Chapter 8, Problem 117QAP

To determine

(a)

Average angular speed of the diver when he fell towards water having body tucked in

Answer to Problem 117QAP

Average angular speed = $14.18{\text{ rads}}^{\text{-1}}$

Explanation of Solution

Given info:

Turns made by the diver while he dives = $3.5$

Maximum height reached by the diver = $2.0\text{ m}$

Height to the platform from the water level = $\text{10}\text{.0 m}$

Length of the modeled rod = $2.0\text{ m}$

Diameter of the modeled rod = $0.75\text{ m}$

Formula used:

  $t=\sqrt{\frac{2h}{g}}$

  $t=$ Time taken for the complete the flight

  $h=$ Height

  $g=$ Gravitational acceleration

  $\omega =\text{number of turns*}\frac{2\pi }{t}$

  $\omega =$ Angular momentum

  $t=$ Time taken for the complete the flight

Calculation:

  $\text{Total hiehgt that diver travels}=10.0\text{ m}+2.0\text{ m=12}\text{.0 m}$

  $\begin{array}{l}t=\sqrt{\frac{2h}{g}}\\ t=\sqrt{\frac{2*12\text{ m}}{10{\text{ ms}}^{\text{-2}}}}\\ t=1.55\text{ s}\end{array}$

  $\begin{array}{l}\omega =\text{number of turns*}\frac{2\pi }{t}\\ \omega =\text{3}\text{.5*}\frac{2*3.14}{1.55\text{ s}}\\ \omega =14.18{\text{ rads}}^{\text{-1}}\end{array}$

Conclusion:

Average angular speed = $14.18{\text{ rads}}^{\text{-1}}$

To determine

(b)

Angular speed just after the diver stretched out

Answer to Problem 117QAP

Angular speed just after the diver stretched out= $2.99{\text{ rads}}^{\text{-1}}$

Explanation of Solution

Given info:

Turns made by the diver while he dives = $3.5$

Maximum height reached by the diver = $2.0\text{ m}$

Height to the platform from the water level = $\text{10}\text{.0 m}$

Length of the modeled rod = $2.0\text{ m}$

Diameter of the modeled cylinder = $0.75\text{ m}$

Formula used:

  $I_1{\omega }_1=I_2{\omega }_2$

  $I_1=$ Initial moment of inertia

  ${\omega }_1=$ Initial angular momentum

  $I_2=$ Final moment of inertia

  ${\omega }_2=$ Final angular momentum

  $I_2=\frac{1}{12}m{l}^2$

  $I_1=$ Final moment of inertia

  $m=$ Mass of the diver

  $r=$ Length of the modeled rod

  $I_1=\frac{1}{2}m{r}^2$

  $I_1=$ Initial moment of inertia

  $m=$ Mass of the diver

  $r=$ Radius of the modeled cylinder

Calculation:

  $\begin{array}{l}{I}_1{\omega }_1=I_2{\omega }_2\\ {\omega }_2=\frac{I_1{\omega }_1}{I_2}\\ {\omega }_2=\frac{\frac{1}{2}m{r}^2*14.18{\text{ rads}}^{\text{-1}}}{\frac{1}{12}m{l}^2}\\ {\omega }_2=\frac{\frac{1}{2}*{(0.375\text{ m)}}^{\text{2}}*14.18{\text{ rads}}^{\text{-1}}}{\frac{1}{12}{(2.0\text{ m)}}^2}\\ {\omega }_2=2.99{\text{ rads}}^{\text{-1}}\end{array}$

Conclusion:

Angular speed just after the diver stretched out= $2.99{\text{ rads}}^{\text{-1}}$

To determine

(c)

Rotational kinetic energy difference

Answer to Problem 117QAP

Rotational kinetic energy difference= $418\text{ J}$

Explanation of Solution

Given info:

Turns made by the diver while he dives = $3.5$

Maximum height reached by the diver = $2.0\text{ m}$

Height to the platform from the water level = $\text{10}\text{.0 m}$

Length of the modeled rod = $2.0\text{ m}$

Diameter of the modeled cylinder = $0.75\text{ m}$

Formula used:

  $E=\frac{1}{2}I{\omega }^2$

  $E=$ Rotational kinetic energy

  $I=$ Moment of inertia

  $\omega =$ Angular momentum

  $I_2=\frac{1}{12}m{l}^2$

  $I_1=$ Final moment of inertia

  $m=$ Mass of the diver

  $r=$ Length of the modeled rod

  $I_1=\frac{1}{2}m{r}^2$

  $I_1=$ Initial moment of inertia

  $m=$ Mass of the diver

  $r=$ Radius of the modeled cylinder

Calculation:

  $E=\frac{1}{2}I{\omega }^2$

Kinetic energy difference in the dive,

  $\begin{array}{l}dE=\frac{1}{2}{I}_2{\omega }_2{}^2-\frac{1}{2}{I}_1{\omega }_1{}^2\\ dE=\frac{1}{2}*\frac{1}{2}*75\text{ kg}*{(}^{0.375}*{(}^{14.18}-\frac{1}{2}*75\text{ kg}*\frac{1}{12}{(}^{2.0}{(}^{2.99}\\ dE=530J-112J\\ dE=418J\end{array}$

Conclusion:

Rotational kinetic energy difference= $418\text{ J}$