Question

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to “take up” the angular momentum (In figure). In $\mathbf{0}\mathbf{.700}\mathbf{}\mathbf{s}$, one arm sweeps throughrole="math" localid="1660995016580" $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{rev}$and the other arms weeps through$\mathbf{1.000}\mathbf{rev}$. Treat each arm as a thin rod of mass$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$and length$\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}$rotating around one end. In the athlete’s reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Windmill motion of the arms during a long jump helps maintain body orientation for a proper landing.

Short answer

Magnitude of the total angular momentum of the arms of athlete is$\mathrm{L}=6.46\left({\text{kg.m}}^{\text{2}}\right)\text{/s}$.

Step-by-step solution

Step 1: Given

$\begin{array}{l}\mathrm{L}=0.6\text{m}\\ {\mathrm{n}}_1=0.5\text{rev}\\ {\mathrm{n}}_2=1.0\text{rev}\end{array}$

Determining the concept

Calculate the angular speed of arms by using revolutions and time. Then, calculate angular momentum due to each arm to find the total angular momentum.

Formula are as follow:

1. $\mathbf{\omega }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{}\mathbf{\times }\mathbf{}\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{revolutions}}{\mathbf{T}}$
   
2. $\mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{=}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}$
   

Determining the magnitude of the total angular momentum of the arms of athlete

Angular velocity is equal to the angular displacement per unit time. For each revolution, arm would cover $2\mathrm{\pi }$rad. So, if the number of revolutions of the arm are known in known time, the angular velocity can be found as,

$\mathrm{\omega }=\frac{2\mathrm{\pi }\times \mathrm{Number}\mathrm{of}\mathrm{revolutions}}{\mathrm{T}}$

Therefore, angular speed of arms,

$\begin{array}{l}{\mathrm{\omega }}_1=\frac{2{\mathrm{\pi n}}_1}{0.7}=\frac{2\mathrm{\pi }\times 0.5}{0.7}=4.49\frac{\text{rad}}{\text{s}}\\ {\mathrm{\omega }}_2=\frac{2{\mathrm{\pi n}}_2}{0.7}=\frac{2\mathrm{\pi }\times 1.0}{0.7}=8.98\frac{\text{rad}}{\text{s}}\end{array}$

Angular momentum of each arm,

$\begin{array}{l}{\mathrm{L}}_1={\mathrm{I}}_1{\mathrm{\omega }}_1=4\times {0.6}^2\times 4.49=6.46{\text{kgm}}^{\text{2}}\text{/s}\\ {\mathrm{L}}_1={\mathrm{I}}_1{\mathrm{\omega }}_1=4\times {0.6}^2\times 8.98=12.92{\text{kgm}}^{\text{2}}\text{/s}\end{array}$

Here, one arm rotates clockwise and other rotates anticlockwise. Hence, total angular momentum about the common rotation axis through the shoulders is,

$\mathrm{L}={\mathrm{L}}_2-{\mathrm{L}}_1=12.92-6.46=6.46\text{kg}\frac{{\text{m}}^{\text{2}}}{\text{s}}$

Hence,the magnitude of the total angular momentum of the arms of athlete is $\mathrm{L}=6.46\left({\text{kg.m}}^{\text{2}}\right)\text{/s}$.

Therefore, using the formula for the angular velocity and law of conservation of angular momentum, the total angular momentum of two arms can be found with reference to athlete’s body.