Chapter 17.2, Problem 17.93P

To determine

The velocity of the rod AB relative to cylinder DE when end B of the rod strikes end E of the cylinder.

Answer to Problem 17.93P

Velocity of the rod relative to cylinder: ${\left(v_r\right)}_2=7.45m\text{/}s$

Explanation of Solution

Given information:

The mass of the rod: $m=3kg$

Length of the rod is 800 mm.

Length of cylinder DE is 240 mm.

Angular velocity of the assembly: $\omega =40rad/s$

Velocity of the end B of the rod towards the cylinder is 75 mm/s.

The centroidal moment of inertia of the cylinder about a vertical axis is 0.025 kg.m².

Calculations:

From the schematic shown above:

$\begin{array}{l}\text{At initial position:}\\ v_1=\left(0.04m\right){\omega }_1\\ v_1=\left(0.04m\right)40rad/s\\ v_1=1.6m/s\\ \text{Also, at final position:}\\ v_2=\left(0.28m\right){\omega }_2\end{array}$

$\begin{array}{l}\text{The moment of inertia of the rod:}\\ {\overline{I}}_{AB}=\frac{1}{12}m{k}^2=\frac{1}{12}\left(3kg\right){\left(0.8m\right)}^2\\ {\overline{I}}_{AB}=0.16kg\cdot m^2\\ \text{Now,}\text{taking moment about point C,}\\ {\overline{I}}_{AB}{\omega }_1+m{\overline{v}}_1\left(0.04m\right)+{\overline{I}}_{DE}{\omega }_1={\overline{I}}_{AB}{\omega }_2\left(0.04m\right)+{\overline{I}}_{DE}{\omega }_2\\ substitut\text{i}\text{n}gvalues;\end{array}$

$\begin{array}{l}\left\{\begin{array}{l}\left(0.16kg\cdot m^2\right)\left(40rad\text{/}s\right)+\\ \left(3kg\right)\left(1.6m/s\right)\left(0.04m\right)+\\ \left(0.025kg\cdot m^2\right)\left(40rad\text{/}s\right)\end{array}\right\}=\left\{\begin{array}{l}\left(0.16kg\cdot m^2\right){\omega }_2+\\ \left(3kg\right)\left(\text{0}\text{.18 }{\omega }_2\right)\left(0.28\right)+\\ \left(0.025kg\cdot m^2\right){\omega }_2\end{array}\right\}\\ 6.4+0.192+1=\left(0.16+0.2352+0.025\right){\omega }_2\\ 7.592=0.4202{\omega }_2\\ {\omega }_2=\frac{7.592}{0.4202}\\ {\omega }_2=18.068rad\text{/}s\end{array}$

$\begin{array}{l}\text{Velocity of rod at position 1:}{\left(v_r\right)}_1=0.075m/s\\ \text{At initial position:}\\ Potentialenergy:V_1=0\\ k\text{i}\text{n}eticenergy:T_1=\frac{1}{2}{\overline{I}}_{DE}{\omega }_1^2+\frac{1}{2}{\overline{I}}_{AB}{\omega }_1^2+\frac{1}{2}{m}_{AB}{v}_1^2+\frac{1}{2}{m}_{AB}{\left(v_r\right)}_1^2\\ \text{Substituting values,}\\ T_1=\left\{\begin{array}{l}\frac{1}{2}\left(0.025\right){\left(40rad/s\right)}^2+\frac{1}{2}\left(0.16\right){\left(40rad/s\right)}^2+\\ \frac{1}{2}\left(3kg\right){\left(1.6\right)}^2+\frac{1}{2}\left(3kg\right){\left(0.075\right)}^2\end{array}\right\}\\ T_1=20+128+3.84+0.0084\\ T_1=151.85J\\ \\ \text{At final position:}\\ v_2=\left(0.28m\right){\omega }_2\\ v_2=\left(0.28m\right)\left(18.068\right)\\ v_2=5.059m/s\end{array}$

$\begin{array}{l}Potentialenergy:V_2=0\\ k\text{i}\text{n}eticenergy:T_2=\frac{1}{2}{\overline{I}}_{DE}{\omega }_2^2+\frac{1}{2}{\overline{I}}_{AB}{\omega }_2^2+\frac{1}{2}{m}_{AB}{v}_2^2+\frac{1}{2}{m}_{AB}{\left(v_r\right)}_2^2\\ Substitut\text{i}\text{n}gvalues;\\ T_2=\left\{\begin{array}{l}\frac{1}{2}\left(0.025\right){\left(18.068rad/s\right)}^2+\frac{1}{2}\left(0.16\right){\left(18.068rad/s\right)}^2+\\ \frac{1}{2}\left(3kg\right){\left(5.059\right)}^2+\frac{1}{2}\left(3kg\right){\left(v_r\right)}_2^2\end{array}\right\}\\ T_2=4.081J+26.116J+38.39J+1.5{\left(v_r\right)}_2^2\\ T_2=68.578J+1.5{\left(v_r\right)}_2^2\end{array}$

$\begin{array}{l}Now,apply\text{i}\text{n}gthelawofconversationofenergy,\\ T_1+V_1=T_2+V_2\\ 151.85+0=68.587+1.5{\left(v_r\right)}_2^2\\ or,\\ 1.5{\left(v_r\right)}_2^2=83.263\end{array}$

$\begin{array}{l}{\left(v_r\right)}_2=\sqrt{\frac{83.263}{1.5}}\\ {\left(v_r\right)}_2=7.45m\text{/}s\\ Therefore,\text{the velocity is }\\ \Rightarrow {\left(v_r\right)}_2=7.45m\text{/}s\end{array}$

Conclusion:

The velocity of the rod AB relative to cylinder DE when end B of the rod strikes end E of the cylinder is ${\left(v_r\right)}_2=7.45m\text{/}s$.