Chapter 17.2, Problem 17.94P

To determine

Find the velocity of the tube relative to the rod as the tube strikes end E of the assembly.

Answer to Problem 17.94P

The velocity of the tube relative to the rod as the tube strikes end E of the assembly is $\underset{\_}{1.540\text{m/s}}$.

Explanation of Solution

Given information:

The mass $\left(m_{AB}\right)$ of tube AB is 1.6 kg.

The initial angular velocity $\left({\omega }_1\right)$ of the assembly is 5 rad/s.

The moment of inertia of the rod and bracket $\left({\overline{I}}_{DCE}\right)$ about the vertical axis of rotation is $0.30\text{kg}\cdot {\text{m}}^2$.

The centroidal moment of inertia of the tube $\left({\overline{I}}_{AB}\right)$ about a vertical axis is $0.0025\text{kg}\cdot {\text{m}}^2$

Calculation:

Refer the system.

Find the radius of the tube AB at initial position $\left(r_{{\left(G/A\right)}_1}\right)$.

$r_{{\left(G/A\right)}_1}=r_{{\left(G/C\right)}_1}$

Here, $r_{{\left(G/C\right)}_1}$ is the radius of the tube AB at initial position at point C.

Substitute 125 mm for $r_{{\left(G/C\right)}_1}$.

$\begin{array}{c}{r}_{{\left(G/A\right)}_1}=\frac{1}{2}\left(125\text{mm}\right)\\ =62.5\text{mm}\end{array}$

Find the radius of the tube AB at final position $\left(r_{{\left(G/A\right)}_2}\right)$.

$\begin{array}{c}{r}_{{\left(G/A\right)}_2}=r_{{\left(G/C\right)}_2}\\ =r_{DA}-r_{{\left(G/A\right)}_1}\end{array}$

Here, $r_{{\left(G/C\right)}_2}$ is the radius of the tube AB at final position at point C.

Substitute 500 mm for $r_{DA}$ and 62.5 mm for $r_{{\left(G/A\right)}_1}$.

$\begin{array}{c}{r}_{{\left(G/A\right)}_2}=500-62.5\text{mm }\\ =437.5\text{mm}\end{array}$

Write the equation of the velocity of the tube ${\left(v_G\right)}_{\theta }$ at the center G.

$\begin{array}{c}{\left(v_G\right)}_{\theta }={\overline{v}}_{\theta }\\ =r_{G/C}\omega \end{array}$

Here, ${\overline{v}}_{\theta }$ is the centroidal velocity of the tube and $r_{\left(G/C\right)}$ is the radius of the tube AB at point C.

Consider ${\overline{v}}_r$ be the velocity of the tube relative to the rod.

At initial position the system is at rest. Therefore, the velocity of the tube relative to the rod at initial position ${\left({\overline{v}}_r\right)}_1$ will be zero.

Consider the impulse and momentum principle.

Sketch the impulse and momentum diagram of the system as shown in Figure (1).

Refer Figure (1),

Take moment about C (Anti-clockwise as positive).

$\begin{array}{c}{\overline{I}}_{AB}{\omega }_1+{\overline{I}}_{DCE}{\omega }_1+m_{AB}{\left({\overline{v}}_{\theta }\right)}_1{\left(r_{G/C}\right)}_1+0={\overline{I}}_{AB}{\omega }_2+{\overline{I}}_{DCE}{\omega }_2+m_{AB}{\left({\overline{v}}_{\theta }\right)}_2{\left(r_{G/C}\right)}_2\\ {\overline{I}}_{AB}{\omega }_1+{\overline{I}}_{DCE}{\omega }_1+m_{AB}{\left({\overline{v}}_{\theta }\right)}_1{\left(r_{G/C}\right)}_1={\overline{I}}_{AB}{\omega }_2+{\overline{I}}_{DCE}{\omega }_2+m_{AB}{\left({\overline{v}}_{\theta }\right)}_2{\left(r_{G/C}\right)}_2\end{array}$

Here, ${\omega }_1$ is the angular velocity at initial position, ${\omega }_2$ is the angular velocity at final position, ${\left({\overline{v}}_{\theta }\right)}_1$ is the velocity of the tube at the center G at initial position, and ${\left({\overline{v}}_{\theta }\right)}_2$ is the velocity of the tube at the center G at final position.

Substitute ${\left(r_{G/C}\right)}_1{\omega }_1$ for ${\left({\overline{v}}_{\theta }\right)}_1$ and ${\left(r_{G/C}\right)}_2{\omega }_2$ for ${\left({\overline{v}}_{\theta }\right)}_2$.

$\begin{array}{l}{\overline{I}}_{AB}{\omega }_1+{\overline{I}}_{DCE}{\omega }_1+m_{AB}\left({\left(r_{G/C}\right)}_1{\omega }_1\right){\left(r_{G/C}\right)}_1={\overline{I}}_{AB}{\omega }_2+{\overline{I}}_{DCE}{\omega }_2+m_{AB}\left({\left(r_{G/C}\right)}_2{\omega }_2\right){\left(r_{G/C}\right)}_2\\ {\overline{I}}_{AB}{\omega }_1+{\overline{I}}_{DCE}{\omega }_1+m_{AB}{\left(r_{G/C}\right)}_1{}^2{\omega }_1={\overline{I}}_{AB}{\omega }_2+{\overline{I}}_{DCE}{\omega }_2+m_{AB}{\left(r_{G/C}\right)}_2{}^2{\omega }_2\\ \left({\overline{I}}_{AB}+{\overline{I}}_{DCE}+m_{AB}{\left(r_{G/C}\right)}_1{}^2\right){\omega }_1=\left({\overline{I}}_{AB}+{\overline{I}}_{DCE}+m_{AB}{\left(r_{G/C}\right)}_2{}^2\right){\omega }_2\end{array}$

Substitute $0.0025\text{kg}\cdot {\text{m}}^{\text{2}}$ for ${\overline{I}}_{AB}$, $0.30\text{kg}\cdot {\text{m}}^{\text{2}}$ for ${\overline{I}}_{DCE}$, 1.6 kg for $m_{AB}$, $5\text{rad}/\text{s}$ for ${\omega }_1$, $62.5\text{mm}$ for $r_{{\left(G/C\right)}_1}$ and $\text{437}\text{.5}\text{mm}$ for $r_{{\left(G/C\right)}_2}$.

$\begin{array}{l}\left[\begin{array}{l}0.0025+0.30\\ +1.6{\left(62.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\end{array}\right]\left(5\right)=\left[\begin{array}{l}0.0025+0.30\\ +1.6{\left(437.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\end{array}\right]{\omega }_2\\ 1.54375=0.60875{\omega }_2\\ {\omega }_2=2.5359\text{rad/s}\end{array}$

Find the kinetic energy of the system $\left(T_1\right)$ at initial position using the equation:

$\begin{array}{c}{T}_1=\frac{1}{2}{\overline{I}}_{AB}{\omega }_1^2+\frac{1}{2}{\overline{I}}_{DCE}{\omega }_1^2+\frac{1}{2}{m}_{AB}{\left(r_{G/C}\right)}_1^2{\omega }_1{}^2+\frac{1}{2}{m}_{AB}{\left({\overline{v}}_r\right)}_1^2\\ =\frac{1}{2}\left({\overline{I}}_{AB}+{\overline{I}}_{DCE}+m_{AB}{\left(r_{G/C}\right)}_1^2\right){\omega }_1{}^2+\frac{1}{2}{m}_{AB}{\left({\overline{v}}_r\right)}_1^2\end{array}$

Substitute $0.0025\text{kg}\cdot {\text{m}}^{\text{2}}$ for ${\overline{I}}_{AB}$, $0.30\text{kg}\cdot {\text{m}}^{\text{2}}$ for ${\overline{I}}_{DCE}$, 1.6 kg for $m_{AB}$, $5\text{rad}/\text{s}$ for ${\omega }_1$, $62.5\text{mm}$ for $r_{{\left(G/C\right)}_1}$ and 0 for ${\left({\overline{v}}_r\right)}_1$.

$\begin{array}{c}{T}_1=\frac{1}{2}\left[\begin{array}{l}0.0025+0.30\\ +1.6{\left(62.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\end{array}\right]{\left(5\right)}^2+\frac{1}{2}\left(1.6\right)\left(0\right)\\ =\frac{1}{2}\left(0.30875\right){\left(5\right)}^2+0\\ =3.8594\text{J}\end{array}$

Find the kinetic energy of the system $\left(T_2\right)$ at final position using the equation:

$\begin{array}{c}{T}_2=\frac{1}{2}{\overline{I}}_{AB}{\omega }_2^2+\frac{1}{2}{\overline{I}}_{DCE}{\omega }_2^2+\frac{1}{2}{m}_{AB}{\left(r_{G/C}\right)}_2^2{\omega }_2^2+\frac{1}{2}{m}_{AB}{\left({\overline{v}}_r\right)}_2^2\\ =\frac{1}{2}\left({\overline{I}}_{AB}+{\overline{I}}_{DCE}+m_{AB}{\left(r_{G/C}\right)}_2^2\right){\omega }_2^2+\frac{1}{2}{m}_{AB}{\left({\overline{v}}_r\right)}_2^2\end{array}$

Substitute $0.0025\text{kg}\cdot {\text{m}}^{\text{2}}$ for ${\overline{I}}_{AB}$, $0.30\text{kg}\cdot {\text{m}}^{\text{2}}$ for ${\overline{I}}_{DCE}$, 1.6 kg for $m_{AB}$, $2.539\text{rad}/\text{s}$ for ${\omega }_2$, and $437.5\text{mm}$ for $r_{{\left(G/C\right)}_2}$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(\begin{array}{l}0.0025+0.30\\ +1.6{\left(437.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\end{array}\right){\left(2.539\right)}^2+\frac{1}{2}\left(1.6\right){\left({\overline{v}}_r\right)}_2^2\\ =\frac{1}{2}\left(0.60875\right){\left(2.539\right)}^2+\frac{1}{2}\left(1.6\text{ kg}\right){\left({\overline{v}}_r\right)}_2^2\\ =1.9621\text{+0}\text{.8}{\left({\overline{v}}_r\right)}_2^2\end{array}$

The bearing reaction at point C is zero. This leads to frictionless sliding contact between rod and tube. Therefore, the work done $\left(U_{1\to 2}\right)$ of the system is zero.

$U_{1\to 2}=0$

Consider the Principle of work and energy.

Find the velocity ${\left({\overline{v}}_r\right)}_2$ of the tube relative to the rod as the tube strikes end E of the assembly.

$T_1+U_{1y2}=T_2$

Substitute $3.8594\text{J}$ for $T_1$, $1.9621\text{+0}\text{.8}{\left({\overline{v}}_r\right)}_2^2$ for $T_2$ and 0 for $U_{1\to 2}$

$\begin{array}{l}3.8594+0=1.9621\text{+0}\text{.8}{\left({\overline{v}}_r\right)}_2^2\\ \text{0}\text{.8}{\left({\overline{v}}_r\right)}_2{}^2=3.8594-1.9621\\ \text{0}\text{.8}{\left({\overline{v}}_r\right)}_2{}^2=1.8973\\ {\left({\overline{v}}_r\right)}_2{}^2=\frac{1.8973}{0.8}\\ {\left({\overline{v}}_r\right)}_2{}^2=2.3716\\ {\left({\overline{v}}_r\right)}_2=\sqrt{2.3716}\\ {\left({\overline{v}}_r\right)}_2=1.540\text{m}/\text{s}\end{array}$

Thus, the velocity of the tube relative to the rod as the tube strikes end E of the assembly is $\underset{\_}{1.540\text{m/s}}$.