Chapter 16.1, Problem 16.36P

To determine

(a)

The angular acceleration of the gear $A$.

Answer to Problem 16.36P

The angular acceleration of the gear $A$ is $6.057\text{rad}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given Information:

The mass of the gear $A$ is $9\text{kg}$, the radius of gyration of the gear $A$ is $200\text{mm}$, the mass of the gear $B$ is $9\text{kg}$, the radius of gyration of the gear $B$ is $200\text{mm}$, the mass of the gear $C$ is $3\text{kg}$, the radius of gyration of the gear $C$ is $75\text{mm}$, the magnitude of the couple $\text{M}$ is $5\text{N}\cdot \text{m}$, the radius of gear $A$ is $25\text{0}\text{mm}$, the radius of gear $B$ is $25\text{0}\text{mm}$ and the radius of gear $C$ is $\text{100}\text{mm}$.

Write the expression for the tangential acceleration of the gear teeth.

${\alpha }_t=\alpha r$ ...... (I)

Here, the angular acceleration is $\alpha$ and the radius of the gear is $r$.

Write the expression for the tangential acceleration of the gear $A$.

${\alpha }_{tA}=r_A{\alpha }_A$ ...... (II)

Here, the angular acceleration of gear $A$ is ${\alpha }_A$ and the radius of the gear $A$ is $r_A$.

Write the expression for the tangential acceleration of the gear $B$.

${\alpha }_{tB}=r_B{\alpha }_B$ ...... (III)

Here, the angular acceleration of gear $B$ is ${\alpha }_B$ and the radius of the gear $B$ is $r_B$.

Write the expression for the tangential acceleration of the gear $C$.

${\alpha }_{tC}=r_C{\alpha }_C$ ...... (IV)

Here, the angular acceleration of gear $C$ is ${\alpha }_C$ and the radius of the gear $C$ is $r_C$.

Draw the free body diagram for the gear $B$.

Figure-(1)

Draw the kinetic diagram of the gear $B$.

Figure-(2)

Write the expression for the moment of inertia gear $B$.

$I_B=m_B{k}_B^2$ ...... (V)

Here, the mass of the gear $B$ is $m_B$ and the radius of gyration of the gear $B$ is $k_B$.

Write the expression for the external moment at $B$ using the Figure-(1).

$\sum M_B=r_B{F}_{BC}$ ...... (VI)

Here, the tangential force exerted by gear $B$ on gear $A$ is $F_{BC}$.

Write the expression for the effective forces using the Figure-(2).

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

Since, the system of external forces is equivalent to system of effective forces hence $\sum {\left(M_B\right)}_{eff}=\sum \left(M_B\right)$.

Substitute $I_B{\alpha }_B$ for $\sum M_B$ in Equation (VI).

$\begin{array}{l}{I}_B{\alpha }_B=r_B{F}_{BC}\\ F_{BC}=\frac{I_B{\alpha }_B}{r_B}\end{array}$ ...... (VIII)

Draw the free body diagram for the gear $C$.

Figure-(3)

Draw the kinetic diagram for the gear $C$.

Figure-(4)

Write the expression for the moment of inertia gear $C$.

$I_C=m_C{k}_C^2$ ...... (IX)

Here, the mass of the gear is $C$

$m_C$ and the radius of gyration of the gear $C$ is $k_C$.

Write the expression for the external moment at $C$ using the Figure-(3).

$\sum M_C=-F_{BC}{r}_C+F_{AC}{r}_C$ ...... (X)

Here, the tangential force exerted by gear $A$ on gear $C$ is $F_{AC}$.

Write the expression for the effective forces using the Figure-(4).

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

Since, the system of external forces is equivalent to system of effective forces hence $\sum {\left(M_C\right)}_{eff}=\sum \left(M_C\right)$.

Substitute $I_C{\alpha }_C$ for $\sum \left(M_C\right)$ in Equation (VI).

$\begin{array}{l}{I}_C{\alpha }_C=-F_{BC}{r}_C+F_{AC}{r}_C\\ F_{AC}=\frac{I_C{\alpha }_C+F_{BC}{r}_C}{r_C}\end{array}$ ...... (XII)

Draw the free body diagram for the gear $A$.

Figure-(5)

Draw the kinetic diagram of the gear $A$.

Figure-(6) Write the expression for the moment of inertia gear $A$.

$I_A=m_A{k}_A^2$ ...... (XIII)

Here, the mass of the gear is $m_A$ and the radius of gyration of the gear $A$ is $k_A$.

Write the expression for the external moment at $A$ using the Figure-(5).

$\sum M_A=\text{M}-r_A{F}_{AC}$ ...... (XIV)

Here, the couple force applied on the gear $A$ is $\text{M}$.

Write the expression for the effective forces using the Figure-(6).

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

Since, the system of external forces is equivalent to system of effective forces hence $\sum {\left(M_A\right)}_{eff}=\sum \left(M_A\right)$.

Substitute $I_A{\alpha }_A$ for $\sum \left(M_A\right)$ in Equation (VI).

$\begin{array}{l}{I}_A{\alpha }_A=\text{M}-r_A{F}_{AC}\\ F_{AC}=\frac{\text{M}-I_A{\alpha }_A}{r_A}\end{array}$ ...... (XVI)

Calculation:

Substitute $250\text{mm}$ for $r_A$ in Equation (II).

$\begin{array}{c}{\alpha }_{tA}=\left(250\text{mm}\right){\alpha }_A\\ =\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right){\alpha }_A\\ =\left(0.25\text{m}\right){\alpha }_A\end{array}$ ...... (XVII)

Substitute $250\text{mm}$ for $r_B$ in Equation (III).

$\begin{array}{c}{\alpha }_{tB}=\left(250\text{mm}\right){\alpha }_B\\ =\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right){\alpha }_B\\ =\left(0.25\text{m}\right){\alpha }_B\end{array}$ ...... (XVIII)

Substitute $100\text{mm}$ for $r_B$ in Equation (IV).

$\begin{array}{c}{\alpha }_{tC}=\left(100\text{mm}\right){\alpha }_C\\ =\left(100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right){\alpha }_C\\ =\left(0.1\text{m}\right){\alpha }_C\end{array}$ ...... (XIX)

The tangential acceleration of the gear $A$ and $B$ is same that is ${\alpha }_A={\alpha }_B$.

Substitute $\left(0.25\text{m}\right){\alpha }_B$ for ${\alpha }_{tA}$ in Equation (XVII).

$\begin{array}{l}\left(0.25\text{m}\right){\alpha }_A=\left(0.25\text{m}\right){\alpha }_B\\ {\alpha }_A={\alpha }_B\end{array}$

The tangential acceleration of the gear $A$ and $C$ is same that is ${\alpha }_A={\alpha }_C$.

Substitute $\left(0.1\text{m}\right){\alpha }_C$ for ${\alpha }_{tA}$ in Equation (XVII).

$\begin{array}{l}\left(0.25\text{m}\right){\alpha }_A=\left(0.1\text{m}\right){\alpha }_C\\ {\alpha }_C=2.5{\alpha }_A\end{array}$

Substitute $9\text{kg}$ for $m_B$ and $200\text{mm}$ for $k_B$ in Equation (V).

$\begin{array}{c}{I}_B=\left(9\text{kg}\right){\left(200\text{mm}\right)}^2\\ =\left(9\text{kg}\right){\left(200\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\\ =\left(9\text{kg}\right){\left(0.2\text{m}\right)}^2\\ =0.36\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute ${\alpha }_A$ for ${\alpha }_B$, $0.36\text{kg}\cdot {\text{m}}^2$ for $I_B$, $250\text{mm}$ for $r_B$ in Equation (VIII).

$\begin{array}{c}{F}_{BC}=\frac{\left(0.36\text{kg}\cdot {\text{m}}^2\right){\alpha }_A}{\left(250\text{mm}\right)}\\ =\frac{\left(0.36\text{kg}\cdot {\text{m}}^2\right){\alpha }_A}{\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\left(1.44\text{kg}\cdot \text{m}\right){\alpha }_A\end{array}$

Substitute $9\text{kg}$ for $m_A$ and $200\text{mm}$ for $k_A$ in Equation (XIII).

$\begin{array}{c}{I}_A=\left(9\text{kg}\right){\left(200\text{mm}\right)}^2\\ =\left(9\text{kg}\right){\left(200\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\\ =\left(9\text{kg}\right){\left(0.2\text{m}\right)}^2\\ =0.36\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $5\text{N}\cdot \text{m}$ for $\text{M}$, $0.36\text{kg}\cdot {\text{m}}^2$ for $I_A$, $250\text{mm}$ for $r_A$ in Equation (XVI).

$\begin{array}{c}{F}_{AC}=\frac{5\text{N}\cdot \text{m}-\left(0.36\text{kg}\cdot {\text{m}}^2\right){\alpha }_A}{\left(250\text{mm}\right)}\\ =\frac{5\text{N}\cdot \text{m}-\left(0.36\text{kg}\cdot {\text{m}}^2\right){\alpha }_A}{\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =5\text{N}\cdot \text{m}-\left(1.44\text{kg}\cdot \text{m}\right){\alpha }_A\end{array}$

Substitute $3\text{kg}$ for $m_C$ and $75\text{mm}$ for $k_C$ in Equation (IX).

$\begin{array}{c}{I}_C=\left(3\text{kg}\right){\left(75\text{mm}\right)}^2\\ =\left(3\text{kg}\right){\left(75\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\\ =\left(3\text{kg}\right){\left(0.075\text{m}\right)}^2\\ =0.01687\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $\frac{I_B{\alpha }_A}{r_B}$ for $F_{BC}$, $2.5{\alpha }_A$ for ${\alpha }_C$, $100\text{mm}$ for $r_C$ and $250\text{mm}$ for $r_B$ in Equation (XII).

$\begin{array}{l}{F}_{AC}=\frac{I_C\left(2.5{\alpha }_A\right)+\left(\frac{I_B{\alpha }_A}{\left(250\text{mm}\right)}\right)\left(100\text{mm}\right)}{\left(100\text{mm}\right)}\\ =\frac{I_C\left(2.5{\alpha }_A\right)+\left(\frac{I_B{\alpha }_A}{\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\right)\left(100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}{\left(100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\frac{1}{0.1\text{m}}\left(2.5I_C+0.4I_B\right){\alpha }_A\end{array}$ ....... (XX)

Substitute $\frac{1}{0.1\text{m}}\left(2.5I_C+0.4I_B\right){\alpha }_A$ for $F_{AC}$ and $250\text{mm}$ for $r_A$ in Equation (XVI).

$\begin{array}{l}\frac{1}{0.1\text{m}}\left(2.5I_C+0.4I_B\right){\alpha }_A=\frac{\text{M}-I_A{\alpha }_A}{\left(250\text{mm}\right)}\\ \frac{1}{0.1\text{m}}\left(2.5I_C+0.4I_B\right){\alpha }_A=\frac{\text{M}-I_A{\alpha }_A}{\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ \text{M}=I_A{\alpha }_A+\left(6.25I_C+I_B\right){\alpha }_A\\ {\alpha }_A=\frac{\text{M}}{\left(I_A+6.25I_C+I_B\right)}\end{array}$ ...... (XXI)

Substitute $0.01687\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_C$, $5\text{N}\cdot \text{m}$ for $\text{M}$, $0.36\text{kg}\cdot {\text{m}}^2$ for $I_A$ and $0.36\text{kg}\cdot {\text{m}}^2$ for $I_B$ in Equation (XXI).

$\begin{array}{c}{\alpha }_A=\frac{5\text{N}\cdot \text{m}}{\left(0.36\text{kg}\cdot {\text{m}}^2+6.25\left(0.01687\text{kg}\cdot {\text{m}}^{\text{2}}\right)+0.36\text{kg}\cdot {\text{m}}^2\right)}\\ =\frac{5\text{N}\cdot \text{m}\left(\frac{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{\text{1}\text{N}}\right)}{0.8254\text{kg}\cdot {\text{m}}^2}\\ =6.057\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

The angular acceleration of the gear $A$ is $6.057\text{rad}/{\text{s}}^{\text{2}}$.

To determine

(b)

The tangential force exerted by gear $C$ on gear $A$.

Answer to Problem 16.36P

The tangential force exerted by gear $C$ on gear $A$ is $11.27\text{N}$.

Explanation of Solution

Write the expression for the tangential force exerted by gear $C$ on gear $A$.

$F_{AC}=\frac{1}{0.1\text{m}}\left(2.5I_C+0.4I_B\right){\alpha }_A$ ...... (XXII).

Calculation:

Substitute $6.057\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$, $0.36\text{kg}\cdot {\text{m}}^2$ for $I_B$, $0.01687\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_C$ in Equation (XXII).

$\begin{array}{c}{F}_{AC}=\frac{1}{0.1\text{m}}\left(2.5\left(0.01687\text{kg}\cdot {\text{m}}^{\text{2}}\right)+0.4\left(0.36\text{kg}\cdot {\text{m}}^2\right)\right)\left(6.057\text{rad}/{\text{s}}^{\text{2}}\right)\\ =\left(\left(0.4217\text{kg}\cdot \text{m}\right)+\left(1.44\text{kg}\cdot \text{m}\right)\right)\left(6.057\text{rad}/{\text{s}}^{\text{2}}\right)\\ =1.8617\text{kg}\cdot \text{m}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(6.057\text{rad}/{\text{s}}^{\text{2}}\right)\\ =11.27\text{N}\end{array}$

Conclusion:

The tangential force exerted by gear $C$ on gear $A$ is $11.27\text{N}$.