Chapter 17.2, Problem 17.83P

(a)

To determine

Find the angular velocity of the assembly after the tube has moved to end E.

Answer to Problem 17.83P

The angular velocity of the assembly after the tube has moved to end E is $\underset{\_}{2.54\text{rad/s}}$.

Explanation of Solution

Given information:

The mass (m) of the tube AB is 1.6 kg.

The initial angular velocity $\left({\omega }_1\right)$ of the system 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.$0.30\text{kg}\cdot {\text{m}}^{\text{2}}$

The centroidal moment of inertia of the tube about a vertical axis is $0.0025\text{kg}\cdot {\text{m}}^{\text{2}}$.

Calculation:

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

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

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

$\begin{array}{c}{r}_{{\left(G/A\right)}_2}=r_{{\left(G/C\right)}_2}\\ =500-62.5\\ =\text{437}\text{.5}\text{mm}\end{array}$

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

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

Consider the principal of impulse and momentum.

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

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

Refer Figure (1).

Find the angular velocity of the assembly after the tube has moved to end E.

Take moment about C.

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

Here, ${\left({\overline{v}}_{\theta }\right)}_1$ 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 $r_{{\left(G/C\right)}_2}$.

$\begin{array}{l}\left[\begin{array}{l}0.0025+0.30\\ +1.6{\left(62.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\end{array}\right]\left(5\right)=\left[\begin{array}{l}0.0025+0.30\\ +1.6{\left(\text{437}\text{.5}\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\end{array}\right]{\omega }_2\\ \left(0.30875\right)\left(5\right)=0.60875{\omega }_2\\ 0.60875{\omega }_2=1.54375\\ {\omega }_2=\frac{1.54375}{0.60875}\\ {\omega }_2=2.54\text{rad}/\text{s}\end{array}$

Thus, the angular velocity of the assembly after the tube has moved to end E is $\underset{\_}{2.54\text{rad/s}}$.

(b)

To determine

Find the energy lost during the plastic impact at E.

Answer to Problem 17.83P

The energy lost during the plastic impact at E.is $\underset{\_}{1.902\text{J}}$.

Explanation of Solution

Calculation:

Write the equation of the kinetic energy $\left(T_1\right)$ of the system at initial position.

$T_1=\frac{1}{2}{\overline{I}}_{AB}{\omega }_1{}^2+\frac{1}{2}{\overline{I}}_{DCE}{\omega }_1{}^2+\frac{1}{2}{m}_{AB}{\left({\overline{v}}_{\theta }\right)}_1$

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

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

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 ${\omega }_1$, and $62.5\text{mm}$ for $r_{{\left(G/C\right)}_1}$.

$\begin{array}{c}{T}_1=\frac{1}{2}\left[0.0025+0.30+1.6{\left(62.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\right]{\left(5\right)}^2\\ =3.859375\text{J}\end{array}$

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

$T_2=\frac{1}{2}{\overline{I}}_{AB}{\omega }_2{}^2+\frac{1}{2}{\overline{I}}_{DCE}{\omega }_2{}^2+\frac{1}{2}{m}_{AB}{\left({\overline{v}}_{\theta }\right)}_2$

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

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

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 $m_{AB}$, $2.54\text{rad}/\text{s}$ for ${\omega }_2$, and $437.5\text{mm}$ for $r_{{\left(G/C\right)}_2}$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left[0.0025+0.30+1.6{\left(437.5\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\right]{\left(2.54\right)}^2\\ =1.9573\text{J}\end{array}$

Find the energy lost during plastic impact using the equation:

$\text{Energy lost}=T_1-T_2$

Substitute $3.8593\text{J}$ for $T_1$ and $1.9573\text{J}$ for $T_2$.

$\begin{array}{c}\text{Energy lost}=3.8593-1.9573\\ =1.902\text{J}\end{array}$

Thus, the speed of the ball at time $\left(t_1\right)$.is $\underset{\_}{16.95\text{ft/s}}$.