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). part d what would be the time for one revolution if the passenger’s apparent weight at the highest point were zero? express your answer in seconds.

Explanation:

Step1: Recall centripetal acceleration formula

At the highest point, apparent weight \( W' = 0 \). The apparent weight is given by \( W' = mg - F_c \), where \( F_c \) is the centripetal force. If \( W' = 0 \), then \( mg = F_c \). The centripetal force \( F_c = m\frac{v^2}{r} \), so \( mg = m\frac{v^2}{r} \), which simplifies to \( v = \sqrt{gr} \). The radius \( r=\frac{d}{2}=\frac{100\,\text{m}}{2} = 50\,\text{m} \), and \( g = 9.8\,\text{m/s}^2 \). So \( v=\sqrt{9.8\times50}=\sqrt{490}\approx22.14\,\text{m/s} \).

Step2: Relate velocity to period

The circumference of the Ferris wheel is \( C = 2\pi r=2\pi\times50 = 100\pi\,\text{m} \). The period \( T \) (time for one revolution) is related to velocity \( v \) by \( v=\frac{C}{T} \), so \( T=\frac{C}{v} \). Substituting \( C = 100\pi \) and \( v\approx22.14 \), we get \( T=\frac{100\pi}{22.14}\approx\frac{314.16}{22.14}\approx14.19\,\text{s} \). Wait, actually, from \( mg = m\frac{4\pi^2r}{T^2} \) (since \( v=\frac{2\pi r}{T} \), so \( v^2=\frac{4\pi^2r^2}{T^2} \), and substituting into \( mg = m\frac{v^2}{r} \) gives \( g=\frac{4\pi^2r}{T^2} \)), so solving for \( T \): \( T = 2\pi\sqrt{\frac{r}{g}} \).

Step3: Calculate period using correct formula

Using \( T = 2\pi\sqrt{\frac{r}{g}} \), with \( r = 50\,\text{m} \), \( g = 9.8\,\text{m/s}^2 \). First, \( \frac{r}{g}=\frac{50}{9.8}\approx5.102 \). Then \( \sqrt{5.102}\approx2.259 \). Then \( T = 2\pi\times2.259\approx4.518\pi\approx14.19\,\text{s} \). Wait, but let's do it precisely. \( T = 2\pi\sqrt{\frac{50}{9.8}} \). Calculate \( \sqrt{\frac{50}{9.8}}=\sqrt{\frac{500}{98}}=\sqrt{\frac{250}{49}}=\frac{\sqrt{250}}{7}=\frac{5\sqrt{10}}{7}\approx\frac{5\times3.1623}{7}\approx\frac{15.8115}{7}\approx2.2588 \). Then \( T = 2\times3.1416\times2.2588\approx6.2832\times2.2588\approx14.19\,\text{s} \).

Wait, maybe I made a mistake earlier. Let's rederive. At highest point, apparent weight is zero, so normal force \( N = 0 \). The net force towards the center is \( mg = m\frac{v^2}{r} \), so \( v = \sqrt{gr} \). The period \( T \) is the time to go around the circle, so \( T=\frac{2\pi r}{v} \). Substituting \( v = \sqrt{gr} \), we get \( T = \frac{2\pi r}{\sqrt{gr}}=2\pi\sqrt{\frac{r}{g}} \), which is the same as the formula for the period of a circular motion where the centripetal acceleration is \( g \). So plugging in \( r = 50\,\text{m} \), \( g = 9.8\,\text{m/s}^2 \):

\( T = 2\pi\sqrt{\frac{50}{9.8}} \)

Calculate \( \frac{50}{9.8} \approx 5.10204 \)

\( \sqrt{5.10204} \approx 2.25877 \)

\( 2\pi\times2.25877 \approx 6.28319\times2.25877 \approx 14.2 \) seconds (approximate). Let's do it more accurately:

\( \sqrt{\frac{50}{9.8}} = \sqrt{\frac{500}{98}} = \sqrt{\frac{250}{49}} = \frac{\sqrt{250}}{7} = \frac{5\sqrt{10}}{7} \)

\( \sqrt{10} \approx 3.16227766 \)

So \( 5\sqrt{10} \approx 15.8113883 \)

\( \frac{15.8113883}{7} \approx 2.25876976 \)

\( 2\pi\times2.25876976 \approx 4.51753952\pi \approx 14.19 \) seconds, which is approximately 14.2 seconds. But let's check the formula again. The key is that when apparent weight is zero, the centripetal force is provided by gravity, so \( mg = m\frac{4\pi^2r}{T^2} \), so \( T = 2\pi\sqrt{\frac{r}{g}} \), which is correct. So with \( r = 50\,\text{m} \), \( g = 9.8\,\text{m/s}^2 \), we calculate \( T \approx 14.2\,\text{s} \).

Wait, but maybe the problem is expecting using \( g = 9.8 \) and precise calculation. Let's compute \( T = 2\pi\sqrt{\frac{50}{9.8}} \):

First, \( 50/9.8 = 5.102040816 \)

Square root of that is \( \sqrt{5.102040816} = 2.25877 \)

Multiply by \( 2\pi \): \( 2.25877 \times 6.28319 = 14.19 \) seconds. So approximately 14.2 seconds, or more precisely, let's use a calculator:

\( T = 2 \times \pi \times \sqrt{\frac{50}{9.8}} \)

\( \sqrt{\frac{50}{9.8}} \approx 2.25876976 \)

\( 2 \times \pi \times 2.25876976 \approx 14.19 \) seconds. So the time for one revolution is approximately 14.2 seconds.

Answer:

\boxed{14.2} (or more precisely, around 14.2 seconds; the exact value is approximately \( 2\pi\sqrt{\frac{50}{9.8}} \approx 14.2 \) s)