Chapter 16.1, Problem 16.35P

Each of the gears A and B has a mass of 9 kg and has a radius of gyration of 200 mm; gear C has a mass of 3 kg and has a radius of gyration of 75 mm. if a couple M of constant magnitude 5 N-m is applied to gear C, determine (a) the angular acceleration of gear A, (b) the tangential force that gear C exerts on gear A.

To determine

(a)

The angular acceleration of the gear $A$.

Answer to Problem 16.35P

The angular acceleration of the gear $A$ is $15.143\text{rad}/{\text{s}}^2$.

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 $A$.

Figure-(1)

Draw the kinetic diagram of the gear $A$.

Figure-(2)

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

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

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-(1).

$\sum M_A=r_{A}F$ ...... (VI)

Here, the tangential force is $F$.

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

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

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).

$\begin{array}{l}{I}_A{\alpha }_A=r_{A}F\\ F=\frac{I_A{\alpha }_A}{r_A}\end{array}$ ...... (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).

$\sum M_C=M-2r_{C}F$ ...... (X)

Here, the tangential force is $F$.

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).

$\begin{array}{l}{I}_C{\alpha }_C=\text{M}-2r_{C}F\\ F=\frac{\text{M}-I_C{\alpha }_C}{2r_C}\end{array}$ ...... (XII)

Calculation:

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

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

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

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

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

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

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 (XIII).

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

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 (XIII).

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

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

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

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

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

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

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

Substitute $\left(2.5\right){\alpha }_A$ for ${\alpha }_C$, $5\text{N}\cdot \text{m}$ for $\text{M}$, $0.01687\text{kg}\cdot {\text{m}}^2$ for $I_C$, $100\text{mm}$ for $r_C$ in Equation (XII).

$\begin{array}{c}F=\frac{\left(5\text{N}\cdot \text{m}\right)-\left(0.01687\text{kg}\cdot {\text{m}}^2\right)\left(\left(2.5\right){\alpha }_A\right)}{2\left(100\text{mm}\right)}\\ =\frac{\left(5\text{N}\cdot \text{m}\right)-\left(0.04217\text{kg}\cdot {\text{m}}^2\right)\left({\alpha }_A\right)}{2\left(100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\frac{\left(5\text{N}\cdot \text{m}\right)-\left(0.04217\text{kg}\cdot {\text{m}}^2\right)\left({\alpha }_A\right)}{0.2\text{m}}\end{array}$ ...... (XVI)

Substitute $\left(1.44\text{kg}\cdot \text{m}\right){\alpha }_A$ for $F$ in Equation (XVI).

$\begin{array}{l}\left(1.44\text{kg}\cdot \text{m}\right){\alpha }_A=\frac{\left(5\text{N}\cdot \text{m}\right)-\left(0.04217\text{kg}\cdot {\text{m}}^2\right)\left({\alpha }_A\right)}{0.2\text{m}}\\ \left(0.288\text{kg}\cdot {\text{m}}^2\right){\alpha }_A=\left(5\text{N}\cdot \text{m}\right)-\left(0.04217\text{kg}\cdot {\text{m}}^2\right)\left({\alpha }_A\right)\\ {\alpha }_A=\frac{\left(5\text{N}\right)\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{\text{N}}\right)}{0.3301\text{kg}\cdot \text{m}}\\ {\alpha }_A=15.143\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of the gear $A$ is $15.143\text{rad}/{\text{s}}^2$.

To determine

(b)

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

Answer to Problem 16.35P

The tangential force exerted by gear $C$ on gear $A$ is $21.80\text{N}$.

Explanation of Solution

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

$F=\frac{I_A{\alpha }_A}{r_A}$ ...... (XVII).

Calculation:

Substitute $15.143\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$, $0.36\text{kg}\cdot {\text{m}}^2$ for $I_A$, $250\text{mm}$ for $r_A$ in Equation (XVII).

$\begin{array}{c}F=\frac{\left(0.36\text{kg}\cdot {\text{m}}^2\right)\left(15.143\text{rad}/{\text{s}}^{\text{2}}\right)}{\left(250\text{mm}\right)}\\ =\frac{\left(0.36\text{kg}\cdot {\text{m}}^2\right)\left(15.143\text{rad}/{\text{s}}^{\text{2}}\right)}{\left(250\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\left(1.44\text{kg}\cdot \text{m}\right)\left(15.143\text{rad}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =21.80\text{N}\end{array}$

Conclusion:

The tangential force exerted by gear $C$ on gear $A$ is $21.80\text{N}$.