Chapter 17.1, Problem 17.20P

To determine

(a)

The angular velocity and force exerted on the hands of the gymnast after he has rotated through 90°.

Answer to Problem 17.20P

${\omega }_2=3.94rad/s$

$R_x=270.35lb$

Explanation of Solution

Given:

Calculation:

Position 1: Directly above the bar.

Elevation: $h_1=3.5ft$

Potential energy: $V_1=W{h}_1=\left(160\right)\left(3.5\right)=560ft.lb$

Speeds: ${\omega }_1=0,v_1=0$

Kinetic energy: $T_1=0$

Position 2: Body at level of bar after rotating 90°.

Elevation: $h_2=0$

Potential energy: $V_2=0$

Speeds: $v_2=3.5{\omega }_2$

Kinetic energy: $T_2=\frac{1}{2}m{v}_2^2+\frac{1}{2}m{k}^2{\omega }_2^2$

$\begin{array}{l}{T}_2=\frac{1}{2}\times \left(\frac{160}{32.2}\right){\left(3.5{\omega }_2\right)}^2+\frac{1}{2}\times \left(\frac{160}{32.2}\right){\left(1.5\right)}^2{\omega }_2^2\\ T_2=\frac{1}{2}\times \left(\frac{160}{32.2}\right)\left[{\left(3.5{\omega }_2\right)}^2+{\left(1.5\right)}^2{\omega }_2^2\right]\\ T_2=2.4844\left[12.25{\omega }_2^2+2.25{\omega }_2^2\right]\\ T_2=36.025{\omega }_2^2\end{array}$

Applying Principle of conservation of energy

$\begin{array}{l}{T}_1+V_1=T_2+V_2\\ 0+560=36.025{\omega }_2^2+0\\ 560=36.025{\omega }_2^2\\ {\omega }_2^2=\frac{560}{36.025}\\ {\omega }_2^2=15.545\\ {\omega }_2=\sqrt{15.545}=3.94rad/s\end{array}$

From the kinematics of figure:

$\begin{array}{l}{a}_t=3.5\alpha \\ a_n=3.5{\omega }_2^2=3.5\left(15.545\right)=54.407ft/s^2\end{array}$

Taking moment about point O,

$\begin{array}{l}\Sigma M_0=\Sigma {\left(M_0\right)}_{eff}\\ 3.5\left(160\right)=\left(\frac{160}{32.2}\right)\left(3.5\right)\left(3.5\alpha \right)+\left(\frac{160}{32.2}\right){\left(1.5\right)}^2\alpha \\ 560=\left(\frac{160}{32.2}\right)\left[12.25\alpha +2.25\alpha \right]\\ 560=72.05\alpha \\ \alpha =\frac{560}{72.05}=7.772rad/s^2\end{array}$

Now,

$\begin{array}{l}{a}_t=3.5\alpha \\ a_t=3.5\left(7.772\right)\\ a_t=27.203ft/s^2\downarrow \end{array}$

Taking horizontal reaction, $\sum F_x=m{a}_n$

$\begin{array}{l}{R}_x=\left(\frac{160}{32.2}\right)\times 54.407\\ R_x=270.35lb\end{array}$

Vertical reaction, $\sum F_y=-m{a}_t$

$\begin{array}{l}{R}_y-160=-\left(\frac{160}{32.2}\right)\times 27.203\\ R_y-160=-135.17\\ R_y=24.83lb\end{array}$

Conclusion:

The angular velocity and force exerted on the hands of the gymnast is ${\omega }_2=3.94rad/s$ and $R_x=270.35lb,$ respectively.

To determine

(b)

The angular velocity and force exerted on the hands of the gymnast after he has rotated through 180°.

Answer to Problem 17.20P

${\omega }_2=3.94rad/s$

$R_x=270.35lb$

Explanation of Solution

Given:

Calculation:

Position 3: Directly below the bar after rotating 180°.

Elevation: $h_3=-3.5ft$

Potential energy: $V_3=W{h}_3=\left(160\right)\left(-3.5\right)=-560ft.lb$

Speeds: $v_3=3.5{\omega }_3$

Kinetic energy: $T_3=\frac{1}{2}m{v}_3^2+\frac{1}{2}m{k}^2{\omega }_3^2$

$\begin{array}{l}{T}_3=\frac{1}{2}\times \left(\frac{160}{32.2}\right){\left(3.5{\omega }_3\right)}^2+\frac{1}{2}\times \left(\frac{160}{32.2}\right){\left(1.5\right)}^2{\omega }_3^2\\ T_3=\frac{1}{2}\times \left(\frac{160}{32.2}\right)\left[{\left(3.5{\omega }_3\right)}^2+{\left(1.5\right)}^2{\omega }_3^2\right]\\ T_3=2.4844\left[12.25{\omega }_3^2+2.25{\omega }_3^2\right]\\ T_3=36.025{\omega }_3^2\end{array}$

Applying Principle of conservation of energy

$\begin{array}{l}{T}_1+V_1=T_3+V_3\\ 0+560=36.025{\omega }_3^2-560\\ 1120=36.025{\omega }_3^2\\ {\omega }_3^2=\frac{1120}{36.025}\\ {\omega }_3^2=31.09\\ {\omega }_3=\sqrt{31.09}=5.58rad/s\end{array}$

From the kinematics of figure:

$\begin{array}{l}{a}_t=3.5\alpha \\ a_n=3.5{\omega }_3^2=3.5\left(31.09\right)=108.81ft/s^2\end{array}$

Taking moment about point O,

$\begin{array}{l}\Sigma M_0=\Sigma {\left(M_0\right)}_{eff}\\ \alpha =0\end{array}$

Now,

$\begin{array}{l}{a}_t=3.5\alpha \\ a_t=0\end{array}$

Taking horizontal reaction, $\sum F_x=0$

$R_x=0$

Vertical reaction, $\sum F_y=m{a}_n$

$\begin{array}{l}{R}_y-160=\left(\frac{160}{32.2}\right)\times 108.81\\ R_y-160=540.67\\ R_y=700.67lb\end{array}$

Conclusion:

The angular velocity and force exerted on the hands of the gymnast is ${\omega }_2=5.58rad/s$ and $R_y=700.67lb,$ respectively.