Chapter 18.1, Problem 18.27P

To determine

(a)

The velocity of the mass centre $G$.

Answer to Problem 18.27P

The velocity of the mass centre $G$ is $-\left(0.3\text{m}/\text{s}\right)\text{k}$.

Explanation of Solution

Given information:

The mass of the each plate is $4\text{kg}$ and the corresponding impulse is $-\left(2.4\text{N}\cdot \text{s}\right)\text{k}$.

Write the expression for moment of the inertia about the x- axis.

$I_x=m\left(\frac{1}{4}{r}^2+d^2\right)$ ...... (I).

Here, the distance of mass centre from the circular plate is $d$ and the radius of the circular plate is $r$.

Write the expression for moment of the inertia about the y- axis.

$I_y=\frac{3}{2}m{r}^2$ ...... (II)

Write the expression for moment of the inertia about the z- axis.

$I_z=m\left(\frac{5}{4}{r}^2+d^2\right)$ ...... (III)

Write expression for the product moment of inertia about the plane $xy$.

$I_{xy}=-mrd$ ...... (IV)

Write expression for the product moment of inertia about the plane $yz$.

$I_{yz}=0$

Write expression for the product moment of inertia about the plane $zx$.

$I_{zx}=0$

The figure below shows the effective kinetic diagram of the system.

Figure-(1)

Write the expression for the impulse about point $D$.

$\left(\text{F}\Delta t\right)\text{k}+\left(A\Delta t\right)\text{j}=2m\left(v_x\text{i}+v_z\text{k}\right)$ ...... (V)

Write the expression for the velocity of the mass centre of the system.

$v=v_x\text{i}+v_y\text{j}+v_z\text{k}$ ....... (VI)

Here, the velocity of the mass centre in x- direction is $v_x$, the velocity of the mass centre in y- direction is $v_y$ and the velocity of the mass centre in z- direction is $v_z$.

Calculation:

Substitute $4\text{kg}$ for $m$, $180\text{mm}$ for $r$ and $150\text{mm}$ for $d$ in Equation (I).

$\begin{array}{c}{I}_x=\left(4\text{kg}\right)\left(\frac{1}{4}{\left(180\text{mm}\right)}^2+{\left(150\text{mm}\right)}^2\right)\\ =\left(4\text{kg}\right)\left(\frac{1}{4}{\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2+{\left(150\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\right)\\ =\left(4\text{kg}\right)\left(0.25\left(0.180{\text{m}}^2\right)+\left(0.150{\text{m}}^2\right)\right)\\ =0.1224\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Substitute $4\text{kg}$ for $m$ and $180\text{mm}$ for $r$ in Equation (II).

$\begin{array}{c}{I}_y=\frac{3}{2}\left(4\text{kg}\right){\left(180\text{mm}\right)}^2\\ =\frac{3}{2}\left(4\text{kg}\right){\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\\ =\frac{3}{2}\left(4\text{kg}\right)\left(0.180{\text{m}}^2\right)\\ =0.1944\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Substitute $4\text{kg}$ for $m$, $180\text{mm}$ for $r$ and $150\text{mm}$ for $d$ in Equation (III).

$\begin{array}{c}{I}_z=4\text{kg}\left(\frac{5}{4}{\left(180\text{mm}\right)}^2+{\left(150\text{mm}\right)}^2\right)\\ =4\text{kg}\left(\frac{5}{4}{\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2+{\left(150\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\right)\\ =4\text{kg}\left(\frac{5}{4}\left(0.180{\text{m}}^2\right)+\left(0.150{\text{m}}^2\right)\right)\\ =0.252\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Substitute $4\text{kg}$ for $m$, $180\text{mm}$ for $r$ and $150\text{mm}$ for $d$ in Equation (IV).

$\begin{array}{c}{I}_{xy}=-\left(4\text{kg}\right)\left(180\text{mm}\right)\left(150\text{mm}\right)\\ =-\left(4\text{kg}\right)\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\left(150\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\\ =-\left(4\text{kg}\right)\left(0.180\text{m}\right)\left(0.150\text{m}\right)\\ =-0.108\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $4\text{kg}$ for $m$ and $-\left(2.4\text{N}\cdot \text{s}\right)\text{k}$ for $\text{F}\Delta t$ in Equation (V).

$\begin{array}{l}\left(-\left(2.4\text{N}\cdot \text{s}\right)\text{k}\right)\text{k}+\left(A\Delta t\right)\text{j}=2\left(4\text{kg}\right)\left(v_x\text{i}+v_z\text{k}\right)\\ \left(-\left(2.4\text{N}\cdot \text{s}\right)\text{k}\right)\text{k}+\left(A\Delta t\right)\text{j}=\left(\left(8\text{kg}\right)v_x\text{i}+\left(8\text{kg}\right)v_z\text{k}\right)\end{array}$ ...... (VII)

Equate coefficient of $\text{i}$ in Equation (VII).

$\begin{array}{l}\left(8\text{kg}\right)v_x\text{i}=0\\ v_x=0\end{array}$

Equate coefficient of $\text{j}$ in Equation (VII).

$\begin{array}{l}\left(A\Delta t\right)\text{j}=0\\ A\Delta t=0\end{array}$

Equate coefficient of $\text{k}$ in Equation (VII).

$\begin{array}{l}\left(8\text{kg}\right)v_z\text{k}=\left(-\left(2.4\text{N}\cdot \text{s}\right)\text{k}\right)\\ v_z=\frac{\left(-\left(2.4\text{N}\cdot \text{s}\right)\right)}{\left(8\text{kg}\right)}\\ v_z=-0.3\text{N}\cdot \text{s}/\text{kg}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ v_z=-0.3\text{m}/\text{s}\end{array}$

Substitute $-0.3\text{m}/\text{s}$ for $v_z$, $0$ for $v_y$ and $0$ for $v_x$ in Equation (VI).

$\begin{array}{c}v=\left(0\right)\text{i}+\left(0\right)\text{j}+\left(-0.3\text{m}/\text{s}\right)\text{k}\\ =-\left(0.3\text{m}/\text{s}\right)\text{k}\end{array}$

Conclusion:

The velocity of the mass centre $G$ is $-\left(0.3\text{m}/\text{s}\right)\text{k}$.

To determine

(b)

The angular velocity of the assembly.

Answer to Problem 18.27P

The angular velocity of the assembly is $\omega =\left(-0.9154\text{rad}/\text{s}\right)\text{i}+\left(-0.5769\text{rad}/\text{s}\right)\text{j}$.

Explanation of Solution

Write the expression for the moments about centre point $G$.

$\left[-r\text{i}+d\text{j}\right]\times \text{F}\Delta t=H_x\text{i}+H_y\text{j}+H_z\text{k}$ ...... (VIII)

Here, the angular momentum about x- direction is $H_x$, the angular momentum about y- direction is $H_y$ and the angular momentum about z- direction is $H_z$.

Write the expression for the angular momentum in x- direction.

$H_x=I_x{\omega }_x-I_{xy}{\omega }_y-I_{xz}{\omega }_z$ ...... (IX)

Write the expression for the angular momentum in y- direction.

$H_y=-I_{xy}{\omega }_x+I_y{\omega }_y-I_{xz}{\omega }_z$ ...... (X)

Write the expression for the angular momentum in z- direction.

$H_z=-I_{xy}{\omega }_x+I_{yz}{\omega }_y-I_z{\omega }_z$ ...... (XI)

Write the expression for the angular velocity of the assembly.

$\overline{\omega }={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$ ...... (XII)

Calculation:

Substitute $-\left(2.4\text{N}\cdot \text{s}\right)\text{k}$ for $\text{F}\Delta t$, $180\text{mm}$ for $r$ and $150\text{mm}$ for $d$ in Equation (VIII).

$\begin{array}{l}\left[-\left(180\text{mm}\right)\text{i}+\left(150\text{mm}\right)\text{j}\right]\times \left(-\left(2.4\text{N}\cdot \text{s}\right)\text{k}\right)=H_x\text{i}+H_y\text{j}+H_z\text{k}\\ \left\{\begin{array}{l}\left[-\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\text{i}+\left(150\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\text{j}\right]\times \left(-\left(2.4\text{N}\cdot \text{s}\right)\text{k}\right)=\\ H_x\text{i}+H_y\text{j}+H_z\text{k}\end{array}\right\}\\ \left[-\left(0.180\text{m}\right)\text{i}+\left(0.150\text{m}\right)\text{j}\right]\times \left(-\left(2.4\text{N}\cdot \text{s}\right)\text{k}\right)=H_x\text{i}+H_y\text{j}+H_z\text{k}\end{array}$ .... (XIII)

Equate coefficient of $\text{i}$ in Equation (XIII).

$H_x=-0.360\text{kg}\cdot {\text{m}}^{\text{2}}$

Equate coefficient of $\text{j}$ in Equation (XIII).

$H_x=-0.432\text{kg}\cdot {\text{m}}^{\text{2}}$

Equate coefficient of $\text{k}$ in Equation (XIII).

$H_z=0$

Substitute $-0.360\text{kg}\cdot {\text{m}}^{\text{2}}$ for $H_x$, $0.2448\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_x$, $0$ for $I_{zx}$ and $0.216\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_{xy}$ in Equation (IX).

$\begin{array}{c}-0.360\text{kg}\cdot {\text{m}}^{\text{2}}=\left(0.2448\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_x-\left(0.2448\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_y-\left(0\right){\omega }_z\\ -0.360\text{kg}\cdot {\text{m}}^{\text{2}}=\left(0.2448\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_x-\left(0.2448\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_y\end{array}$ ...... (XIV)

Substitute $-0.432\text{kg}\cdot {\text{m}}^{\text{2}}$ for $H_y$, $0.3888\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_y$, $0.216\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_{xy}$ and $0$ for $I_{xz}$ in Equation (IX).

$\begin{array}{l}-0.432\text{kg}\cdot {\text{m}}^{\text{2}}=\left(0.216\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_x+\left(0.3888\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_y-\left(0\right){\omega }_z\\ -0.432\text{kg}\cdot {\text{m}}^{\text{2}}=\left(0.216\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_x+\left(0.3888\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_y\\ {\omega }_x=\frac{-0.432\text{kg}\cdot {\text{m}}^{\text{2}}-\left(0.3888\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_y}{\left(0.216\text{kg}\cdot {\text{m}}^{\text{2}}\right)}\end{array}$ ..... (XV)

The angular velocity in the z- direction is zero that is ${\omega }_z=0$.

Substitute $\frac{-0.432\text{kg}\cdot {\text{m}}^{\text{2}}-\left(0.3888\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }_y}{\left(0.216\text{kg}\cdot {\text{m}}^{\text{2}}\right)}$ for ${\omega }_x$ in Equation (XIV).

Substitute $-0.5769\text{rad}/\text{s}$ for ${\omega }_y$ in Equation (XV).

$\begin{array}{c}{\omega }_x=\frac{-0.432\text{kg}\cdot {\text{m}}^{\text{2}}-\left(0.3888\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(-0.5769\text{rad}/\text{s}\right)}{\left(0.216\text{kg}\cdot {\text{m}}^{\text{2}}\right)}\\ =\frac{-0.432\text{kg}\cdot {\text{m}}^{\text{2}}\left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}\left(\text{m}/\text{s}\right)\left(\text{rad}/\text{s}\right)}{\text{1}\text{N}\cdot \text{m}}\right)-\left(-0.2242\text{rad}/\text{s}\right)}{\left(0.216\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}\left(\text{m}/\text{s}\right)\left(\text{rad}/\text{s}\right)}{\text{1}\text{N}\cdot \text{m}}\right)}\\ =-0.9154\text{rad}/\text{s}\end{array}$

Substitute $0$ for ${\omega }_z$, $-0.9154\text{rad}/\text{s}$ for ${\omega }_x$ and $-0.5769\text{rad}/\text{s}$ for ${\omega }_y$ in Equation (X).

$\begin{array}{c}\omega =\left(-0.9154\text{rad}/\text{s}\right)\text{i}+\left(-0.5769\text{rad}/\text{s}\right)\text{j}+0\text{k}\\ =\left(-0.9154\text{rad}/\text{s}\right)\text{i}+\left(-0.5769\text{rad}/\text{s}\right)\text{j}\end{array}$

Conclusion:

The angular velocity of the assembly is $\omega =\left(-0.9154\text{rad}/\text{s}\right)\text{i}+\left(-0.5769\text{rad}/\text{s}\right)\text{j}$.