Chapter 17.1, Problem 17.12P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

Answer to Problem 17.12P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $\underset{\_}{2.54\text{rev}}$.

Explanation of Solution

Given information:

The mass $\left(m_A\right)$ of gear A is 2.4 kg.

The mass $\left(m_B\right)$ of gear B is 2.4 kg.

The mass $\left(m_C\right)$ of gear C is 12 kg.

The radius of gyration $\left(k_A\right)$ of gear A is 60 mm.

The radius of gyration $\left(k_B\right)$ of gear B is 60 mm.

The radius of gyration $\left(k_C\right)$ of gear C is 150 mm.

The radius $\left(r_A\right)$ of gear A is 80 mm.

The radius $\left(r_B\right)$ of gear B is 80 mm.

The radius $\left(r_C\right)$ of gear C is 200 mm.

The initial angular velocity ${\left({\omega }_C\right)}_1$ of gear C is 100 rpm.

The final angular velocity ${\left({\omega }_C\right)}_2$ of gear C is 450 rpm.

The couple acting on gear B is $10\text{N}\cdot \text{m}$.

Calculation:

Find the mass moment of inertia $\left(I_A\right)$ of gear A using the equation:

$I_A=m_A{k}_A^2$

Substitute 2.4 kg for $m_A$ and 60 mm for $k_A$.

$\begin{array}{c}{I}_A=\left(2.4\right){\left(60\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the mass moment of inertia $\left(I_B\right)$ of gear B using the equation:

$I_B=m_B{k}_B^2$

Substitute 2.4 kg for $m_B$ and 60 mm for $k_B$.

$\begin{array}{c}{I}_B=\left(2.4\right){\left(60\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the mass moment of inertia $\left(I_C\right)$ of gear C using the equation:

$I_C=m_C{k}_C^2$

Substitute 12 kg for $m_C$ and 150 mm for $k_A$.

$\begin{array}{c}{I}_C=\left(12\right){\left(150\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =270\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the initial angular velocity ${\left({\omega }_A\right)}_1$ of gear A using the kinematics:

$\begin{array}{l}{r}_A{\left({\omega }_A\right)}_1=r_C{\left({\omega }_C\right)}_1\\ {\left({\omega }_A\right)}_1=\frac{r_C}{r_A}{\left({\omega }_C\right)}_1\end{array}$

Substitute 200 mm for $r_C$, 80 mm for $r_A$, and 100 rpm for ${\left({\omega }_C\right)}_1$.

$\begin{array}{c}{\left({\omega }_A\right)}_1=\frac{200}{80}\left(100\frac{\text{rev}}{\min }\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\right)\\ =26.18\text{rad/s}\end{array}$

Find the initial angular velocity ${\left({\omega }_B\right)}_1$ of gear B using the kinematics:

$\begin{array}{l}{r}_B{\left({\omega }_B\right)}_1=r_C{\left({\omega }_C\right)}_1\\ {\left({\omega }_B\right)}_1=\frac{r_C}{r_B}{\left({\omega }_C\right)}_1\end{array}$

Substitute 200 mm for $r_C$, 80 mm for $r_B$, and 100 rpm for ${\left({\omega }_C\right)}_1$.

$\begin{array}{c}{\left({\omega }_B\right)}_1=\frac{200}{80}\left(100\frac{\text{rev}}{\min }\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\right)\\ =26.18\text{rad/s}\end{array}$

Find the initial kinetic energy $\left(T_1\right)$ of the system using the equation:

$\begin{array}{c}{T}_1={\left(T_1\right)}_A+{\left(T_1\right)}_B+{\left(T_1\right)}_C\\ =\frac{1}{2}{I}_A{\left({\omega }_A\right)}_1^2+\frac{1}{2}{I}_B{\left({\omega }_B\right)}_1^2+\frac{1}{2}{I}_C{\left({\omega }_C\right)}_1^2\end{array}$

Here, ${\left(T_1\right)}_A$ is the initial angular velocity of gear A, ${\left(T_1\right)}_B$ is the initial angular velocity of gear B, and ${\left(T_1\right)}_C$ is the initial angular velocity of gear C.

Substitute $8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_A$, $26.18\text{rad/s}$ for ${\left({\omega }_A\right)}_1$, $8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_B$, $26.18\text{rad/s}$ for ${\left({\omega }_B\right)}_1$, $270\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_C$, and 100 rpm for ${\omega }_C$.

$\begin{array}{c}{T}_1=\left\{\frac{1}{2}\left[\begin{array}{l}\left(8.64\times {10}^{-3}\right){\left(26.18\right)}^2+\frac{1}{2}\left(8.64\times {10}^{-3}\right){\left(26.18\right)}^2\\ +\frac{1}{2}\left(270\times {10}^{-3}\right)\end{array}\right]{\left(100\frac{\text{rev}}{\min }\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\right)}^2\right\}\\ =2.9609+2.9609+14.8044\\ =20.726\text{J}\end{array}$

Find the final angular velocity ${\left({\omega }_A\right)}_2$ of gear A using the kinematics:

$\begin{array}{l}{r}_A{\left({\omega }_A\right)}_2=r_C{\left({\omega }_C\right)}_2\\ {\left({\omega }_A\right)}_2=\frac{r_C}{r_A}{\left({\omega }_C\right)}_2\end{array}$

Substitute 200 mm for $r_C$, 80 mm for $r_A$, and 450 rpm for ${\left({\omega }_C\right)}_2$.

$\begin{array}{c}{\left({\omega }_A\right)}_2=\frac{200}{80}\left(450\frac{\text{rev}}{\min }\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\right)\\ =117.81\text{rad/s}\end{array}$

Find the final angular velocity ${\left({\omega }_B\right)}_2$ of gear B using the kinematics:

$\begin{array}{l}{r}_B{\left({\omega }_B\right)}_2=r_C{\left({\omega }_C\right)}_2\\ {\left({\omega }_B\right)}_2=\frac{r_C}{r_B}{\left({\omega }_C\right)}_2\end{array}$

Substitute 200 mm for $r_C$, 80 mm for $r_B$, and 450 rpm for ${\left({\omega }_C\right)}_2$.

$\begin{array}{c}{\left({\omega }_B\right)}_2=\frac{200}{80}\left(450\frac{\text{rev}}{\min }\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\right)\\ =117.81\text{rad/s}\end{array}$

Find the initial kinetic energy $\left(T_2\right)$ of the system using the equation:

$\begin{array}{c}{T}_2={\left(T_2\right)}_A+{\left(T_2\right)}_B+{\left(T_2\right)}_C\\ =\frac{1}{2}{I}_A{\left({\omega }_A\right)}_2^2+\frac{1}{2}{I}_B{\left({\omega }_B\right)}_2^2+\frac{1}{2}{I}_C{\left({\omega }_C\right)}_2^2\end{array}$

Here, ${\left(T_2\right)}_A$ is the final angular velocity of gear A, ${\left(T_2\right)}_B$ is the final angular velocity of gear B, and ${\left(T_2\right)}_C$ is the final angular velocity of gear C.

Substitute $8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_A$, $117.81\text{rad/s}$ for ${\left({\omega }_A\right)}_1$, $8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_B$, $117.81\text{rad/s}$ for ${\left({\omega }_B\right)}_1$, $270\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_C$, and 450 rpm for ${\omega }_C$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left[\begin{array}{l}\left(8.64\times {10}^{-3}\right){\left(117.81\right)}^2+\frac{1}{2}\left(8.64\times {10}^{-3}\right){\left(117.81\right)}^2\\ +\frac{1}{2}\left(270\times {10}^{-3}\right)\end{array}\right]{\left(450\frac{\text{rev}}{\min }\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\right)}^2\\ =59.958+59.958+299.8\\ =419.7\text{J}\end{array}$

Find the work done $\left(U_{1\to 2}\right)$ due to couple acting on gear B using the equation:

$U_{1\to 2}=M_B{\theta }_B$

Substitute $10\text{N}\cdot \text{m}$ for $M_B$.

$U_{1\to 2}=10{\theta }_B$

Find the rotation of gear B.

Apply Principle of work and energy for system.

$T_1+U_{1\to 2}=T_2$

Substitute $20.726\text{J}$ for $T_1$, 419.7 J for $T_2$, and $10{\theta }_B$ for $U_{1\to 2}$.

$\begin{array}{l}20.726+10{\theta }_B=419.7\\ {\theta }_B=\frac{419.7-20.726}{10}\\ {\theta }_B=39.897\text{rad}\end{array}$

Find the number of revolutions of gear C $\left({\theta }_C\right)$ required for its angular velocity to increase from 100 to 450 rpm.

$\begin{array}{l}{r}_C{\theta }_C=r_B{\theta }_B\\ {\theta }_C=\frac{r_B}{r_C}{\theta }_B\end{array}$

Substitute 80 mm for $r_B$, 200 mm for $r_C$, and $39.897\text{rad}$ for ${\theta }_B$

$\begin{array}{c}{\theta }_C=\frac{80}{200}\left(39.897\right)\\ =15.96\text{rad}\times \frac{1\text{rev}}{2\pi \text{rad}}\\ =2.54\text{rev}\end{array}$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $\underset{\_}{2.54\text{rev}}$.

(b)

To determine

Find the tangential force acting on gear A.

Answer to Problem 17.12P

The tangential force acting on gear A is $\underset{\_}{17.86\text{N}}$.

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

${\theta }_A={\theta }_B$

Substitute $39.897\text{rad}$ for ${\theta }_B$.

${\theta }_A=39.897\text{rad}$

Find the tangential force $\left(F_t\right)$ acting on gear A.

Apply Principle of work and energy for gear A.

$\begin{array}{l}{\left(T_1\right)}_A+M_A{\theta }_A={\left(T_2\right)}_A\\ \frac{1}{2}{I}_A{\left({\omega }_A\right)}_1^2+M_A{\theta }_A=\frac{1}{2}{I}_A{\left({\omega }_A\right)}_2^2\end{array}$

Substitute $8.64\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_A$, $26.18\text{rad/s}$ for ${\left({\omega }_A\right)}_1$, 39.897 rad/s for ${\theta }_A$, and $117.81\text{rad/s}$ for ${\left({\omega }_A\right)}_2$.

$\begin{array}{l}\frac{1}{2}\left(8.64\times {10}^{-3}\right){\left(26.18\right)}^2+M_A\left(39.897\right)=\frac{1}{2}\left(8.64\times {10}^{-3}\right){\left(117.81\right)}^2\\ 2.9609+39.897M_A=59.958\\ M_A=1.4286\text{N}\cdot \text{m}\end{array}$

Find the tangential force $\left(F_t\right)$ acting on gear A using the equation:

$F_t=\frac{M_A}{r_A}$

Substitute $1.4286\text{N}\cdot \text{m}$ for $M_A$ and 80 mm for $r_A$.

$\begin{array}{c}{F}_t=\frac{1.4286}{\left(80\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}\\ =17.86\text{N}\end{array}$

Thus, the tangential force acting on gear A is $\underset{\_}{17.86\text{N}}$.