Chapter 17.3, Problem 17.130P

a)

To determine

Find the angular velocity of the bar after the impact.

Answer to Problem 17.130P

The angular velocity of the bar after the impact is $\underset{\_}{1.286\text{rad}/\text{s}}$.

Explanation of Solution

Given information:

The weight $\left(W_S\right)$ of the sphere is 3 lb.

The radius $\left(r\right)$ of the sphere is 3 in..

The weight $\left(W_{AB}\right)$ of the uniform rod is 2 lb.

The length $\left(L\right)$ of the uniform rod is 10 in..

The angular velocity $\left({\omega }_0\right)$ of the rod before the impact is 3 rad/s.

The velocity $\left(v_0\right)$ of the sphere before the impact is 2 ft/s.

Calculation:

Consider G is the mass center of rod AB.

Find the mass $\left(m_{AB}\right)$ of the rod AB using the equation:

$m_{AB}=\frac{W_{AB}}{g}$

Substitute 2 lb for $W_{AB}$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_{AB}=\frac{2}{32.2}\\ =62.11\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Find the moment of inertia $\left(I_{AB}\right)$ of the rod AB using the equation:

$I_{AB}=\frac{1}{12}{m}_{AB}{L}^2$

Substitute $62.11\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_{AB}$ and 10 in. for L.

$\begin{array}{c}{I}_{AB}=\frac{1}{12}\left(62.11\times {10}^{-3}\right){\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}}\right)}^2\\ =3.594\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Consider C be the mass center of the sphere.

Find the mass of the sphere $\left(m_S\right)$ using the equation:

$m_S=\frac{W_S}{g}$

Substitute 3 lb for $W_S$ and $32.2\text{ft}/{\text{s}}^2$ for g

$\begin{array}{c}{m}_S=\frac{3\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =93.17\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Find the moment of inertia $\left(I_G\right)$ of the sphere using the equation:

$I_G=\frac{2}{5}{m}_S{r}^2$

Substitute $93.17\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_S$ and 3 in. for r.

$\begin{array}{c}{I}_G=\frac{2}{5}\left(93.17\times {10}^{-3}\right){\left(3\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}}\right)}^2\\ =2.329\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

The rod and sphere have same angular velocity $\left(\omega \right)$ after the impact.

Sketch the geometry of the sphere and the uniform rod as shown in Figure (1).

Refer Figure (1),

Find the distance R.

$R=\sqrt{L^2+r^2}$

Substitute 10 in. for L and 3 in. for r.

$\begin{array}{c}R=\sqrt{{\left(10\text{in}\text{.}\right)}^2+{\left(3\text{in}\text{.}\right)}^2}\\ =\sqrt{109\text{in}{\text{.}}^2}\\ =10.4403\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =0.87\text{ft}\end{array}$

Find the angle $\theta$.

$\tan \theta =\frac{r}{L}$

Substitute 10 in. for L and 3 in. for r.

$\begin{array}{l}\tan \theta =\frac{3\text{in}\text{.}}{10\text{in}\text{.}}\\ \tan \theta =0.3\\ \theta ={\tan }^{-1}\left(0.3\right)\\ \theta =16.7°\end{array}$

Find the velocity $\left(v_C\right)$ of the uniform rod before impact using kinematics.

$v_C=\frac{L}{2}{\omega }_0$

Substitute 10 in. for L and $3\text{rad}/\text{s}$ for ${\omega }_0$.

$\begin{array}{c}{v}_C=\frac{\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}\times \left(3\right)\\ =1.25\text{ft}/\text{s}\end{array}$

Write the velocity $\left({v^{\prime }}_C\right)$ of the uniform rod after the impact using kinematics.

${v^{\prime }}_C=\frac{L}{2}\omega$

Write the velocity $\left({v^{\prime }}_C\right)$ of the sphere after the impact using kinematics.

${v^{\prime }}_G=R\omega$

Consider the conservation of momentum principle.

${\text{Syst Momenta}}_{\text{1}}+{\text{ Syst Ext Imp}}_{\text{1y2 }}={\text{ Syst Momenta}}_{\text{2}}$

Sketch the impulse and momentum diagram of the system as shown in Figure (2).

Refer Figure (2).

Take moment about A (positive sign in clockwise direction).

$\begin{array}{l}{m}_S{v}_{0}L-I_{AB}{\omega }_0-m_{AB}{v}_C\frac{L}{2}+0=I_G\omega +m_S{v}_{G}R+I_{AB}\omega +m_{AB}{v}_C\frac{L}{2}\\ m_S{v}_{0}L-I_{AB}{\omega }_0-m_{AB}{v}_C\frac{L}{2}=I_G\omega +m_S{v}_{G}R+I_{AB}\omega +m_{AB}{v}_C\frac{L}{2}\end{array}$

Substitute $R\omega$ for ${v^{\prime }}_G$, and $\frac{L}{2}\omega$ for ${v^{\prime }}_C$.

$\begin{array}{c}{m}_S{v}_{0}L-I_{AB}{\omega }_0-m_{AB}{v}_C\frac{L}{2}=I_G\omega +m_S\left(R\omega \right)R+I_{AB}\omega +m_{AB}\left(\frac{L}{2}\omega \right)\frac{L}{2}\\ =I_G\omega +m_S{R}^2\omega +I_{AB}\omega +\frac{1}{4}{m}_{AB}{L}^2\omega \\ =\left(I_G+m_S{R}^2+I_{AB}+\frac{1}{4}{m}_{AB}{L}^2\right)\omega \end{array}$

Substitute $93.17\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_S$, $2\text{ft}/\text{s}$ for $v_0$, 10 in. for L, $3.594\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_{AB}$, $3\text{rad}/\text{s}$ for ${\omega }_0$, $62.11\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_{AB}$, $1.25\text{ft}/\text{s}$ for $v_C$, $2.329\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_G$, and 0.87 ft for R.

$\begin{array}{l}\left\{\begin{array}{l}\left(93.17\times {10}^{-3}\right)\left(2\right)\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\\ -\left(3.594\times {10}^{-3}\right)\left(3\right)\\ -\left(62.11\times {10}^{-3}\right)\left(1.25\right)\frac{\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}\end{array}\right\}=\left(\begin{array}{l}2.329\times {10}^{-3}\\ \text{+}\left(93.17\times {10}^{-3}\right){\left(0.8700\right)}^2\\ +3.594\times {10}^{-3}\\ +\frac{1}{4}\left(62.11\times {10}^{-3}\right){\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\end{array}\right)\omega \\ 0.1122=0.0872\omega \\ \omega =\frac{0.1122}{0.0872}\\ \omega =1.286\text{rad}/\text{s}\end{array}$

Thus, the angular velocity of the bar after the impact is $\underset{\_}{1.286\text{rad}/\text{s}}$.

b)

To determine

Find the components of the reactions at A.

Answer to Problem 17.130P

The horizontal component of the reactions at A is $\underset{\_}{0.719\text{lb}}$.

The vertical component of the reactions at A is $\underset{\_}{1.006\text{lb}}$.

Explanation of Solution

Calculation:

Find the normal acceleration ${\left(a_C\right)}_n$ of the uniform rod using the equation.

${\left(a_C\right)}_n=\frac{L}{2}{\omega }^2$

Substitute 10 in. for L and $1.286\text{rad}/\text{s}$ for $\omega$.

$\begin{array}{c}{\left(a_C\right)}_n=\frac{\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}{\left(1.286\right)}^2\\ =0.6889\text{ft}/{\text{s}}^2\end{array}$

Write the equation of tangential acceleration ${\left(a_C\right)}_t$ of the uniform rod using the equation.

${\left(a_C\right)}_t=\frac{L}{2}\alpha$

Here,$\alpha$ is the angular acceleration of the uniform rod and sphere.

Substitute 10 in. for L.

$\begin{array}{c}{\left(a_C\right)}_t=\frac{\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}\alpha \\ =0.4167\alpha \end{array}$ (1)

Find the normal acceleration ${\left(a_G\right)}_n$ of the sphere using the equation.

${\left(a_G\right)}_n=R{\omega }^2$

Substitute 0.8700 ft for R and $1.286\text{rad}/\text{s}$ for $\omega$.

$\begin{array}{c}{\left(a_G\right)}_n=\left(0.8700\right){\left(1.286\right)}^2\\ =1.4384\text{ft}/{\text{s}}^2\end{array}$

Write the equation of tangential acceleration ${\left(a_G\right)}_t$ of the sphere using the equation:

${\left(a_G\right)}_t=R\alpha$

Substitute 0.8700 ft for R.

$\begin{array}{c}{\left(a_G\right)}_t=\left(0.8700\text{ft}\right)\alpha \\ =0.87\alpha \end{array}$ (2)

Sketch the free body diagram of the uniform rod and sphere as shown in Figure (1).

Refer Figure (3).

Take moment about A (positive sign in clockwise direction).

$W_{AB}\frac{L}{2}+W_{S}L=I_{AB}\alpha +\frac{L}{2}{m}_{AB}{\left(a_C\right)}_t+I_G\alpha +m_S{\left(a_G\right)}_{t}R$

Substitute $\frac{L}{2}\alpha$ for ${\left(a_C\right)}_t$ and $R\alpha$ for ${\left(a_G\right)}_t$.

$\begin{array}{c}{W}_{AB}\frac{L}{2}+W_{S}L=I_{AB}\alpha +\frac{L}{2}{m}_{AB}\left(\frac{L}{2}\alpha \right)+I_G\alpha +m_S\left(R\alpha \right)R\\ =I_{AB}\alpha +\frac{1}{4}{m}_{AB}{L}^{}\alpha +I_G\alpha +m_S{R}^2\alpha \\ =\left(I_{AB}+\frac{1}{4}{m}_{AB}{L}^2+I_G+m_S{R}^2\right)\alpha \end{array}$

Substitute 2 lb for $W_{AB}$, 3 lb for $W_S$, 10 in. for L, $3.594\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_{AB}$, $62.11\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_{AB}$, $2.329\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for $I_G$, $93.17\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_S$, and 0.8700 ft for R.

$\begin{array}{l}\left(2\right)\frac{\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}+\left(3\right)\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)=\left[\begin{array}{l}3.594\times {10}^{-3}\\ +\frac{1}{4}\left(62.11\times {10}^{-3}\right){\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ +2.329\times {10}^{-3}\\ +\left(93.17\times {10}^{-3}\right){\left(0.8700\right)}^2\end{array}\right]\alpha \\ 0.83333+2.5=\left(3.594\times {10}^{-3}\right)+0.010783+\left(2.329\times {10}^{-3}\right)+0.07052\\ 3.3333=0.0872\alpha \\ \alpha =38.214{\text{rad/s}}^2\end{array}$

Find the tangential acceleration ${\left(a_C\right)}_t$ of the uniform rod using Equation (1).

${\left(a_C\right)}_t=0.4167\alpha$

Substitute $38.214{\text{rad}/\text{s}}^{\text{2}}$ for $\alpha$.

$\begin{array}{c}{\left(a_C\right)}_t=0.4167\left(38.214{\text{rad}/\text{s}}^{\text{2}}\right)\\ =15.923{\text{ft}/\text{s}}^{\text{2}}\downarrow \end{array}$

Find the tangential acceleration ${\left(a_G\right)}_t$ of the sphere using Equation (2).

${\left(a_G\right)}_t=0.87\alpha$

Substitute $38.214{\text{rad}/\text{s}}^{\text{2}}$ for $\alpha$

$\begin{array}{c}{\left(a_G\right)}_t=0.87\left(38.214{\text{rad}/\text{s}}^{\text{2}}\right)\\ =33.247{\text{ft}/\text{s}}^{\text{2}}\end{array}$

Refer Figure (3),

Consider the horizontal component forces.

$A_x=-m_{AB}{\left(a_C\right)}_n-m_S{\left(a_G\right)}_n\cos 16.7°+m_S{\left(a_G\right)}_t\sin 16.7°$

Here, $A_x$ is the horizontal component of the reaction.

Substitute $62.11\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_{AB}$, $0.6889\text{ft}/{\text{s}}^2$ for ${\left(a_C\right)}_n$, $93.17\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_S$, $1.4384\text{ft}/{\text{s}}^2$ for ${\left(a_G\right)}_n$, and $33.247{\text{ft}/\text{s}}^{\text{2}}$ for ${\left(a_G\right)}_t$.

$\begin{array}{c}{A}_x=\left[\begin{array}{l}-\left(62.11\times {10}^{-3}\right)\left(0.6889\right)\\ -\left(93.17\times {10}^{-3}\right)\left(1.4384\right)\cos 16.7°\\ +\left(93.17\times {10}^{-3}\right)\left({33.247}^{\text{2}}\right)\sin 16.7°\end{array}\right]\\ =-0.0428-0.1284+0.8901\\ =0.719\text{lb}\to \end{array}$

Thus, the horizontal component of the reactions at A is $\underset{\_}{0.719\text{lb}}$.

Consider vertical components of forces.

$A_y-W_{AB}-W_S=m_{AB}{\left(a_G\right)}_t-m_S{\left(a_G\right)}_t\cos 16.7°-m_S{\left(a_G\right)}_n\sin 16.7°$

Here, $A_y$ is the vertical component of the reaction.

Substitute 2 lb for $W_{AB}$, 3 lb for $W_S$, $62.11\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_{AB}$, $93.17\times {10}^{-3}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_S$, $1.4384\text{ft}/{\text{s}}^2$ for ${\left(a_G\right)}_n$, and $33.247{\text{ft}/\text{s}}^{\text{2}}$ for ${\left(a_G\right)}_t$.

$\begin{array}{l}{A}_y-\left(2\right)-\left(3\right)=\left\{\begin{array}{c}-\left(62.11\times {10}^{-3}\right)\left(15.923\right)\\ -\left(93.17\times {10}^{-3}\right)\left(33.247\right)\cos 16.7°\\ -\left(93.17\times {10}^{-3}\right)\left(1.4384\text{ft}/{\text{s}}^2\right)\sin 16.7°\end{array}\right\}\\ A_y-5=-0.9889-2.9669-0.0385\\ A_y=5\text{lb}-0.9889-2.9669-0.0385\\ A_y=1.006\text{lb}\uparrow \end{array}$

Thus, the vertical component of the reactions at A is $\underset{\_}{1.006\text{lb}}$.