Question

During a jump to his partner, an aerialist is to make a quadruple somersault lasting a time $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.87}\mathbf{}\mathbf{s}\mathbf{.}$For the first and last quarter-revolution, he is in the extended orientation shown in Figure, with rotational inertia${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{19.9}\mathbf{}\mathbf{kg}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}\mathbf{.}$around his center of mass (the dot). During the rest of the flight he is in a tight tuck, with rotational inertia$I_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.93}\mathbf{}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$.What must be his angular speedv₂around his center of mass during the tuck?

Short answer

Angular speed around the center of mass is$\mathrm{\omega }=3.23\text{rev/s.}$

Step-by-step solution

Step 1: Given

$\begin{array}{l}{\mathrm{I}}_1=19.9\text{kg}.{\text{m}}^2\\ {\mathrm{I}}_2=3.93\text{kg}.{\text{m}}^2\\ \mathrm{t}=1.87\text{s}\end{array}$

Determining the concept

Find the final angular speed using law of conservation of angular momentum.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula is as follow:

$\mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{=}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}$

Determining the angular speed around the center of mass

According to law of conservation of angular momentum:

$\text{Initialangularmomentum}=\text{Finalangularmomentum}$

$\begin{array}{c}{\mathrm{L}}_1={\mathrm{L}}_2\\ {\mathrm{I}}_1{\mathrm{\omega }}_1={\mathrm{I}}_2{\mathrm{\omega }}_2\\ {\mathrm{\omega }}_1={\mathrm{I}}_2\frac{{\mathrm{\omega }}_2}{{\mathrm{I}}_1}\end{array}$

$\text{Totaltime}\left(\mathrm{t}\right)={\mathrm{t}}_1+{\mathrm{t}}_2$

$\begin{array}{c}\mathrm{t}=\frac{{\mathrm{\theta }}_1}{{\mathrm{\omega }}_1}+\frac{{\mathrm{\theta }}_2}{{\mathrm{\omega }}_2}\\ \mathrm{t}=\frac{{\mathrm{\theta }}_1}{{\mathrm{I}}_2\frac{{\mathrm{\omega }}_2}{{\mathrm{I}}_1}}+\frac{{\mathrm{\theta }}_2}{{\mathrm{\omega }}_2}\\ 1.87=\frac{0.5}{3.93\times \frac{{\mathrm{\omega }}_2}{19.9}}+\frac{3.5}{{\mathrm{\omega }}_2}\end{array}$

By solving above equation for ${\mathrm{\omega }}_2$,

${\mathrm{\omega }}_2=3.23\text{rev}/\text{s}$

Hence,angular speed around the center of mass is$\mathrm{\omega }=3.23\text{rev/s.}$

Therefore, by using the total time and conservation of momentum principle, the final angular speed can be found.