Chapter 16.1, Problem 16.36P

Solve Prob. 16.35, assuming that the couple M is applied to disk A.

To determine

i.

The angular acceleration of gear A.

Answer to Problem 16.36P

Angular acceleration of gear A = 6.057 rad/s²

Explanation of Solution

Given:

Mass of Gear A,m_a = 9 kg

radius of gyration of gear A, $\overline{k_a}$ = 200 mm = 0.2 m

Mass of gear B, m_b = 9 kg

radius of gyration of gear B, $\overline{k_b}$ = 200 mm = 0.2 m

Mass of Gear C,m_c = 3 kg

radius of gyration of gear C, $\overline{k_c}$ = 75 mm = 0.075 m

Magnitude of Couple applied on gear A, M = 5N-m

Concept used:

Mass moment of acceleration is given by-

$\text{I = m}\overline{{\text{k}}^{\text{2}}},\text{ where }\overline{k}\text{ is the radius of gyration }$

The tangential force acting on a gear will provide the angular acceleration to the gear. Therefore,

$\begin{array}{l}\text{Summation of moments applied on it = Mass Moment of inertia }\times \text{ angular acceleration}\\ \Sigma \text{M = Iα }\end{array}$

The free body diagram of the three gears is as following-

Calculation:

$\begin{array}{l}\text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ \text{Therefore, α = }\frac{\text{a}}{\text{r}}\end{array}$

The tangential component of two gears in mesh will be equal, therefore,

$\begin{array}{l}{\text{a}}_{\text{t}}{\text{= r}}_{\text{a}}{\text{α}}_{\text{a}}{\text{ = r}}_{\text{b}}{\text{α}}_{\text{b}}{\text{ = r}}_{\text{c}}{\text{α}}_{\text{c}}\\ {\text{a}}_{\text{t}}\text{= 0}{\text{.25α}}_{\text{a}}\text{ = 0}{\text{.1α}}_{\text{c}}\text{ = 0}{\text{.25α}}_{\text{b}}\\ {\text{α}}_{\text{c}}\text{ = 2}{\text{.5α}}_{\text{a}}\text{ = 2}{\text{.5α}}_{\text{b}}\text{ -(1)}\end{array}$

Since, ${\text{a}}_{\text{t}}\text{= 0}{\text{.25α}}_{\text{a}}\text{ = 0}{\text{.25α}}_{\text{b}}$

$\therefore {\text{α}}_{\text{a}}{\text{ = α}}_{\text{b}}$

For gear B,

$\begin{array}{l}{\text{I}}_A{\text{= I}}_B\text{ = m}\overline{{\text{k}}^{\text{2}}}\text{ = 9×(0}{\text{.2)}}^{\text{2}}\\ {\text{I}}_B\text{=0}\text{.36kg}\cdot {\text{m}}^{\text{2}}\end{array}$

$\begin{array}{l}{\text{ΣM}}_B{\text{ = I}}_B{\text{α}}_B\\ F_{BC}\times r_B={\text{ I}}_B{\alpha }_B\\ {\text{F}}_{BC}\text{×(0}\text{.25 m) = 0}{\text{.36 kg×m}}^{\text{2}}{\text{ × α}}_B\\ {\text{F}}_{BC}\text{= 1}{\text{.44α}}_B-(2)\end{array}$

Since, gear a A and B are of same size, therefore because of symmetry, gear C will exert same force on gear B as on gear A.

For gear C,

$\begin{array}{l}{\text{I}}_C\text{ = m}\overline{{\text{k}}^{\text{2}}}\text{ = 3×(0}{\text{.075)}}^{\text{2}}\\ {\text{I}}_C\text{=0}\text{.016875 kg}\cdot {\text{m}}^{\text{2}}\end{array}$

$\begin{array}{l}{\text{ΣM}}_{\text{c}}{\text{ = I}}_{\text{c}}{\text{α}}_{\text{c}}\\ {\text{F}}_{\text{AC}}{\text{(r}}_{\text{C}}{\text{) - F}}_{\text{BC}}{\text{(r}}_{\text{c}}\text{) = }\overline{{\text{I}}_{\text{c}}}{\text{α}}_{\text{c}}\\ {\text{Substituting F}}_{\text{BC}}\text{ from equation (2) }\\ {\text{F}}_{\text{AC}}\text{(0}\text{.1) - (1}{\text{.44α}}_{\text{B}}\text{)(0}\text{.1) = 0}{\text{.016875kg×m}}^{\text{2}}{\text{×α}}_{\text{c}}\\ {\text{Substituting thevalue of α}}_{\text{c}}\text{ from ed (1)}\\ {\text{F}}_{\text{AC}}\text{= (0}\text{.016875×(2}{\text{.5α}}_{\text{B}}\text{) + }{\text{.144α}}_{\text{B}}\text{)}\frac{\text{1}}{\text{0}\text{.1}}\\ {\text{F}}_{\text{AC}}\text{= (0}\text{.016875×(2}\text{.5) + }\text{.144)}\frac{\text{1}}{\text{0}\text{.1}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{AC}}\text{ = 1}{\text{.862 α}}_{\text{B}}{\text{ rad/s}}^{\text{2}}\text{ - (3) }\end{array}$

For gear A,

$\begin{array}{l}{\text{ΣM}}_A{\text{ = I}}_A{\text{α}}_A\\ {\text{M - F}}_{\text{AC}}{\text{(r}}_A\text{) = }\overline{{\text{I}}_A}{\text{α}}_A\\ {\text{Substituting F}}_{\text{AC}}\text{ from equation (3) }\\ \text{5N}\cdot \text{m - (1}\text{.862}{\alpha }_B\text{)(0}\text{.25) = 0}{\text{.36kg×m}}^{\text{2}}{\text{×α}}_A\text{ (}\because {\alpha }_A\text{=}{\alpha }_B\text{)}\\ \text{5N}\cdot \text{m - (1}\text{.862}{\alpha }_B\text{)(0}\text{.25) = 0}{\text{.36kg×m}}^{\text{2}}{\text{×α}}_B\\ {\text{Substituting thevalue of α}}_{\text{c}}\text{ from ed (1)}\\ \text{5 N-m = 0}{\text{.8255 α}}_{\text{B}}\\ {\alpha }_B=6.057{\text{ rad/s}}^2\end{array}$

Angular acceleration of gear A, ${\text{α}}_{\text{A}}{\text{ = α}}_{\text{B}}\text{=6}{\text{.057 rad/s}}^{\text{2}}$

Conclusion:

Angular acceleration of gear A = 6.057 rad/s²

To determine

ii.

The tangential force that gear A exerts on gear C.

Answer to Problem 16.36P

Force exerted by gear C on gear A = 11.278 N

Explanation of Solution

Given:

Mass of Gear A,m_a = 9 kg

radius of gyration of gear A, $\overline{k_a}$ = 200 mm = 0.2 m

mass of gear B, m_b = 9 kg

radius of gyration of gear B, $\overline{k_b}$ = 200 mm = 0.2 m

mass of Gear C,m_c = 3 kg

radius of gyration of gear C, $\overline{k_c}$ = 75 mm = 0.075 m

Magnitude of Couple applied on gear A, M = 5N-m

Concept used:

Mass moment of acceleration is given by-

$\text{I = m}\overline{{\text{k}}^{\text{2}}},\text{ where }\overline{k}\text{ is the radius of gyration }$

The tangential force acting on a gear will provide the angular acceleration to the gear. Therefore,

$\begin{array}{l}\text{Summation of moments applied on it = Mass Moment of inertia }\times \text{ angular acceleration}\\ \Sigma \text{M = Iα }\end{array}$

The free body diagram of the three gears is as following-

Calculation:

Angular acceleration of gear A = 6.057 rad/s²

$\begin{array}{l}\text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ \text{Therefore, α = }\frac{\text{a}}{\text{r}}\end{array}$

The tangential component of two gears in mesh will be equal, therefore,

$\begin{array}{l}{\text{a}}_{\text{t}}{\text{= r}}_{\text{a}}{\text{α}}_{\text{a}}{\text{ = r}}_{\text{b}}{\text{α}}_{\text{b}}{\text{ = r}}_{\text{c}}{\text{α}}_{\text{c}}\\ {\text{a}}_{\text{t}}\text{= 0}{\text{.25α}}_{\text{a}}\text{ = 0}{\text{.1α}}_{\text{c}}\text{ = 0}{\text{.25α}}_{\text{b}}\\ {\text{α}}_{\text{c}}\text{ = 2}{\text{.5α}}_{\text{a}}\text{ = 2}{\text{.5α}}_{\text{b}}\text{ -(1)}\end{array}$

Since, ${\text{a}}_{\text{t}}\text{= 0}{\text{.25α}}_{\text{a}}\text{ = 0}{\text{.25α}}_{\text{b}}$

$\therefore {\text{α}}_{\text{a}}{\text{ = α}}_{\text{b}}$

For gear B,

$\begin{array}{l}{\text{I}}_A{\text{= I}}_B\text{ = m}\overline{{\text{k}}^{\text{2}}}\text{ = 9×(0}{\text{.2)}}^{\text{2}}\\ {\text{I}}_B\text{=0}\text{.36kg}\cdot {\text{m}}^{\text{2}}\end{array}$

$\begin{array}{l}{\text{ΣM}}_B{\text{ = I}}_B{\text{α}}_B\\ F_{BC}\times r_B={\text{ I}}_B{\alpha }_B\\ {\text{F}}_{BC}\text{×(0}\text{.25 m) = 0}{\text{.36 kg×m}}^{\text{2}}{\text{ × α}}_B\\ {\text{F}}_{BC}\text{= 1}{\text{.44α}}_B-(2)\end{array}$

Since gear a A and B are of same size, therefore because of symmetry, gear C will exert same force on gear B as on gear A.

For gear C,

$\begin{array}{l}{\text{I}}_C\text{ = m}\overline{{\text{k}}^{\text{2}}}\text{ = 3×(0}{\text{.075)}}^{\text{2}}\\ {\text{I}}_C\text{=0}\text{.016875 kg}\cdot {\text{m}}^{\text{2}}\end{array}$

$\begin{array}{l}{\text{ΣM}}_{\text{c}}{\text{ = I}}_{\text{c}}{\text{α}}_{\text{c}}\\ {\text{F}}_{\text{AC}}{\text{(r}}_{\text{C}}{\text{) - F}}_{\text{BC}}{\text{(r}}_{\text{c}}\text{) = }\overline{{\text{I}}_{\text{c}}}{\text{α}}_{\text{c}}\\ {\text{Substituting F}}_{\text{BC}}\text{ from equation (2) }\\ {\text{F}}_{\text{AC}}\text{(0}\text{.1) - (1}{\text{.44α}}_{\text{B}}\text{)(0}\text{.1) = 0}{\text{.016875kg×m}}^{\text{2}}{\text{×α}}_{\text{c}}\\ {\text{Substituting thevalue of α}}_{\text{c}}\text{ from ed (1)}\\ {\text{F}}_{\text{AC}}\text{= (0}\text{.016875×(2}{\text{.5α}}_{\text{B}}\text{) + }{\text{.144α}}_{\text{B}}\text{)}\frac{\text{1}}{\text{0}\text{.1}}\\ {\text{F}}_{\text{AC}}\text{= (0}\text{.016875×(2}\text{.5) + }\text{.144)}\frac{\text{1}}{\text{0}\text{.1}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{AC}}\text{ = 1}{\text{.862 α}}_{\text{B}}{\text{ rad/s}}^{\text{2}}\text{ - (3) }\end{array}$

Angular acceleration of gear A, ${\text{α}}_{\text{A}}{\text{ = α}}_{\text{B}}\text{=6}{\text{.057 rad/s}}^{\text{2}}$

Tangential force on gear A, on gearC-

$\begin{array}{l}{\text{F}}_{AC}\text{=1}{\text{.862α}}_{\text{a}}\\ {\text{F}}_{AC}\text{=1}\text{.862×6}\text{.057}\\ {\text{F}}_{AC}\text{= 11}\text{.278 N}\end{array}$

Conclusion:

Force exerted by gear C on gear A = 11.278 N