Chapter 17.1, Problem 17.20P

(a)

To determine

The angular velocity of gymnast and the force exerted on his hands after he has rotated through $\theta =90°$.

Answer to Problem 17.20P

The angular velocity of gymnast and the force exerted on his hands after he has rotated through $\theta =90°$ is $\underset{\_}{3.94\text{rad/s}}$.

The force exerted on gymnast’s hands after he has rotated through $\theta =90°$ is $\underset{\_}{271\text{lb}}$.

Explanation of Solution

Given information:

The weight (W) of the gymnast is 160 lb.

The centroidal radius of gyration $\left(\overline{k}\right)$ of gymnast’s body is 1.5 ft.

Calculation:

Find the mass (m) of the gymnast using the equation:

$m=\frac{W}{g}$

Here, g is the acceleration due to gravity.

Substitute 160 lb for W.

$\begin{array}{c}m=\frac{160}{32.2}\\ =4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}\end{array}$

Find the mass moment of inertia $\left(\overline{I}\right)$ using the equation:

$\overline{I}=\frac{1}{2}m{k}^2$

Substitute $4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}$ for m and 1.5 ft for k.

$\begin{array}{c}\overline{I}=\frac{1}{2}\left(4.9689\right){\left(1.5\right)}^2\\ =5.59\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Consider position 1 of the gymnast is directly above the bar.

The elevation $\left(h_1\right)$ of at position 1 is 3.5 ft.

Find the potential energy $\left(V_1\right)$ of the gymnast at position 1 using the equation:

$V_1=W{h}_1$

Substitute 160 lb for W, and 3.5 ft for $h_1$.

$\begin{array}{c}{V}_1=\left(160\right)\left(3.5\right)\\ =560\text{ft}\cdot \text{lb}\end{array}$

At initial position, the gymnast is at rest. Therefore, the velocity $\left({\overline{v}}_1\right)$, angular velocity $\left({\omega }_1\right)$, and kinetic energy $\left(T_1\right)$ of the gymnast are zero.

Consider position 2 of the gymnast in which the body of gymnast at level of bar after rotating $90°$.

Sketch the free body diagram and kinetic diagram of the position 2 as shown in Figure (1).

Refer Figure (1),

At the position 2, the body is in horizontal. Therefore, potential energy $\left(V_1\right)$ at position 2 is zero.

Find the equation of velocity $\left({\overline{v}}_2\right)$ of gymnast at position 2.

${\overline{v}}_2=h_1{\omega }_2$

Substitute 3.5 ft for $h_1$.

${\overline{v}}_2=3.5{\omega }_2$

Write the equation of kinetic energy $\left(T_2\right)$ of the gymnast at position 2.

$T_2=\frac{1}{2}m{\overline{v}}_2^2+\overline{I}{\omega }_2^2$

Substitute $4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}$ for m, $3.5{\omega }_2$ for ${\overline{v}}_2$ and $5.59\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $\overline{I}$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(4.9689\right){\left(3.5{\omega }_2\right)}^2+\left(5.59\right){\omega }_2^2\\ =36.025{\omega }_2^2\end{array}$

Apply the Principle of conservation of energy.

$T_1+V_1=T_2+V_2$

Substitute 0 for $T_1$, $560\text{lb}\cdot \text{ft}$ for $V_1$, $36.025{\omega }_2^2$ for $T_2$, and 0 for $V_2$.

$\begin{array}{l}0+560=36.025{\omega }_2^2+0\\ {\omega }_2=3.94\text{rad/s}\end{array}$

Thus, the angular velocity of gymnast and the force exerted on his hands after he has rotated through $\theta =90°$ is $\underset{\_}{3.94\text{rad/s}}$.

Refer Figure (2),

Find the acceleration $\left(a_t\right)$ of gymnastic along horizontal axis using kinematics.

$a_t=h\alpha$

Here, $\alpha$ is the angular acceleration of gymnast.

Substitute 3.5 ft for h.

$a_t=3.5\alpha$

Find the acceleration $\left(a_t\right)$ of gymnastic along vertical axis using kinematics.

$a_n=h{\omega }_2^2$

Substitute 3.5 ft for h and $3.94\text{rad/s}$ for ${\omega }_2$.

$\begin{array}{c}{a}_n=\left(3.5\right){\left(3.94\right)}^2\\ =54.3326{\text{ft/s}}^2\end{array}$

Take moment about mass canter O.

$\begin{array}{l}\Sigma M_O=\Sigma {\left(M_0\right)}_{eff}\\ mgh=m{a}_{t}h+m{k}^2\alpha \\ gh=a_{t}h+k^2\alpha \end{array}$

Substitute $32.2{\text{ft/s}}^2$ fo g, $3.5\alpha$ for $a_t$, 3.5 ft for h, and 1.5 ft for $\alpha$.

$\begin{array}{l}32.2\left(3.5\right)=\left(3.5\alpha \right)\left(3.5\right)+{\left(1.5\right)}^2\alpha \\ 112.7=12.25\alpha +2.25\alpha \\ \alpha =7.7724{\text{rad/s}}^2\end{array}$

Find the horizontal force $\left(R_x\right)$ exerted on gymnast’s hands after he has rotated through $\theta =90°$.

Consider equilibrium along horizontal axis.

$R_x=m{a}_n$

Substitute $4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}$ for m and $54.3326{\text{ft/s}}^2$ for $a_n$.

$\begin{array}{c}{R}_x=\left(4.9689\right)\left(54.3326\right)\\ =270\text{lb}\end{array}$

Find the vertical force $\left(R_x\right)$ exerted on gymnast’s hands after he has rotated through $\theta =90°$.

Consider equilibrium along vertical axis.

$R_y-mg=-m{a}_t$

Substitute $3.5\alpha$ for $a_t$.

$R_y-mg=-m\left(3.5\alpha \right)$

Substitute $4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}$ for m, $32.2{\text{ft/s}}^2$ for g, and $7.7724{\text{rad/s}}^2$ for $\alpha$.

$\begin{array}{l}{R}_y-\left(4.9689\right)\left(32.2\right)=-\left(4.9689\right)\left(3.5\times 7.7724\right)\\ R_y=24.83\text{lb}\end{array}$

Find the force $\left(R\right)$ exerted on gymnast’s hands after he has rotated through $\theta =90°$ using the equation:

$R=\sqrt{R_x^2+R_y^2}$

Substitute $270\text{lb}$ for $R_x$ and $24.83\text{lb}$ for $R_y$.

$\begin{array}{c}R=\sqrt{{270}^2+{24.83}^2}\\ =271\text{lb}\end{array}$

Thus, the force exerted on gymnast’s hands after he has rotated through $\theta =90°$ is $\underset{\_}{271\text{lb}}$.

(b)

To determine

The angular velocity of gymnast and the force exerted on his hands after he has rotated through $\theta =180°$.

Answer to Problem 17.20P

The angular velocity of gymnast and the force exerted on his hands after he has rotated through $\theta =180°$ is $\underset{\_}{5.58\text{rad/s}}$.

The force exerted on gymnast’s hands after he has rotated through $\theta =180°$ is $\underset{\_}{701\text{lb}}$.

Explanation of Solution

Calculation:

Consider position 3 of the gymnast is directly below the bar after rotating $180°$.

The elevation $\left(h_1\right)$ of at position 3 is $-3.5\text{ft}$.

Find the potential energy $\left(V_1\right)$ of the gymnast at position 3 using the equation:

$V_3=W{h}_3$

Substitute 160 lb for W, and $-3.5\text{ft}$ for $h_1$.

$\begin{array}{c}{V}_3=\left(160\right)\left(-3.5\right)\\ =-560\text{ft}\cdot \text{lb}\end{array}$

At initial position, the gymnast is at rest. Therefore, the velocity $\left({\overline{v}}_1\right)$, angular velocity $\left({\omega }_1\right)$, and kinetic energy $\left(T_1\right)$ of the gymnast are zero.

Consider position 2 of the gymnast in which the body of gymnast at level of bar after rotating $90°$.

Sketch the free body diagram and kinetic diagram of the position 3 as shown in Figure 2.

Refer Figure (2),

Find the equation of velocity $\left({\overline{v}}_2\right)$ of gymnast at position 3.

${\overline{v}}_3=h_3{\omega }_3$

Substitute 3.5 ft for $h_3$.

${\overline{v}}_2=3.5{\omega }_2$

Write the equation of kinetic energy $\left(T_2\right)$ of the gymnast at position 2.

$T_2=\frac{1}{2}m{\overline{v}}_2^2+\overline{I}{\omega }_2^2$

Substitute $4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}$ for m, $3.5{\omega }_2$ for ${\overline{v}}_2$ and $5.59\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $\overline{I}$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(4.9689\right){\left(3.5{\omega }_2\right)}^2+\left(5.59\right){\omega }_2^2\\ =36.025{\omega }_2^2\end{array}$

Find the angular velocity $\left({\omega }_3\right)$ of gymnast and the force exerted on his hands after he has rotated through $\theta =180°$.

Apply the Principle of conservation of energy.

$T_1+V_1=T_3+V_3$

Substitute 0 for $T_1$, $560\text{lb}\cdot \text{ft}$ for $V_1$, $36.025{\omega }_2^2$ for $T_3$, and $-560\text{ft}\cdot \text{lb}$ for $V_3$.

$\begin{array}{l}0+560=36.025{\omega }_3^2-560\\ {\omega }_3=5.58\text{rad/s}\end{array}$

Thus, the angular velocity of gymnast and the force exerted on his hands after he has rotated through $\theta =180°$ is $\underset{\_}{5.58\text{rad/s}}$.

Refer Figure (2),

The acceleration $\left(a_t\right)$ of gymnastic along horizontal axis at position 3 is zero.

$a_t=0$

Find the acceleration $\left(a_t\right)$ of gymnastic along vertical axis using kinematics.

$a_n=h{\omega }_3^2$

Substitute 3.5 ft for h and $5.58\text{rad/s}$ for ${\omega }_2$.

$\begin{array}{c}{a}_n=\left(3.5\right){\left(5.58\right)}^2\\ =108.98{\text{ft/s}}^2\end{array}$

Find the horizontal force $\left(R_x\right)$ exerted on gymnast’s hands after he has rotated through $\theta =180°$

Consider equilibrium along horizontal axis.

$R_x=m{a}_t$

Substitute 0 for $a_t$.

$\begin{array}{c}{R}_x=m\left(0\right)\\ =0\end{array}$

Find the vertical force $\left(R_x\right)$ exerted on gymnast’s hands after he has rotated through $\theta =180°$

Consider equilibrium along vertical axis.

$R_y-mg=m{a}_n$

Substitute $4.9689\text{lb}\cdot {\text{s}}^2\text{/ft}$ for m, $32.2{\text{ft/s}}^2$ for g, and $108.98{\text{ft/s}}^2$ for $a_n$.

$\begin{array}{l}{R}_y-\left(4.9689\right)\left(32.2\right)=\left(4.9689\right)\left(108.98\right)\\ R_y\simeq 701\text{lb}\end{array}$

Find the force $\left(R\right)$ exerted on gymnast’s hands after he has rotated through $\theta =180°$ using the equation:

$R=\sqrt{R_x^2+R_y^2}$

Substitute 0 for $R_x$ and $701\text{lb}$ for $R_y$.

$\begin{array}{c}R=\sqrt{0^2+{701}^2}\\ =701\text{lb}\end{array}$

Thus, the force exerted on gymnast’s hands after he has rotated through $\theta =180°$ is $\underset{\_}{701\text{lb}}$.