Chapter 17.2, Problem 17.88P

To determine

Find the speed of the rod relative to the tube when $x=400\text{mm}$.

Answer to Problem 17.88P

The speed of the rod relative to the tube when $x=400\text{mm}$ is $\underset{\_}{2.51\text{m/s}}$.

Explanation of Solution

Given information:

The mass $\left(m_R\right)$ of the rod AB is 4 kg.

The mass $\left(m_T\right)$ of the tube CD is 6 kg.

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

Calculation:

Consider l is the length of the rod and the tube and O is the point of intersection of the tube and axle.

Write the equation of the centroidal mass moment of inertia $\left({\overline{I}}_T\right)$ of the tube.

${\overline{I}}_T=\frac{1}{12}{m}_T{l}^2$

Write the equation of the centroidal mass moment of inertia $\left({\overline{I}}_R\right)$ of the rod.

${\overline{I}}_R=\frac{1}{12}{m}_R{l}^2$

Write the equation of the tangential equation of the tube using kinematics.

${\left({\overline{v}}_{\theta }\right)}_T={\overline{r}}_T\omega$

Substitute $\frac{l}{2}$ for ${\overline{r}}_T$.

${\left({\overline{v}}_{\theta }\right)}_T=\frac{l}{2}\omega$

Write the equation of the tangential equation of the rod using kinematics.

${\left({\overline{v}}_{\theta }\right)}_R={\overline{r}}_R\omega$

Substitute $\frac{l}{2}+x$ for ${\overline{r}}_R$.

${\left({\overline{v}}_{\theta }\right)}_R=\left(\frac{l}{2}+x\right)\omega$

Find the equation of angular momentum about point O.

$H_O=m_T{\overline{r}}_T{\left({\overline{v}}_{\theta }\right)}_T+{\overline{I}}_T\omega +m_R{\overline{r}}_R{\left({\overline{v}}_{\theta }\right)}_R+{\overline{I}}_R\omega$

Substitute $\frac{l}{2}$ for ${\overline{r}}_T$, $\frac{l}{2}\omega$ for ${\left({\overline{v}}_{\theta }\right)}_T$, $\frac{1}{12}{m}_T{l}^2$ for ${\overline{I}}_T$, $\frac{l}{2}+x$ for ${\overline{r}}_R$, $\left(\frac{l}{2}+x\right)\omega$ for ${\left({\overline{v}}_{\theta }\right)}_R$, and $\frac{1}{12}{m}_R{l}^2$ for ${\overline{I}}_R$.

$\begin{array}{c}{H}_O=m_T\frac{l}{2}\left(\frac{l}{2}\omega \right)+\frac{1}{12}{m}_T{l}^2\omega +m_R\left(\frac{l}{2}+x\right)\left(\frac{l}{2}+x\right)\omega +\frac{1}{12}{m}_R{l}^2\omega \\ =\frac{4}{12}{m}_T{l}^2\omega +m_R\left(\frac{l^2}{4}+2\frac{l}{2}x+x^2+\frac{1}{12}{l}^2\right)\omega \\ =\frac{1}{3}{m}_T{l}^2\omega +m_R\left(\frac{4l^2}{12}+lx+x^2\right)\omega \\ =\left[\frac{1}{3}{m}_T{l}^2+m_R\left(\frac{1}{3}{l}^2+lx+x^2\right)\right]\omega \end{array}$ (1)

Find the equation of kinetic energy.

$T=\frac{1}{2}{m}_T{\left({\overline{v}}_{\theta }\right)}_T^2+\frac{1}{2}{\overline{I}}_T{\omega }^2+\frac{1}{2}{m}_R{\left({\overline{v}}_{\theta }\right)}_R^2+\frac{1}{2}{m}_R{v}_r^2+\frac{1}{2}{\overline{I}}_R{\omega }^2$

Substitute $\frac{l}{2}\omega$ for ${\left({\overline{v}}_{\theta }\right)}_T$, $\frac{1}{12}{m}_T{l}^2$ for ${\overline{I}}_T$, $\left(\frac{l}{2}+x\right)\omega$ for ${\left({\overline{v}}_{\theta }\right)}_R$, and $\frac{1}{12}{m}_R{l}^2$ for ${\overline{I}}_R$.

$\begin{array}{c}T=\frac{1}{2}{m}_T{\left(\frac{l}{2}\omega \right)}^2+\frac{1}{2}\left(\frac{1}{12}{m}_T{l}^2\right){\omega }^2+\frac{1}{2}{m}_R{\left(\frac{l}{2}+x\right)}^2{\omega }^2+\frac{1}{2}{m}_R{v}_r^2+\frac{1}{2}\left(\frac{1}{12}{m}_R{l}^2\right){\omega }^2\\ =\frac{1}{2}\left[m_T\frac{l^2}{4}{\omega }^2+\frac{1}{12}{m}_T{l}^2{\omega }^2+m_R\left(\frac{l^2}{4}+lx+x^2\right){\omega }^2+\frac{1}{12}{m}_R{l}^2{\omega }^2\right]+\frac{1}{2}{m}_R{v}_r^2\\ =\frac{1}{2}\left[\frac{1}{3}{m}_T{l}^2+m_R\left(\frac{1}{3}{l}^2+lx+x^2\right)\right]{\omega }^2+\frac{1}{2}{m}_R{v}_r^2\end{array}$ (2)

Substitute Equation (1) in Equation (2),

$T=\frac{1}{2}{H}_O\omega +\frac{1}{2}{m}_R{v}_r^2$ (3)

All the motion in the system is horizontal. Therefore, the potential energy is zero.

Consider the initial position. $\left(x=0\right)$.

At the initial position, the initial angular velocity of the system is 5 rad/s and the radial velocity is zero.$\left(\omega ={\omega }_1=5\text{rad/s}\text{and}{v}_r=0\right)$.

Find the angular momentum at the initial position using equation (1).

${\left(H_O\right)}_1=\left[\frac{1}{3}{m}_T{l}^2+m_R\left(\frac{1}{3}{l}^2+lx+x^2\right)\right]{\omega }_1$

Substitute 0 for x.

$\begin{array}{c}{\left(H_O\right)}_1=\left\{\frac{1}{3}{m}_T{l}^2+m_R\left[\frac{1}{3}{l}^2+l\left(0\right)+{\left(0\right)}^2\right]\right\}{\omega }_1\\ =\left(\frac{1}{3}{m}_T{l}^2+\frac{1}{3}{m}_R{l}^2\right){\omega }_1\\ =\frac{1}{3}\left(m_T+m_R\right)l^2{\omega }_1\end{array}$

Substitute 6 kg for $m_T$, 4 kg for $m_R$, 0.8 m for $l$, and 5 rad/s for ${\omega }_1$.

$\begin{array}{c}{\left(H_O\right)}_1=\frac{1}{3}\left(6+4\right){\left(0.8\right)}^2\left(5\right)\\ =10.6667\text{kg}\cdot {\text{m}}^2\text{/s}\end{array}$

Find the kinetic energy at initial position using equation (3).

$T_1=\frac{1}{2}{\left(H_O\right)}_1{\omega }_1+\frac{1}{2}{m}_R{v}_r^2$

Substitute $10.6667\text{kg}\cdot {\text{m}}^2\text{/s}$ for $H_O$, 5 rad/s for ${\omega }_1$, and 0 for $v_r$.

$\begin{array}{c}{T}_1=\frac{1}{2}\left(10.6667\right)\left(5\right)+\frac{1}{2}{m}_R{\left(0\right)}^2\\ =26.6667\text{J}\end{array}$

Consider the final position. $\left(x=0.4\text{m}\right)$.

Find the angular momentum at the final position using equation (1).

${\left(H_O\right)}_2=\left[\frac{1}{3}{m}_T{l}^2+m_R\left(\frac{1}{3}{l}^2+lx+x^2\right)\right]{\omega }_1$

Substitute 0.4 m for x.

${\left(H_O\right)}_2=\left\{\frac{1}{3}{m}_T{l}^2+m_R\left[\frac{1}{3}{l}^2+lx+x^2\right]\right\}{\omega }_2$

Substitute 6 kg for $m_T$, 0.8 m for l, 4 kg for $m_R$, and 0.4 m for x.

$\begin{array}{c}{\left(H_O\right)}_2=\left\{\frac{1}{3}\left(6\right){\left(0.8\right)}^2+4\left[\frac{1}{3}{\left(0.8\right)}^2+\left(0.8\right)\left(0.4\right)+{\left(0.4\right)}^2\right]\right\}{\omega }_2\\ =4.0533{\omega }_2\end{array}$

Find the kinetic energy at final position using equation (3).

$T_2=\frac{1}{2}{\left(H_O\right)}_2{\omega }_2+\frac{1}{2}{m}_R{v}_r^2$

Substitute $4.0533{\omega }_2$ for ${\left(H_O\right)}_2$ and 4 kg for $m_R$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(4.0533{\omega }_2\right){\omega }_2+\frac{1}{2}\left(4\right)v_r^2\\ =2.02665{\omega }_2^2+2v_r^2\end{array}$

Consider the conservation of angular momentum.

${\left(H_O\right)}_1={\left(H_O\right)}_2$

Substitute $10.6667\text{kg}\cdot {\text{m}}^2\text{/s}$ for ${\left(H_O\right)}_1$ and $4.0533{\omega }_2$ for ${\left(H_O\right)}_2$.

$\begin{array}{l}10.6667=4.0533{\omega }_2\\ {\omega }_2=2.6316\text{rad/s}\end{array}$

Find the speed of the rod relative to the tube when $x=400\text{mm}$.

Consider the conservation of energy.

$T_1+V_1=T_2+V_2$

Substitute $26.6667\text{J}$ for $T_1$, 0 for $V_1$, $2.02665{\omega }_2^2+2v_r^2$ for $T_2$, and 0 for $V_2$.

$\begin{array}{l}26.6667+0=2.02665{\omega }_2^2+2v_r^2+0\\ 2.02665{\omega }_2^2+2v_r^2=26.6667\end{array}$

Substitute $2.6316\text{rad/s}$ for ${\omega }_2$.

$\begin{array}{l}2.02665{\left(2.6316\right)}^2+2v_r^2=26.6667\\ v_r=2.51\text{m/s}\end{array}$

Thus, the speed of the rod relative to the tube when $x=400\text{mm}$ is $\underset{\_}{2.51\text{m/s}}$.