Chapter 18.1, Problem 18.15P

To determine

(a)

The angular momentum of the assembly about point $A$

Answer to Problem 18.15P

The angular momentum of the assembly about point $A$ is $5.6535i-1.885j+12.56k\text{kg}\cdot {\text{m}}^2/\text{s}$ .

Explanation of Solution

Given information:

The mass of each L-shaped arm is $5\text{kg}$, speed of assembly is $360\text{rpm}$.

The below figure represents x’, y’, and z’ axis parallel to the x, y, and z axis.

Figure (1)

Write the expression for angular velocity in x’-direction.

${\omega }_{x\text{'}}=0$ …… (I)

Write the expression for angular velocity in y’-direction.

${\omega }_{y\text{'}}=0$ …… (II)

Write the expression for mass of each segment of arm

$m^{\text{'}}=\frac{m}{2}$ ……. (III)

Here, $m$ is mass each L-shaped arm.

Write the expression for angular velocity in z-direction.

${\omega }_z=\frac{2\pi N}{60}$ ……. (IV)

Here, speed of assembly is $N$.

Write the expression for moment of inertia for part (1) in x-z direction

${\left(I_{xz}\right)}_1=-\frac{1}{2}m\text{'}{l}^2$ …… (V)

Here mass moment of inertia along x-y direction is ${\left(I_{xz}\right)}_1$ and the length of each segment is $l$.

Write the expression for moment of inertia for part (1) in y-z direction.

${\left(I_{yz}\right)}_1=\frac{1}{4}m\text{'}{l}^2$ …… (VI)

Here mass moment of inertia along y-z direction is ${\left(I_{yz}\right)}_1$.

Write the expression for moment of inertia for part (1) in z direction

${\left(I_z\right)}_1=\frac{4}{3}m\text{'}{l}^2$ …… (VII)

Here mass moment of inertia along z direction is ${\left(I_z\right)}_1$.

Write the expression for moment of inertia for part (2) in x-z direction.

${\left(I_{xz}\right)}_2=-\frac{1}{2}m\text{'}{l}^2$ …… (VIII)

Here, mass moment of inertia along x-y direction is ${\left(I_{xz}\right)}_2$.

Write the expression for moment of inertia for part (2) in y-z direction.

${\left(I_{yz}\right)}_2=0$ …… (IX)

Here mass moment of inertia along y-z direction is ${\left(I_{yz}\right)}_2$.

Write the expression for moment of inertia for part (2) in z direction.

${\left(I_z\right)}_2=\frac{1}{3}m\text{'}{l}^2$ …… (X)

Here mass moment of inertia along z direction is ${\left(I_z\right)}_2$.

Write the expression for moment of inertia for part (3) in x-z direction.

${\left(I_{xz}\right)}_3=-\frac{1}{4}m\text{'}{l}^2$ …… (XI)

Here, mass moment of inertia along x-z direction is ${\left(I_{xz}\right)}_3$.

Write the expression for moment of inertia for part (3) in y-z direction.

${\left(I_{yz}\right)}_3=0$ …... (XII)

Here, mass moment of inertia along y-z direction is ${\left(I_{yz}\right)}_3$.

Write the expression for moment of inertia for part (3) in z direction.

${\left(I_z\right)}_3=\frac{1}{3}m\text{'}{l}^2$ …… (XIII)

Here mass moment of inertia along z direction is ${\left(I_z\right)}_3$.

Write the expression for moment of inertia for part (4) x-z direction.

${\left(I_{xz}\right)}_4=-\frac{1}{4}m\text{'}{l}^2$ …… (XIV)

Here, mass moment of inertia along x-y direction is ${\left(I_{xz}\right)}_4$.

Write the expression for moment of inertia for part (4) y-z direction.

${\left(I_{yz}\right)}_4=\frac{1}{4}m\text{'}{l}^2$ …… (XV)

Here, mass moment of inertia along y-z direction is ${\left(I_{yz}\right)}_4$.

Write the expression for moment of inertia for part (4) z direction.

${\left(I_z\right)}_4=\frac{4}{3}m\text{'}{l}^2$ …… (XVI)

Here mass moment of inertia along z direction is ${\left(I_z\right)}_2$.

Write the expression for total mass moment of inertia in x-y direction.

$\sum I_{xz}={\left(I_{xz}\right)}_1+{\left(I_{xz}\right)}_2+{\left(I_{xz}\right)}_3+{\left(I_{xz}\right)}_4$ …… (XVII)

Write the expression for total mass moment of inertia in y-z direction.

$\sum I_{yz}={\left(I_{yz}\right)}_1+{\left(I_{yz}\right)}_2+{\left(I_{yz}\right)}_3+{\left(I_{yz}\right)}_4$ …… (XVIII)

Write the expression for total mass moment of inertia in z direction.

$\sum I_z={\left(I_z\right)}_1+{\left(I_z\right)}_2+{\left(I_z\right)}_3+{\left(I_z\right)}_4$ …… (XIX)

Substitute $-\frac{1}{2}m\text{'}{l}^2$ for ${\left(I_{xz}\right)}_1$, $-\frac{1}{4}m\text{'}{l}^2$ for ${\left(I_{xz}\right)}_2$, $-\frac{1}{4}m\text{'}{l}^2$ for ${\left(I_{xz}\right)}_3$ and $-\frac{1}{2}m\text{'}{l}^2$ for ${\left(I_{xz}\right)}_4$ in Equation(XVII)

$\begin{array}{c}\sum I_{xz}=-\frac{1}{2}m\text{'}{l}^2-\frac{1}{4}m\text{'}{l}^2-\frac{1}{4}m\text{'}{l}^2-\frac{1}{2}m\text{'}{l}^2\\ =-\frac{3}{2}m\text{'}{l}^2\end{array}$ …… (XX)

Substitute $\frac{1}{4}m\text{'}{l}^2$ for ${\left(I_{yz}\right)}_1$, $0$ for ${\left(I_{yz}\right)}_2$, $0$ for ${\left(I_{yz}\right)}_3$ and $\frac{1}{4}m\text{'}{l}^2$ for ${\left(I_{yz}\right)}_4$ in Equation(XVIII)

$\begin{array}{c}\sum I_{yz}=\frac{1}{4}m\text{'}{l}^2+0+0+\frac{1}{4}m\text{'}{l}^2\\ =\frac{1}{2}m\text{'}{l}^2\end{array}$ …… (XXI)

Substitute $\frac{4}{3}m\text{'}{l}^2$ for ${\left(I_z\right)}_1$, $\frac{1}{3}m\text{'}{l}^2$ for ${\left(I_z\right)}_2$, $\frac{1}{3}m\text{'}{l}^2$ for ${\left(I_z\right)}_3$ and $\frac{4}{3}m\text{'}{l}^2$ for ${\left(I_z\right)}_4$ in Equation(XIX)

$\begin{array}{c}\sum I_z=\frac{4}{3}m\text{'}{l}^2+\frac{1}{3}m\text{'}{l}^2+\frac{1}{3}m\text{'}{l}^2+\frac{4}{3}m\text{'}{l}^2\\ =\frac{10}{3}m\text{'}{l}^2\end{array}$ …… (XXII)

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

${\left(H_G\right)}_x=-\sum I_{xz}\omega \theta$ …… (XXIII)

Substitute $-\frac{3}{2}m\text{'}{l}^2$ for $\sum I_{xz}$ in Equation (XXIII)

${\left(H_G\right)}_x=-\left(-\frac{3}{2}m\text{'}{l}^2\right)\omega$ …… (XXIV)

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

${\left(H_G\right)}_y=-\sum I_{yz}\omega$ …… (XXV)

Substitute $\frac{1}{2}m\text{'}{l}^2$ for $\sum I_{yz}$ in Equation (XXV)

${\left(H_G\right)}_y=-\left(\frac{1}{2}m\text{'}{l}^2\right)\omega$ …… (XXVI)

Write the expression for angular momentum in z direction.

${\left(H_G\right)}_z=\sum I_z\omega$ …… (XXVII)

Substitute $\frac{10}{3}m\text{'}{l}^2$ for $\sum I_z$ in Equation (XXVII)

${\left(H_G\right)}_z=\left(\frac{10}{3}m\text{'}{l}^2\right)\omega$ …… (XXVIII)

Write the expression for $H_G$

$\sum H_G={\left(H_G\right)}_x+{\left(H_G\right)}_y+{\left(H_G\right)}_z$ …… (XXIX)

Substitute, $H_A$ for $H_G$ in Equation

$\sum H_A={\left(H_A\right)}_x+{\left(H_A\right)}_y+{\left(H_A\right)}_z$ …… (XXX)

Calculation:

Substitute $360\text{rpm}$ for $N$ in Equation (IV).

$\begin{array}{l}{\omega }_z=\frac{2\pi \times 360\text{rpm}}{60}\\ =37.69\text{rad}/s\end{array}$ …… (XXIXI)

Substitute $2.5\text{kg}$ for $m\text{'}$, $200\text{mm}$ for $l$, $37.69\text{rad}/s$ for $\omega$ in Equation (XXIV).

$\begin{array}{c}{\left(H_A\right)}_x=-\left(-\frac{3}{2}\times 2.5\text{kg}\times {\left(200\text{mm}\right)}^2\right)\times 37.69\text{rad}/s\\ =-\left(-\frac{3}{2}\times 2.5\text{kg}\times {\left(200\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)\times 37.69\text{rad}/s\\ =5.6535\text{kg}\cdot {\text{m}}^2/s\end{array}$

Substitute $2.5\text{kg}$ for $m\text{'}$, $200\text{mm}$ for $l$, $37.69\text{rad}/s$ for $\omega$ in Equation (XXVI).

$\begin{array}{c}{\left(H_A\right)}_y=-\left(\frac{1}{2}\times 2.5\text{kg}\times {\left(200\text{mm}\right)}^2\right)\times 37.69\text{rad}/s\\ =-\left(\frac{1}{2}\times 2.5\text{kg}\times {\left(200\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)\times 37.69\text{rad}/s\\ =-1.885\text{kg}\cdot {\text{m}}^2/s\end{array}$

Substitute $2.5\text{kg}$ for $m\text{'}$, $200\text{mm}$ for $l$, $37.69\text{rad}/s$ for $\omega$ in Equation (XXVIII).

$\begin{array}{c}{\left(H_A\right)}_z=\left(\frac{10}{3}\times 2.5\text{kg}\times {\left(200\text{mm}\right)}^2\right)\times 37.69\text{rad}/s\\ =\left(\frac{10}{3}\times 2.5\text{kg}\times {\left(200\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)\times 37.69\text{rad}/s\\ =12.56\text{kg}\cdot {\text{m}}^2/s\end{array}$

Substitute $5.6535\text{kg}\cdot {\text{m}}^2/s$ for ${\left(H_A\right)}_x$, $-1.885\text{kg}\cdot {\text{m}}^2/s$ for ${\left(H_A\right)}_y$ and $12.56\text{kg}\cdot {\text{m}}^2/s$ for ${\left(H_A\right)}_z$ in Equation (XXX).

$\begin{array}{c}\sum H_A=5.6535\text{kg}\cdot {\text{m}}^2/\text{s}i+\left(-1.885\right)\text{kg}\cdot {\text{m}}^2/\text{s}j+12.56\text{kg}\cdot {\text{m}}^2/\text{s}k\\ =5.6535i-1.885j+12.56k\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Thus angular momentum of the assembly about point $A$ is $5.6535i-1.885j+12.56k\text{kg}\cdot {\text{m}}^2/\text{s}$

To determine

(b)

The angle formed by $H_A$ and $AB$

Answer to Problem 18.15P

The angle formed by $H_A$ and $AB$ is ${25.37}^{°}$

Explanation of Solution

Given information Write the expression for the magnitude of angular momentum in about assembly A

$H_A=\sqrt{{\left(H_A\right)}_x^2+{\left(H_A\right)}_{{}_y}^2+{\left(H_A\right)}_z^2}$ …… (XXXI)

Write the Expression for angle formed by $H_A$ and $AB$

$\cos \theta =\frac{{\left(H_A\right)}_x}{H_A}$ …… (XXXII)

Calculation:

Substitute $5.6535\text{kg}\cdot {\text{m}}^2/s$ for ${\left(H_A\right)}_x$, ${\left(1.885\text{kg}\cdot {\text{m}}^2/\text{s}\right)}^2$ for ${\left(H_A\right)}_y$, ${\left(12.56\text{kg}\cdot {\text{m}}^2/\text{s}\right)}^2$ for ${\left(H_A\right)}_z$ in Equation (XXXI)

$\begin{array}{c}{H}_A=\sqrt{{\left(H_A\right)}_x^2+{\left(H_A\right)}_{{}_y}^2+{\left(H_A\right)}_z^2}\\ =\sqrt{{\left(5.6535\text{kg}\cdot {\text{m}}^2/\text{s}\right)}^2+{\left(1.885\text{kg}\cdot {\text{m}}^2/\text{s}\right)}^2+{\left(12.56\text{kg}\cdot {\text{m}}^2/\text{s}\right)}^2}\\ =13.9085\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Substitute $12.56\text{kg}\cdot {\text{m}}^2/\text{s}$ for ${\left(H_A\right)}_x$ and $13.9085\text{kg}\cdot {\text{m}}^2/\text{s}$ for $H_A$ in Equation (XXXII)

$\begin{array}{c}\cos \theta =\frac{12.56\text{kg}\cdot {\text{m}}^2/\text{s}}{13.9085\text{kg}\cdot {\text{m}}^2/\text{s}}\\ ={\cos }^{-1}\left(\frac{12.56}{13.9085}\right)\\ ={25.37}^{\circ }\end{array}$

Conclusion:

Thus angle formed by $H_A$ and $AB$ is ${25.37}^{°}$