Chapter 16.1, Problem 16.36P

(a)

To determine

The angular acceleration of gear A $\left({\alpha }_A\right)$.

Answer to Problem 16.36P

The angular acceleration of gear A is $\left({\alpha }_A\right)$ $\underset{\_}{6.06\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

The mass of the gear A and B $\left(m_A\text{and}{m}_B\right)$ is 9 kg.

The radius of gyration $\left(k_{A,B}\right)$ is 200 mm.

The mas of the gear C $\left(m_C\right)$ is 3 kg.

The radius of gyration $\left(k_C\right)$ is 75 mm.

The magnitude of couple (M) is 5 N-m.

Calculation:

Find the tangential acceleration $\left(a_t\right)$ of gear teeth:

$a_t=\alpha r$

Here, r radius of gear.

Calculate the tangential acceleration of gear A:

$\begin{array}{c}{a}_t=\left(250\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right){\alpha }_A\\ =0.25{\alpha }_A\end{array}$ (1)

Calculate the tangential acceleration of gear B.

$\begin{array}{c}{a}_t=\left(250\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right){\alpha }_B\\ =0.25{\alpha }_B\end{array}$ (2)

Calculate the tangential acceleration of gear C.

$\begin{array}{c}{a}_t=\left(100\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right){\alpha }_C\\ =0.1{\alpha }_C\end{array}$ (3)

Equate Equation (1) and (2).

$\begin{array}{l}0.25{\alpha }_A=0.25{\alpha }_B\\ {\alpha }_A={\alpha }_B\end{array}$

Equate Equation (1) and (3).

$\begin{array}{l}0.25{\alpha }_A=0.1{\alpha }_C\\ {\alpha }_C=2.5{\alpha }_A\end{array}$

Show the free body diagram and kinetic diagram of Gear B as in Figure (1).

Take moment about point B.

Since, the system of external forces is equivalent to system of effective forces. Therefore, express the system of external moment:

$\sum M_B=\sum {\left(M_B\right)}_{eff}$ (4)

Take counterclockwise moment as positive.

Take moment about point A $\sum M_B$ for external force using Figure (2).

$\sum M_B=F_{BC}{r}_B$

Take moment about point A $\sum {\left(M_B\right)}_{eff}$ for effective force using Figure (2).

$\sum {\left(M_B\right)}_{eff}=I_B{\alpha }_B$

Substitute $F_{BC}{r}_B$ for $\sum M_B$, and $I_B{\alpha }_A$ for $\sum {\left(M_B\right)}_{eff}$ in Equation (4).

$\begin{array}{c}{F}_{BC}{r}_B=I_B{\alpha }_A\\ F_{BC}=\frac{I_B{\alpha }_A}{r_B}\end{array}$

Show the free body diagram and kinetic diagram of Gear C as in Figure (2).

Calculate the moment of inertia $\left(I_C\right)$ of gear C:

$I_C=m_C{k}_c^2$

Substitute 3 kg for $m_C$ and 0.075 m for $k_C$.

$\begin{array}{c}{I}_C=\left(3\text{kg}\right){\left(0.075\text{m}\right)}^2\\ =0.016875\text{kg}\cdot {\text{m}}^2\end{array}$

Since, the system of external forces is equivalent to system of effective forces. Therefore, express the system of external moment:

$\sum M_C=\sum {\left(M_C\right)}_{eff}$ (5)

Take counterclockwise moment as positive.

Take moment about point C $\sum M_C$ for external force using Figure (3).

$\sum M_C=-F_{BC}{r}_C+F_{AC}{r}_C$

Take moment about point C $\sum {\left(M_C\right)}_{eff}$ for effective force using Figure (3).

$\sum {\left(M_C\right)}_{eff}=I_C{\alpha }_C$

Substitute $-F_{BC}{r}_C+F_{AC}{r}_C$ for $\sum M_C$, and $I_C{\alpha }_C$ for $\sum {\left(M_C\right)}_{eff}$ in Equation (5).

$-F_{BC}{r}_C+F_{AC}{r}_C=I_C{\alpha }_C$

Substitute $\frac{I_B{\alpha }_A}{r_B}$ for $F_{BC}$, $2.5{\alpha }_A$ for ${\alpha }_C$, 0.1 m for $r_C$, and 0.25 m for $r_B$.

$\begin{array}{l}-\frac{I_B{\alpha }_A}{r_B}{r}_C+F_{AC}{r}_C=I_C\left(2.5{\alpha }_A\right)\\ F_{AC}{r}_C=\left(2.5I_C{\alpha }_A+\frac{I_B{\alpha }_A}{0.25\text{m}}0.1\text{m}\right)\\ F_{AC}=\frac{1}{r_C}\left(2.5I_C+0.4I_B\right){\alpha }_A\end{array}$ . (6)

Show the free body diagram and kinetic diagram of gear A as in Figure (3).

Since the system of external forces is equivalent to system of effective forces, express the system of external moment:

$\sum M_A=\sum {\left(M_A\right)}_{eff}$ (7)

Take clockwise moment as positive.

Take moment about point C $\sum M_A$ for external force using Figure 4.

$\sum M_A=M_A-F_{AC}{r}_A$

Take moment about point C $\sum {\left(M_A\right)}_{eff}$ for effective force using Figure 4.

$\sum {\left(M_A\right)}_{eff}=I_A{\alpha }_A$

Substitute $0.25F$ for $M_A-F_{AC}{r}_A$ and $I_A{\alpha }_A$ for $\sum {\left(M_A\right)}_{eff}$ in Equation (7).

$M_A-F_{AC}{r}_A=I_A{\alpha }_A$

Substitute $\frac{1}{r_C}\left(2.5I_C+0.4I_B\right){\alpha }_A$ for $F_{AC}$, 0.1 m for $r_C$, and 0.25 m for $r_A$.

$\begin{array}{l}{M}_A-\left(\frac{1}{r_C}\left(2.5I_C+0.4I_B\right){\alpha }_A\right)r_A=I_A{\alpha }_A\\ M_A=I_A{\alpha }_A+\frac{0.25}{0.1}\left(2.5I_C+0.4I_B\right){\alpha }_A\\ M_A=I_A{\alpha }_A+\left(6.25I_C+I_B\right){\alpha }_A\\ M_A=\left(I_A+I_B+6.25I_C\right){\alpha }_A\end{array}$ (8)

Calculate the moment of inertia of gear C.

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

Substitute 3 kg for $m_C$, and 0.075 m for $k_C$.

$\begin{array}{c}{I}_C=\left(3\text{kg}\right){\left(0.075\text{m}\right)}^2\\ =0.016875\text{kg}\cdot {\text{m}}^2\end{array}$

Calculate the moment of inertia $\left(I_A\right)$ of gear A using the relation:

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

Substitute 9 kg for $m_A$, and 0.2 m for $k_A$.

$\begin{array}{c}{I}_A=\left(9\text{kg}\right){\left(0.2\text{m}\right)}^2\\ =0.36\text{kg}\cdot {\text{m}}^2\end{array}$

The moment of inertia of gear A ($I_A$) is equal to moment of inertia of gear B ($I_B$).

Substitute $0.016875\text{kg}\cdot {\text{m}}^2$ for $I_C$, $5\text{N}\cdot \text{m}$ for $M_A$, and $0.36\text{kg}\cdot {\text{m}}^2$ for $I_A$, and $0.36\text{kg}\cdot {\text{m}}^2$ for $I_B$ in Equation (8).

$\begin{array}{l}{M}_A=\left(I_A+I_B+6.25I_C\right){\alpha }_A\\ 5\text{N}\cdot \text{m}=\left(0.36\text{kg}\cdot {\text{m}}^2+0.36\text{kg}\cdot {\text{m}}^2+6.25\left(0.016875\text{kg}\cdot {\text{m}}^2\right)\right){\alpha }_A\\ {\alpha }_A=\frac{5\text{N}\cdot \text{m}}{\left(0.36\text{kg}\cdot {\text{m}}^2+0.36\text{kg}\cdot {\text{m}}^2+6.25\left(0.016875\text{kg}\cdot {\text{m}}^2\right)\right)}\\ {\alpha }_A=6.06\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Thus, the angular acceleration of gear are $\left({\alpha }_A\right)$ $\underset{\_}{6.06\text{rad}/{\text{s}}^{\text{2}}}$.

(b)

To determine

The tangential force $\left(F_A\right)$ which gear C exerts on gear A.

Answer to Problem 16.36P

The tangential force $\left(F_A\right)$ which gear C exerts on gear A is $\underset{\_}{11.28\text{N}}$.

Explanation of Solution

Given information:

The mass of the gear A and B $\left(m_A\text{and}{m}_B\right)$ is 9 kg.

The radius of gyration $\left(k_{A,B}\right)$ is 200 mm.

The mas of the gear C $\left(m_C\right)$ is 3 kg.

The radius of gyration $\left(k_C\right)$ is 75 mm.

The magnitude of couple (M) is 5N-m.

Calculation:

Calculate the tangential force $\left(F_A\right)$ which gear C exerts on gear A:

Substitute $0.0169\text{kg}\cdot {\text{m}}^2$ for $I_C$, $0.36\text{kg}\cdot {\text{m}}^2$ for $I_B$, $6.06\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$, and 0.1 m for $r_C$ in Equation (6).

$\begin{array}{c}{F}_A=\frac{1}{r_C}\left(2.5I_C+0.4I_B\right){\alpha }_A\\ =\frac{1}{0.1\text{m}}\left(2.5\left(0.0169\text{kg}\cdot {\text{m}}^2\right)+0.4\left(0.36\text{kg}\cdot {\text{m}}^2\right)\right)6.06\text{rad}/{\text{s}}^{\text{2}}\\ =\left(60.572\right)\left(0.1862\right)\\ =11.28\text{N}\end{array}$

Therefore, the tangential force $\left(F_A\right)$ which gear C exerts on gear A is $\underset{\_}{11.28\text{N}}$.