Chapter 17.1, Problem 17.12P

Solve Prob. 17.11, assuming that the 6 N·m couple is applied to gear B.

17.11    Each of the gears A and B has a mass of 10 kg and a radius of gyration of 190 mm, while gear C has a mass of 2.5 kg and a radius of gyration of 80 mm. If a couple M of constant magnitude 6 N·m is applied to gear C, determine (a) the number of revolutions of gear C required for its angular velocity to increase from 450 rpm to 1800 rpm, (b) the corresponding tangential force acting on gear A.

Fig. P17.11

To determine

Find the number of revolutions required for gear C.

Answer to Problem 17.12P

The number of revolutions required for the gear C for the work to be done is $\underset{\_}{145.3\text{rev}}$.

Explanation of Solution

Given information:

The mass of the gear A is $m_A=10\text{kg}$.

The mass of the gear B is $m_B=10\text{kg}$.

The radius of gyration of the gear A is $k_A=190\text{mm}$.

The radius of gyration of the gear B is $k_B=190\text{mm}$.

The mass of the gear C is $m_C=2.5\text{kg}$.

The radius of gyration of the gear C is $k_C=80\text{mm}$.

The radius of the gear A is $r_A=250\text{mm}$.

The radius of the gear B is $r_B=250\text{mm}$.

The radius of the gear C is $r_C=100\text{mm}$.

The magnitude of the couple moment applied at point B is $\text{M}=6\text{N}\cdot \text{m}$.

Calculation:

Find the mass moment of inertia of gear A $\left(\overline{I_A}\right)$ using the equation.

$\overline{I_A}=m_A{k}_A^2$

Substitute 10 kg for $m_A$ and 190 mm for $k_A$.

$\begin{array}{c}\overline{I_A}=10\times {\left(190\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.361\text{kg}\cdot {\text{m}}^2\end{array}$

Find the mass moment of inertia of gear B $\left(\overline{I_B}\right)$ using the equation.

$\overline{I_B}=m_B{k}_B^2$

Substitute 10 kg for $m_B$ and 190 mm for $k_B$.

$\begin{array}{c}\overline{I_B}=10\times {\left(190\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.361\text{kg}\cdot {\text{m}}^2\end{array}$

Find the mass moment of inertia of gear C $\left(\overline{I_C}\right)$ using the equation.

$\overline{I_C}=m_C{k}_C^2$

Substitute 2.5 kg for $m_C$ and 80 mm for $k_C$.

$\begin{array}{c}\overline{I_C}=2.5\times {\left(80\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.016\text{kg}\cdot {\text{m}}^2\end{array}$

The gears A and C are in contact.

Use the kinematics concept;

${\omega }_A{r}_A={\omega }_C{r}_C$

Substitute 250 mm for $r_A$ and 100 mm for $r_C$.

$\begin{array}{l}{\omega }_A\times 250={\omega }_C\times 100\\ {\omega }_A=0.4{\omega }_C\end{array}$

The gears B and C are in contact.

Use the kinematics Equation:

${\omega }_B{r}_B={\omega }_C{r}_C$

Substitute 250 mm for $r_B$ and 100 mm for $r_C$.

$\begin{array}{l}{\omega }_B\times 250={\omega }_C\times 100\\ {\omega }_B=0.4{\omega }_C\end{array}$

Find the total kinetic energy $\left(T_1\right)$ using the relation.

$T_1=\frac{1}{2}\overline{I_A}{\omega }_A^2+\frac{1}{2}\overline{I_B}{\omega }_B^2+\frac{1}{2}\overline{I_C}{\omega }_C^2$

Substitute $0.361\text{kg}\cdot {\text{m}}^2$ for $\overline{I_A}$, $0.4{\omega }_C$ for ${\omega }_A$, $0.361\text{kg}\cdot {\text{m}}^2$ for $\overline{I_B}$, $0.4{\omega }_C$ for ${\omega }_B$, and $0.016\text{kg}\cdot {\text{m}}^2$ for $\overline{I_C}$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 0.361\times {\left(0.4{\omega }_C\right)}^2+\frac{1}{2}\times 0.361\times {\left(0.4{\omega }_C\right)}^2+\frac{1}{2}\times 0.016\times {\omega }_C^2\\ =0.02888{\omega }_C^2+0.02888{\omega }_C^2+0.008{\omega }_C^2\\ =0.06576{\omega }_C^2\end{array}$ (1)

When the angular velocity is at 450 rpm:

${\omega }_1={\omega }_C=450\text{rpm}$.

Substitute 450 rpm for ${\omega }_C$ in Equation (1).

$\begin{array}{c}{T}_1=0.06576\times {\left(450\text{rpm}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)}^2\\ =146.03\text{J}\end{array}$

When the angular velocity is at 1800 rpm:

${\omega }_2={\omega }_C=1800\text{rpm}$.

Substitute 1800 rpm for ${\omega }_C$ in Equation (1).

$\begin{array}{c}{T}_2=0.06576\times {\left(1800\text{rpm}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)}^2\\ =2336.49\text{J}\end{array}$

Find the work done $\left(U_{1\to 2}\right)$ for the rigid body using the relation:

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

Here, the number of revolution at gear B is ${\theta }_B$.

Substitute $6\text{N}\cdot \text{m}$ for M.

$U_{1\to 2}=6{\theta }_B$

Write the equation of work and energy for the system using the equation.

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

Substitute 146.03 J for $T_1$, $6{\theta }_B$ for $U_{1\to 2}$, and 2336.49 J for $T_2$

$\begin{array}{c}146.03+6{\theta }_B=2336.49\\ {\theta }_B=365.08\text{rad}\times \frac{1\text{rev}}{2\pi \text{rad}}\\ =58.1\text{rev}\end{array}$

Find the number of revolution at gear C $\left({\theta }_C\right)$ using the relation.

${\theta }_C=2.5{\theta }_B$

Substitute 58.1 rev for ${\theta }_B$.

$\begin{array}{c}{\theta }_C=2.5\times 58.1\\ =145.3\text{rev}\end{array}$

Therefore, the number of revolutions required for the gear C for the work to be done is $\underset{\_}{145.3\text{rev}}$.

To determine

Find the tangential force acting on gear A.

Answer to Problem 17.12P

The tangential force acting on gear A is $\underset{\_}{10.54\text{N}}$.

Explanation of Solution

Given information:

The mass of the gear A is $m_A=10\text{kg}$.

The mass of the gear B is $m_B=10\text{kg}$.

The radius of gyration of the gear A is $k_A=190\text{mm}$.

The radius of gyration of the gear B is $k_B=190\text{mm}$.

The mass of the gear C is $m_C=2.5\text{kg}$.

The radius of gyration of the gear C is $k_C=80\text{mm}$.

The radius of the gear A is $r_A=250\text{mm}$.

The radius of the gear B is $r_B=250\text{mm}$.

The radius of the gear C is $r_C=100\text{mm}$.

The magnitude of the couple moment is $\text{M}=6\text{N}\cdot \text{m}$.

Calculation:

Refer to part (a) calculation;

The number of revolution in the gear A is equal to the number of revolution in the gear B.

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

Substitute 365.08 rad for ${\theta }_B$.

${\theta }_A=365.08\text{rad}$

Find the total kinetic energy $\left(T_1\right)$ using the relation.

$\begin{array}{c}{T}_A=\frac{1}{2}\overline{I_A}{\omega }_A^2\\ =\frac{1}{2}\overline{I_A}{\left(0.4{\omega }_C\right)}^2\end{array}$ (2)

When the angular velocity is at 450 rpm; ${\omega }_1={\omega }_C=450\text{rpm}$.

Substitute $0.361\text{kg}\cdot {\text{m}}^2$ for $\overline{I_A}$ and 450 rpm for ${\omega }_C$ in Equation (2).

$\begin{array}{c}{T}_{1A}=\frac{1}{2}\times 0.361\times {\left(0.4\times 450\text{rpm}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)}^2\\ =64.133\text{J}\end{array}$

When the angular velocity is at 1800 rpm; ${\omega }_2={\omega }_C=1800\text{rpm}$.

Substitute $0.361\text{kg}\cdot {\text{m}}^2$ for $\overline{I_A}$ and 18000 rpm for ${\omega }_C$ in Equation (2).

$\begin{array}{c}{T}_{2A}=\frac{1}{2}\times 0.361\times {\left(0.4\times 1800\text{rpm}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)}^2\\ =1026.123\text{J}\end{array}$

Find the work done $\left(U_{1\to 2}\right)$ for the rigid body using the relation;

$U_{1\to 2}=M_A{\theta }_A$

Here, the magnitude of couple moment at gear A is $M_A$.

Substitute 365.08 rad for ${\theta }_A$.

$U_{1\to 2}=365.08M_A$

Write the equation of work and energy for the system using the equation.

$T_{1A}+U_{1\to 2}=T_{2A}$

Substitute 64.133 J for $T_{1A}$, $365.08M_A$ for $U_{1\to 2}$, and 1026.123 J for $T_{2A}$

$\begin{array}{c}64.133+365.08M_A=1026.123\\ M_A=2.635\text{N}\cdot \text{m}\end{array}$

Find the tangential force $\left(F_t\right)$ using the relation.

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

Substitute $2.635\text{N}\cdot \text{m}$ for $M_A$ and 250 mm for $r_A$.

$\begin{array}{c}{F}_t=\frac{2.635}{250\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =10.54\text{N}\end{array}$

Therefore, the tangential force acting on gear A is $\underset{\_}{10.54\text{N}}$.