Fidget Spinner!

a) Use the equation that relates constant angular acceleration with angular velocity and time.

b) Use the equation that relates constant acceleration, initial angular velocity, angle and time. Convert your answer in radians to laps.

a) The kinematics equation given constant acceleration can be written as:

\begin{equation*}
\alpha=\frac{\omega_f-\omega_i}{\Delta t},
\end{equation*}

which, with numerical values, yields:

\begin{equation*}
\alpha\approx -3.05\,\text{rad/s}^2.
\end{equation*}

b) For angular displacement, we have:

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

which, with numerical values, gives us:

\begin{equation*}
\Delta\theta=22020 \, \text{rad}.
\end{equation*}

The number of spins will be given as \(\Delta \theta / 2\pi \). This means we can write:

\begin{equation*}
N = 3504.6.
\end{equation*}

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

a) We need to find the angular acceleration due to friction inside the fidget spinner. To find the solution to this problem, we must use the kinematic equations for angular motion. In particular the definition of angular acceleration \(\alpha\) in terms of angular velocities and time, namely

\begin{equation}
\alpha=\frac{\omega_f-\omega_i}{\Delta t},
\end{equation}

where \(\omega_i\) is the angular velocity at the beginning of the motion, \(\omega_f\) is the angular velocity at the end of the motion and \(\Delta t \) is the time interval in which the motion occurs. In our case, we are given the interval of time, \(\Delta t=2\,\text{minutes}=120\,\text{s}\) and the initial angular velocity \(\omega_i=3500\,\text{rpm}\approx 366.5\,\text{rad/s}\). Because the fidget spinner stops after 2 minutes, the final angular velocity is zero \(\omega_f=0\). Using these results in the definition of the angular acceleration, we obtain

\begin{equation}
\alpha=\frac{0-366.5\,\text{rad/s}}{120\,\text{s}}
\end{equation}

\begin{equation}
\alpha\approx -3.05\,\text{rad/s}^2.
\end{equation}

b) Now, we want to calculate how many spins the toy is able to perform before it stops. Now that we know the angular acceleration, we will use the kinematic equation for the angular displacement \(\Delta \theta\) that reads

\begin{equation}
\Delta\theta=\omega_i\Delta t+\frac{1}{2}\alpha \Delta t^2,
\end{equation}

which, after using the numerical values given in the prompt and the value for the angular acceleration found previously, becomes

\begin{equation}
\Delta \theta=(366.5\,\text{rad/s})(120\,\text{s})+\frac{1}{2}(-3.05\,\text{rad/s}^2)(120\,\text{s})^2,
\end{equation}

\begin{equation}
\Delta \theta= 22020\,\text{rad}.
\end{equation}

To find the number of spins \(N\), we divide the result by \(2\pi\,\text{rad}\), which is the equivalent in radians of one spin. Then,

\begin{equation}
N=\frac{\Delta \theta}{2\pi\,\text{rad}},
\end{equation}

which numerically is

\begin{equation}
N=\frac{22020\,\text{rad}}{2\pi\,\text{rad}},
\end{equation}

\begin{equation}
N=3504.6
\end{equation}

Thus, the fidget spinner makes 3505 spins before it stops!

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