Ferris Wheel!

a) Use the kinematics equation for angular displacement given a constant angular acceleration.

b) Using the same equation as part (a), but solve for \(t\) and use the angle for 10 laps.

c) Same hint as part (b), but find the time it takes for 9 laps, and then find the difference for the time for 10 laps.

a) The kinematics equation for constant acceleration angular motion gives:

\begin{equation*}
\theta=\theta_i+\omega_it+\frac{1}{2}\alpha t^2,
\end{equation*}

which, after solving for \(\alpha\), we get:

\begin{equation}
\alpha=\frac{2\theta}{t^2},
\end{equation}

Plugging in numerical values yields:

\begin{equation*}
\alpha\approx 7.85\times 10^{-3}\,\text{rad/s}^2.
\end{equation*}

b) Using the same equation as part  (a), but solving for \(t\), we get:

\begin{equation*}
t=\sqrt{\frac{2\theta}{\alpha}},
\end{equation*}

where 10 laps means \(\theta = 10 \cdot 2 \pi\), so:

\begin{equation*}
t_{10} \approx 126.56 \, \text{s}.
\end{equation*}

c) The same equation for the time as part (b), for 9 laps we get:

\begin{equation*}
t_{9} \approx 120 \, \text{s}.
\end{equation*}

The time difference is:

\begin{equation*}
\Delta t_{10\,\text{th}}=t_{10}-t_9 = 6.5 \, \text{s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) We’ll use the kinematic equations for constant acceleration angular motion as well as the given values to solve for \(\alpha\), the angular acceleration. The equation of angular displacement \(\theta\) as a function of time \(t\) is

\begin{equation}
\label{angularkinem}
\theta=\theta_i+\omega_it+\frac{1}{2}\alpha t^2,
\end{equation}

where \(\theta_i\) is the initial angle with respect to some coordinate system, \(\omega_i\) is the initial angular speed. Because the attraction starts from rest we can say \(\omega_i=0\). Its initial angular position will also be zero \(\theta_i=0\), while its angular position after a time \(t=40\,\text{s}\) has passed is \(\theta=2\pi\,\text{rad}\). We can then write equation \eqref{angularkinem} as

\begin{equation}
\label{kinemsimp}
\theta=\frac{1}{2}\alpha t^2,
\end{equation}

which, after solving for \(\alpha\), we get

\begin{equation}
\alpha=\frac{2\theta}{t^2}.
\end{equation}

Using the numerical values, we obtain

\begin{equation}
\alpha=\frac{2(2\pi\,\text{rad})}{(40\,\text{s})^2},
\end{equation}

\begin{equation}
\alpha\approx 7.85\times 10^{-3}\,\text{rad/s}^2.
\end{equation}

b) Now, we need to find the time it takes for the Ferris wheel to complete 10 full laps. Since we know the angular acceleration, we can use equation \eqref{kinemsimp} with \(\theta=10(2\pi)\,\text{rad}\) and the value we found for \(\alpha\) to solve for the time \(t\). We get

\begin{equation}
t^2=\frac{2\theta}{\alpha},
\end{equation}

and taking the square-root on both sides

\begin{equation}
\label{time}
t=\sqrt{\frac{2\theta}{\alpha}}.
\end{equation}

Using the numerical values, for the time it takes to complete 10 full laps \(t_{10}\), we get

\begin{equation}
t_{10}=\sqrt{\frac{2(10(2\pi\,\text{rad}))}{7.85\times 10^{-3}\,\text{rad/s}^2}},
\end{equation}

\begin{equation}
t_{10}\approx 126.5\,\text{s}.
\end{equation}

c) In order to find the time it takes to complete the 10th lap, we can find the time it takes to complete 10 full laps and the time it takes to complete 9 full laps and take their difference. From the result of part (b) we already know the time it takes to complete 10 full laps. We can then use equation \eqref{time} and use \(\theta=9(2\pi)\) to find the time it takes to complete 9 full laps. Explicitly,

\begin{equation}
t_9=\sqrt{\frac{2(9(2\pi\,\text{rad}))}{7.85\times10^{-3}\,\text{rad/s}^2}},
\end{equation}

\begin{equation}
t_9 = 120\,\text{s}.
\end{equation}

Thus, the time it takes to make the 10th lap \(\Delta t_{10\,\text{th}}\) is

\begin{equation}
\Delta t_{10\,\text{th}}=t_{10}-t_9,
\end{equation}

which numerically is

\begin{equation}
\Delta t_{10\,\text{th}}=126.5\,\text{s}-120\,\text{s}=6.5\,\text{s}.
\end{equation}

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