtons (rounded to a reasonable number of significant figures).

(Note: If the mass is different, substitute the correct mass into the formula \( w_{\text{highest}} = m(g - \frac{4\pi^2 r}{T^2}) \) to get the accurate result.)

Answer:

To solve the problem of finding the apparent weight at the highest point of the Ferris wheel, we need to consider the following:

Step 1: Identify the given values

• Diameter of the Ferris wheel, \( d = 100 \, \text{m} \), so the radius \( r = \frac{d}{2} = 50 \, \text{m} \)
• Time for one revolution, \( T = 60.0 \, \text{s} \)
• Let's assume the mass of the person is \( m \) (we need to know the mass to calculate the weight. Since it's not provided, we'll assume a typical mass, say \( m = 70 \, \text{kg} \) for illustration purposes. If the mass is different, the answer will change accordingly.)

Step 2: Calculate the angular velocity \( \omega \)

The angular velocity \( \omega \) is given by:
\[
\omega = \frac{2\pi}{T}
\]
Substituting \( T = 60.0 \, \text{s} \):
\[
\omega = \frac{2\pi}{60.0} \approx 0.1047 \, \text{rad/s}
\]

Step 3: Calculate the centripetal acceleration \( a_c \)

The centripetal acceleration \( a_c \) is given by:
\[
a_c = \omega^2 r
\]
Substituting \( \omega \approx 0.1047 \, \text{rad/s} \) and \( r = 50 \, \text{m} \):
\[
a_c = (0.1047)^2 \times 50 \approx 0.548 \, \text{m/s}^2
\]

Step 4: Calculate the apparent weight at the highest point

At the highest point of the Ferris wheel, the apparent weight \( w_{\text{highest}} \) is given by:
\[
w_{\text{highest}} = m(g - a_c)
\]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Substituting \( m = 70 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a_c \approx 0.548 \, \text{m/s}^2 \):
\[
w_{\text{highest}} = 70 \times (9.8 - 0.548) = 70 \times 9.252 \approx 647.64 \, \text{N}
\]

Final Answer

If the mass of the person is \( 70 \, \text{kg} \), the apparent weight at the highest point is approximately \(\boxed{648}\) Newtons (rounded to a reasonable number of significant figures).

(Note: If the mass is different, substitute the correct mass into the formula \( w_{\text{highest}} = m(g - \frac{4\pi^2 r}{T^2}) \) to get the accurate result.)