Chapter 18, Problem 18.150RP

A uniform rod of mass m and length 5a is bent into the shape shown and is suspended from a wire attached at point B. Knowing that the rod is hit at point A in the negative y direction and denoting the corresponding impulse by $-\left(\text{F}\Delta t\right)\text{j}$ , determine immediately after the impact (a) the velocity of the mass center G, (b) the angular velocity of the rod.

  

To determine

The velocity of the mass centre $G$.

Answer to Problem 18.150RP

The velocity of the mass centre $G$ is $0$.

Explanation of Solution

Given information:

The mass of the rod is $m$, the length of the rod is $5a$ and the impulse in the $y$ direction is $-\left(\text{F }\Delta t\right)j$.

The below figure represent the schematic diagram of the rod section.

Figure-(1)

Write the expression of length of the section OA.

$L_1=\frac{L}{5}$       ...........(I)

Here, the total length of the rod is $L$.

Write the expression of length of the section AB.

$L_2=\frac{L}{5}$       ...........(II)

Write the expression of length of the section BC.

$L_3=\frac{L}{5}$       ...........(III)

Write the expression of length of the section CD.

$L_4=\frac{L}{5}$       ...........(IV)

Write the expression of length of the section DE.

$L_5=\frac{L}{5}$       ...........(V)

Write the expression of mass of the section OA.

$m_1=\frac{m}{5}$

Here, the total mass of the rod is $L$.

Write the expression of mass of the section AB.

$m_2=\frac{m}{5}$

Write the expression of mass of the section BC.

$m_3=\frac{m}{5}$

Write the expression of mass of the section CD.

$m_4=\frac{m}{5}$

Write the expression of mass of the section DE.

$m_5=\frac{m}{5}$

Write the expression of moment of inertia of section OA in $x$ direction.

${\left(I_x\right)}_1=m_1\frac{L_1{}^2}{12}+m_1\frac{L_1{}^2}{4}+m_1\frac{L_1{}^2}{4}$ ........ (VI)

Write the expression of moment of inertia of section OA in $y$ direction.

${\left(I_y\right)}_1=m_1\frac{L_1{}^2}{12}+m_1{L}_1{}^2+m_1\frac{L_1{}^2}{4}$ ........ (VII)

Write the expression of moment of inertia of section OA in $z$ direction.

${\left(I_z\right)}_1=m_1{L}_1{}^2+m_1\frac{L_1{}^2}{4}$ ........ (VIII)

Write the expression of moment of inertia of section OA in $xy$ direction.

${\left(I_{xy}\right)}_1=-m_2\frac{L_2{}^2}{2}$ ........ (IX)

Write the expression of moment of inertia of section AB in $x$ direction.

${\left(I_x\right)}_2=m_2\frac{L_2{}^2}{4}$ ........ (X)

Write the expression of moment of inertia of section AB in $y$ direction.

${\left(I_y\right)}_2=m_2\frac{L_2{}^2}{3}$ ........ (XI)

Write the expression of moment of inertia of section AB in $z$ direction.

${\left(I_z\right)}_2=m_2\frac{L_2{}^2}{12}+m_2\frac{L_2{}^2}{4}+m_2\frac{L_2{}^2}{4}$ ........ (XII)

Write the expression of moment of inertia of section AB in $xy$ direction.

${\left(I_{xy}\right)}_2=-m_2\frac{L_2{}^2}{4}$ ........ (XIII)

Write the expression of moment of inertia of section BC in $x$ direction.

${\left(I_x\right)}_3=m_3\frac{L_3{}^2}{4}$ ........ (XIV)

Write the expression of moment of inertia of section BC in $x$ direction.

${\left(I_y\right)}_3=0$

Write the expression of moment of inertia of section BC in $z$ direction.

${\left(I_z\right)}_3=m_3\frac{L_3{}^2}{12}$ ........ (XV)

Write the expression of moment of inertia of section BC in $xy$ direction.

${\left(I_{xy}\right)}_3=0$

Write the expression of moment of inertia of section CD in $x$ direction.

${\left(I_x\right)}_4=m_4\frac{L_4{}^2}{4}$ ........ (XVI)

Write the expression of moment of inertia of section CD in $y$ direction.

${\left(I_y\right)}_4=m_4\frac{L_4{}^2}{3}$ ........ (XVII)

Write the expression of moment of inertia of section CD in $z$ direction.

${\left(I_z\right)}_4=m_4\frac{L_4{}^2}{12}+m_4\frac{L_4{}^2}{4}+m_4\frac{L_4{}^2}{4}$ ........ (XVIII)

Write the expression of moment of inertia of section CD in $xy$ direction.

${\left(I_{xy}\right)}_4=-m_4\frac{L_4{}^2}{4}$ ........ (XIX)

Write the expression of moment of inertia of section DE in $x$ direction.

${\left(I_x\right)}_5=m_5\frac{L_5{}^2}{12}+m_5\frac{L_5{}^2}{4}+m_5\frac{L_5{}^2}{4}$ ........ (XX)

Write the expression of moment of inertia of section DE in $y$ direction.

${\left(I_y\right)}_5=m_5\frac{L_5{}^2}{12}+m_5{L}_5{}^2+m_5\frac{L_5{}^2}{4}$ ........ (XXI)

Write the expression of moment of inertia of section DE in $z$ direction.

${\left(I_z\right)}_5=m_5{L}_5{}^2+m_5\frac{L_5{}^2}{4}$ ........ (XXII)

Write the expression of moment of inertia of section DE in $xy$ direction.

${\left(I_{xy}\right)}_5=-m_5\frac{L_5{}^2}{2}$ ........ (XXIII)

Write the expression of total moment of inertia in $x$ direction.

$I_x={\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_3+{\left(I_x\right)}_4+{\left(I_x\right)}_5$ ........ (XXIV)

Write the expression of total moment of inertia in $y$ direction.

$I_y={\left(I_y\right)}_1+{\left(I_y\right)}_2+{\left(I_y\right)}_3+{\left(I_y\right)}_4+{\left(I_y\right)}_5$ ........ (XXV)

Write the expression of total moment of inertia in $z$ direction.

$I_z={\left(I_z\right)}_1+{\left(I_z\right)}_2+{\left(I_z\right)}_3+{\left(I_z\right)}_4+{\left(I_z\right)}_5$ ........ (XXVI)

Write the expression of total moment of inertia in $xy$ direction.

$I_{xy}={\left(I_{xy}\right)}_1+{\left(I_{xy}\right)}_2+{\left(I_{xy}\right)}_3+{\left(I_{xy}\right)}_4+{\left(I_{xy}\right)}_5$ ........ (XXVII)

Write the expression of total moment of inertia in $xz$ direction.

$I_{xz}=\left(m_1\frac{L_1{}^2}{2}\right)+\left(m_4\frac{L_4{}^2}{2}\right)$       ...........(XXVIII)

Write the expression of total moment of inertia in $yz$ direction.

$I_{yz}=\left(-m_2\frac{L_2{}^2}{4}\right)+\left(m_5\frac{L_5{}^2}{4}\right)$       ...........(XXIX)

Write the expression of angular momentum about $x$ axis.

${\left(H_G\right)}_x=I_x{\omega }_x-I_{xy}{\omega }_y-I_{xz}{\omega }_z$       ...........(XXX)

Write the expression of angular momentum about $y$ axis.

${\left(H_G\right)}_y=I_{xy}{\omega }_x-I_y{\omega }_y-I_{yz}{\omega }_z$       ...........(XXXI)

Write the expression of angular momentum about $z$ axis.

${\left(H_G\right)}_z=-I_{xz}{\omega }_x-I_{yz}{\omega }_y-I_z{\omega }_z$       ...........(XXXII)

Write the expression of velocity of mass centre $G$.

$V=\frac{I+\left(\text{F }\Delta t\right)j}{m}$ ........ (XXXIII)

Here, the force is $\text{F}$, the change in time is $\Delta t$, the total mass is $m$, the impact is $I$ and the unit vector in $y$ direction is $j$.

Calculation:

Substitute $5a$ for $L$ in Equation (I).

$\begin{array}{c}{L}_1=\frac{5a}{5}\\ =a\end{array}$

Substitute $5a$ for $L$ in Equation (II).

$\begin{array}{c}{L}_2=\frac{5a}{5}\\ =a\end{array}$

Substitute $5a$ for $L$ in Equation (III).

$\begin{array}{c}{L}_3=\frac{5a}{5}\\ =a\end{array}$

Substitute $5a$ for $L$ in Equation (IV).

$\begin{array}{c}{L}_4=\frac{5a}{5}\\ =a\end{array}$

Substitute $5a$ for $L$ in Equation (V).

$\begin{array}{c}{L}_5=\frac{5a}{5}\\ =a\end{array}$

Substitute $\frac{m}{5}$ for $m_1$, $a$ for $L_1$ in Equation (VI).

$\begin{array}{c}{\left(I_x\right)}_1=\frac{m}{5}\frac{a^2}{12}+\frac{m}{5}\frac{a^2}{4}+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{60}+\frac{m{a}^2}{20}+\frac{m{a}^2}{20}\\ =\frac{7m{a}^2}{60}\end{array}$

Substitute $\frac{m}{5}$ for $m_1$, $a$ for $L_1$ in Equation (VII).

$\begin{array}{c}{\left(I_y\right)}_1=\frac{m}{5}\frac{a^2}{12}+\frac{m}{5}{a}^2+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{60}+\frac{m{a}^2}{5}+\frac{m{a}^2}{20}\\ =\frac{4m{a}^2}{15}\end{array}$

Substitute $\frac{m}{5}$ for $m_1$, $a$ for $L_1$ in Equation (VIII).

$\begin{array}{c}{\left(I_z\right)}_1=\frac{m}{5}{a}^2+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{5}+\frac{m{a}^2}{20}\\ =\frac{m{a}^2}{4}\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $a$ for $L_2$ in Equation (IX).

$\begin{array}{c}{\left(I_{xy}\right)}_1=-\frac{m}{5}{\frac{a}{2}}^2\\ =-\frac{m{a}^2}{10}\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $a$ for $L_2$ in Equation (X).

$\begin{array}{c}{\left(I_x\right)}_2=\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{20}\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $a$ for $L_2$ in Equation (XI).

$\begin{array}{c}{\left(I_y\right)}_2=\frac{m}{5}\frac{a^2}{3}\\ =\frac{m{a}^2}{15}\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $a$ for $L_2$ in Equation (XII).

$\begin{array}{c}{\left(I_z\right)}_2=\frac{m}{5}\frac{a^2}{12}+\frac{m}{5}\frac{a^2}{4}+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{60}+\frac{m{a}^2}{20}+\frac{m{a}^2}{20}\\ =\frac{7m{a}^2}{60}\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $a$ for $L_2$ in Equation (XIII).

$\begin{array}{c}{\left(I_{xy}\right)}_2=-\frac{m}{5}\frac{a^2}{4}\\ =-\frac{m{a}^2}{20}\end{array}$

Substitute $\frac{m}{5}$ for $m_3$, $a$ for $L_3$ in Equation (XIV).

$\begin{array}{c}{\left(I_x\right)}_3=\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{20}\end{array}$

Substitute $\frac{m}{5}$ for $m_3$, $a$ for $L_3$ in Equation (XV).

$\begin{array}{c}{\left(I_z\right)}_3=\frac{m}{5}\frac{a^2}{12}\\ =\frac{m{a}^2}{60}\end{array}$

Substitute $\frac{m}{5}$ for $m_4$, $a$ for $L_4$ in Equation (X).

$\begin{array}{c}{\left(I_x\right)}_4=\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{20}\end{array}$

Substitute $\frac{m}{5}$ for $m_4$, $a$ for $L_4$ in Equation (XVII).

$\begin{array}{c}{\left(I_y\right)}_4=\frac{m}{5}\frac{a^2}{3}\\ =\frac{m{a}^2}{15}\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $a$ for $L_2$ in Equation (XVIII).

$\begin{array}{c}{\left(I_z\right)}_4=\frac{m}{5}\frac{a^2}{12}+\frac{m}{5}\frac{a^2}{4}+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{60}+\frac{m{a}^2}{20}+\frac{m{a}^2}{20}\\ =\frac{7m{a}^2}{60}\end{array}$

Substitute $\frac{m}{5}$ for $m_4$, $a$ for $L_4$ in Equation (XIX).

$\begin{array}{c}{\left(I_{xy}\right)}_4=-\frac{m}{5}\frac{a^2}{4}\\ =-\frac{m{a}^2}{20}\end{array}$

Substitute $\frac{m}{5}$ for $m_5$, $a$ for $L_5$ in Equation (XX).

$\begin{array}{c}{\left(I_x\right)}_5=\frac{m}{5}\frac{a^2}{12}+\frac{m}{5}\frac{a^2}{4}+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{60}+\frac{m{a}^2}{20}+\frac{m{a}^2}{20}\\ =\frac{7m{a}^2}{60}\end{array}$

Substitute $\frac{m}{5}$ for $m_5$, $a$ for $L_5$ in Equation (XXI).

$\begin{array}{c}{\left(I_y\right)}_5=\frac{m}{5}\frac{a^2}{12}+\frac{m}{5}{a}^2+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{60}+\frac{m{a}^2}{5}+\frac{m{a}^2}{20}\\ =\frac{4m{a}^2}{15}\end{array}$

Substitute $\frac{m}{5}$ for $m_5$, $a$ for $L_5$ in Equation (XXII).

$\begin{array}{c}{\left(I_z\right)}_5=\frac{m}{5}{a}^2+\frac{m}{5}\frac{a^2}{4}\\ =\frac{m{a}^2}{5}+\frac{m{a}^2}{20}\\ =\frac{m{a}^2}{4}\end{array}$

Substitute $\frac{m}{5}$ for $m_5$, $a$ for $L_5$ in Equation (XXIII).

$\begin{array}{c}{\left(I_{xy}\right)}_5=-\frac{m}{5}{\frac{a}{2}}^2\\ =-\frac{m{a}^2}{10}\end{array}$

Substitute $\frac{7m{a}^2}{60}$ for ${\left(I_x\right)}_1$, $\frac{7m{a}^2}{60}$ for ${\left(I_x\right)}_5$, $\frac{m{a}^2}{20}$ for ${\left(I_x\right)}_2$, $\frac{m{a}^2}{20}$ for ${\left(I_x\right)}_4$, and $\frac{m{a}^2}{20}$ for ${\left(I_x\right)}_3$ in Equation (XXIV).

$\begin{array}{c}{I}_x=\left(\frac{7m{a}^2}{60}\right)+\left(\frac{m{a}^2}{20}\right)+\left(\frac{m{a}^2}{20}\right)+\left(\frac{m{a}^2}{20}\right)+\left(\frac{7m{a}^2}{60}\right)\\ =\left(0.23m{a}^2\right)+\left(0.15m{a}^2\right)\\ =0.38m{a}^2\end{array}$

Substitute $\frac{4m{a}^2}{15}$ for ${\left(I_y\right)}_1$, $\frac{m{a}^2}{15}$ for ${\left(I_y\right)}_2$, $0$ for ${\left(I_y\right)}_3$, $\frac{m{a}^2}{15}$ for ${\left(I_y\right)}_4$ and $\frac{4m{a}^2}{15}$ for ${\left(I_y\right)}_5$ in Equation (XXV).

$\begin{array}{c}{I}_y=\left(\frac{4m{a}^2}{15}\right)+\left(\frac{m{a}^2}{15}\right)+\left(0\right)+\left(\frac{m{a}^2}{15}\right)+\left(\frac{4m{a}^2}{15}\right)\\ =\left(0.53m{a}^2\right)+\left(0.13m{a}^2\right)+0\\ =0.68m{a}^2\end{array}$

Substitute $\frac{m{a}^2}{4}$ for ${\left(I_z\right)}_1$, $\frac{7m{a}^2}{60}$ for ${\left(I_z\right)}_2$, $\frac{m{a}^2}{60}$ for ${\left(I_z\right)}_3$, $\frac{7m{a}^2}{60}$ for ${\left(I_z\right)}_4$ and $\frac{m{a}^2}{4}$ for ${\left(I_z\right)}_5$ in Equation (XXVI).

$\begin{array}{c}{I}_z=\left(\frac{m{a}^2}{4}\right)+\left(\frac{7m{a}^2}{60}\right)+\left(\frac{m{a}^2}{60}\right)+\left(\frac{7m{a}^2}{60}\right)+\left(\frac{m{a}^2}{4}\right)\\ =\left(0.5m{a}^2\right)+\left(0.05m{a}^2\right)\\ =0.55m{a}^2\end{array}$

Substitute $-\frac{m{a}^2}{10}$ for ${\left(I_{xy}\right)}_1$, $-\frac{m{a}^2}{20}$ for ${\left(I_{xy}\right)}_2$, $0$ for ${\left(I_{xy}\right)}_3$, $-\frac{m{a}^2}{20}$ for ${\left(I_{xy}\right)}_4$ and $-\frac{m{a}^2}{10}$ for ${\left(I_{xy}\right)}_5$ in Equation (XXVII).

$\begin{array}{c}{I}_{xy}=\left(-\frac{m{a}^2}{10}\right)+\left(-\frac{m{a}^2}{20}\right)+\left(0\right)+\left(-\frac{m{a}^2}{20}\right)+\left(-\frac{m{a}^2}{10}\right)\\ =-\left(0.1m{a}^2\right)-\left(0.05m{a}^2\right)-\left(0.05m{a}^2\right)-\left(0.1m{a}^2\right)\\ =-0.2m{a}^2-0.1m{a}^2\\ =-0.3m{a}^2\end{array}$

Substitute $\frac{m}{5}$ for $m_1$, $\frac{m}{5}$ for $m_4$, $a$ for $L_1$ and $a$ for $L_2$ in Equation (XXVIII).

$\begin{array}{c}{I}_{xz}=\left(\frac{m}{5}\frac{a^2}{2}\right)+\left(\frac{m}{5}\frac{a^2}{2}\right)\\ =\frac{m{a}^2}{10}+\frac{m{a}^2}{10}\\ =0.2m{a}^2\end{array}$

Substitute $\frac{m}{5}$ for $m_2$, $\frac{m}{5}$ for $m_5$, $a$ for $L_2$ and $a$ for $L_5$ in Equation (XXIX).

$\begin{array}{c}{I}_{yz}=-\left(\frac{m}{5}\right)\frac{a^2}{4}-\left(\frac{m}{5}\right)\frac{a^2}{4}\\ =-\frac{m{a}^2}{20}-\frac{m{a}^2}{20}\\ =-0.1m{a}^2\end{array}$

Substitute $0.38m{a}^2$ for $I_x$, $-0.3m{a}^2$ for $I_{xy}$ and $0.2m{a}^2$ for $I_{xz}$ in Equation (XXX).

$\begin{array}{c}{\left(H_G\right)}_x=\left(0.38m{a}^2\right){\omega }_x-\left(-0.3m{a}^2\right){\omega }_y-\left(0.2m{a}^2\right){\omega }_z\\ =0.38m{a}^2{\omega }_x+0.3m{a}^2{\omega }_y-0.2m{a}^2{\omega }_z\end{array}$       ...........(XXXIII)

Substitute $-0.3m{a}^2$ for $I_{xy}$, $0.68m{a}^2$ for $I_y$ and $-0.1m{a}^2$ for $I_{yz}$ in Equation (XXXI)

$\begin{array}{c}{\left(H_G\right)}_y=\left(-0.3m{a}^2\right){\omega }_x-\left(0.68m{a}^2\right){\omega }_y-\left(-0.1m{a}^2\right){\omega }_z\\ =-0.3m{a}^2{\omega }_x-0.68m{a}^2{\omega }_y+0.1m{a}^2{\omega }_z\end{array}$       ...........(XXXIV)

Substitute $0.2m{a}^2$ for $I_{xz}$, $-0.1m{a}^2$ for $I_{yz}$ and $0.55m{a}^2$ for $I_z$ in Equation (XXXII)

$\begin{array}{c}{\left(H_G\right)}_z=-\left(0.2m{a}^2\right){\omega }_x-\left(-0.1m{a}^2\right){\omega }_y-\left(0.55m{a}^2\right){\omega }_z\\ =-0.2m{a}^2{\omega }_x+0.1m{a}^2{\omega }_y-0.55m{a}^2{\omega }_z\end{array}$       ...........(XXXV)

Substitute $-\left(\text{F }\Delta t\right)j$ for $I$ in Equation (XXXIII).

$\begin{array}{c}V=\frac{-\left(\text{F }\Delta t\right)j+\left(\text{F }\Delta t\right)j}{m}\\ =\frac{0}{m}\\ =0\end{array}$

Conclusion:

The velocity of the mass centre $G$ is $0$.

To determine

(b)

The angular velocity of the rod.

Answer to Problem 18.150RP

The angular velocity of the rod is $\frac{\Delta t}{ma}\left(2.5i-1.45j+2.2k\right)$.

Explanation of Solution

Given information:

Write the expression of angular momentum in $z$ direction.

${\left(H_G\right)}_z=L_{1}I$ .      ...........(XXXVI)

Write the expression of angular velocity of the rod.

$\omega =\left({\omega }_x\right)i+\left({\omega }_y\right)j+\left({\omega }_z\right)k$       ...........(XXXVII)

Here, the angular velocity in $x$ direction is ${\omega }_x$, the angular velocity in $y$ direction is ${\omega }_y$ and the angular velocity in $z$ direction is ${\omega }_z$.

Calculation:

Substitute $-\left(\text{F }\Delta t\right)$ for $I$ and $a$ for $L_1$ in Equation (XXXVI).

$\begin{array}{c}{\left(H_G\right)}_z=a\left(-\left(\text{F }\Delta t\right)\right)\\ =-a\left(\text{F }\Delta t\right)\end{array}$

Substitute $0$ for ${\left(H_G\right)}_x$ in Equation (XXXIII).

$0=0.38m{a}^2{\omega }_x+0.3m{a}^2{\omega }_y-0.2m{a}^2{\omega }_z$       ...........(XXXVIII)

Substitute $0$ for ${\left(H_G\right)}_y$ in Equation (XXXIV).

$0=-0.3m{a}^2{\omega }_x-0.68m{a}^2{\omega }_y+0.1m{a}^2{\omega }_z$       ...........(XXXIX)

Substitute $-a\left(\text{F }\Delta t\right)$ for ${\left(H_G\right)}_z$ in Equation (XXXV).

$-a\left(\text{F }\Delta t\right)=-0.2m{a}^2{\omega }_x+0.1m{a}^2{\omega }_y-0.55m{a}^2{\omega }_z$       ...........(XXXX)

Solve Equation (XXXIII), (XXXIX) and (XXXX).

${\omega }_x=\frac{2.5F\Delta t}{ma}$

${\omega }_y=\frac{-1.45F\Delta t}{ma}$

${\omega }_z=2.2\frac{F\Delta t}{ma}$

Substitute $\frac{2.5F\Delta t}{ma}$ for ${\omega }_x$, $\frac{-1.45F\Delta t}{ma}$ for ${\omega }_y$ and $2.2\frac{F\Delta t}{ma}$ for ${\omega }_z$ in Equation (XXXVII).

$\begin{array}{c}\omega =\left(\frac{2.5F\Delta t}{ma}\right)i+\left(\frac{-1.45F\Delta t}{ma}\right)j+\left(2.2\frac{F\Delta t}{ma}\right)k\\ =\frac{\Delta t}{ma}\left(2.5i-1.45j+2.2k\right)\end{array}$

Conclusion:

The angular velocity of the rod is $\frac{\Delta t}{ma}\left(2.5i-1.45j+2.2k\right)$.