Chapter 16.1, Problem 16.37P

Gear A weighs 1 lb and has a radius of gyration of 1.3 in.; gear B weighs 6 lb and has a radius of gyration of 3 in.; gear C weighs 9 lb and has a radius of gyration of 4.3 in. Knowing a couple M of constant magnitude of 40 lb $\cdot$ in. is applied to gear A, determine (a) the angular acceleration of gear C, (b) the tangential force that gear B exerts on gear C.

To determine

i.

The angular acceleration of gear C.

Answer to Problem 16.37P

Angular acceleration of gear C = 129.99 rad/s²

Explanation of Solution

Given:

weight of Gear A, w_a = 1 lb radius of gyration of gear A, $\overline{k_a}$ = 1.3 in. = 1.3/12 ft

radius of gear A = 2 in.

weight of gear B, w_b = 6 lb

radius of gyration of gear B, $\overline{k_b}$ = 3 in. = 0.25 ft

radius of gear B, r₁ = 4 in.; r₂ = 2 in.

weight of Gear C, w_c = 9 lb

radius of gyration of gear C, $\overline{k_c}$ = 4.3 in. = 4.3/12 ft

radius of gear C = 6 in.

Magnitude of Couple applied on gear A, M = 40lb-in. = 40/12 lb-ft = 3.333 lb-ft

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:

Mass of gear A = $\frac{\text{1lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.03106 lb-s}}^{\text{2}}\text{/ft}$

Mass of gear B = $\frac{\text{6lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.18634 lb-s}}^{\text{2}}\text{/ft}$

Mass of gear C = $\frac{\text{9lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.2795 lb-s}}^{\text{2}}\text{/ft}$

Mass moment of inertia of gear A = ${\text{ m}}_{\text{A}}{\text{k}}_{\text{A}}^{\text{2}}\text{ = (0}{\text{.03106 lb-s}}^{\text{2}}\text{/ft)×(}\frac{\text{1}\text{.3}}{\text{12}}{\text{)}}^{\text{2}}\text{ = 3}{\text{.6448×10}}^{\text{-4}}{\text{lb×s}}^{\text{2}}\text{ft}$

Mass moment of inertia of gear B = ${\text{ m}}_B{\text{k}}_B^{\text{2}}\text{ = (0}{\text{.18634 lb-s}}^{\text{2}}\text{/ft)×(}{\text{.25)}}^{\text{2}}\text{ = 0}{\text{.01165 lb×s}}^{\text{2}}\text{ft}$

Mass moment of inertia of gear A = ${\text{ m}}_C{\text{k}}_C^{\text{2}}\text{ = (0}{\text{.2795 lb-s}}^{\text{2}}\text{/ft)×(}\frac{\text{4}\text{.3}}{\text{12}}{\text{)}}^{\text{2}}\text{ = 0}{\text{.3589 lb×s}}^{\text{2}}\text{ft}$

At the point of contact between A and B

$\begin{array}{l}\text{tangential acceleration of gear A is equal to the tangential acceleration of gear B}\\ \text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ {\text{Therefore, a}}_{\text{t}}{\text{ = r}}_{\text{A}}{\text{α}}_{\text{A}}{\text{ = r}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{2α}}_{\text{A}}{\text{ = 4α}}_{\text{B}}\\ {\text{α}}_{\text{A}}{\text{ = 2α}}_{\text{B}}\end{array}$

At the point of contact between B and C

$\begin{array}{l}\text{tangential acceleration of gear B is equal to the tangential acceleration of gear C}\\ \text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ {\text{Therefore, a}}_{\text{t}}{\text{ = r}}_{\text{2}}{\text{α}}_{\text{B}}{\text{ = r}}_{\text{C}}{\text{α}}_{\text{C}}\\ {\text{2α}}_{\text{B}}{\text{ = 6α}}_{\text{C}}\\ {\text{α}}_{\text{B}}{\text{ = 3α}}_{\text{C}}\\ {\text{ Therefore, α}}_{\text{A}}{\text{ = 2α}}_{\text{B}}{\text{ = 6α}}_{\text{C}}\end{array}$

For gear A,

$\begin{array}{l}{\text{ΣM}}_{\text{B}}{\text{ = I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{M-F}}_{\text{AB}}{\text{×r}}_{\text{A}}{\text{= I}}_{\text{A}}{\text{α}}_{\text{A}}\\ {\text{F}}_{\text{AB}}\text{×(2/12 ft) = 3}\text{.333lb×ft - (3}{\text{.6448×10}}^{\text{-4}}{\text{) lb×ft×s}}^{\text{2}}{\text{ × (6α}}_{\text{C}}\text{)}\\ {\text{F}}_{\text{AB}}\text{= 20lb - 0}{\text{.01312α}}_{\text{C}}\text{ lb}\end{array}$

For gear B,

$\begin{array}{l}{\text{ΣM}}_{\text{B}}{\text{ = I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{AB}}{\text{×r}}_{\text{1}}{\text{ - F}}_{\text{BC}}{\text{×r}}_{\text{2}}{\text{= I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{BC}}{\text{×(2/12 ft) = F}}_{\text{AB}}\text{×(4/12 ft) - (0}{\text{.01165) lb×ft×s}}^{\text{2}}{\text{ × (3α}}_{\text{C}}\text{)}\\ {\text{F}}_{\text{BC}}\text{ = ((20lb - 0}{\text{.01312α}}_{\text{C}}\text{ lb)×(4/12 ft) - (0}{\text{.01165) lb×ft×s}}^{\text{2}}{\text{ × (3α}}_{\text{C}}\text{))×}\frac{\text{1}}{\text{2/12}}\\ {\text{F}}_{\text{BC}}\text{ = 40lb - 0}{\text{.23594α}}_{\text{C}}\end{array}$

For gear C,

$\begin{array}{l}{\text{ΣM}}_{\text{C}}{\text{ = I}}_{\text{C}}{\text{α}}_{\text{C}}\\ {\text{F}}_{\text{BC}}{\text{×r}}_{\text{C}}{\text{= I}}_{\text{c}}{\text{α}}_{\text{C}}\\ \text{(40lb - 0}{\text{.23594α}}_{\text{C}}\text{)×(6/12 ft) = (0}{\text{.03589) lb×ft×s}}^{\text{2}}{\text{ × (α}}_{\text{C}}\text{)}\\ \text{20lb = 0}{\text{.15386α}}_{\text{C}}\\ {\text{α}}_{\text{C}}\text{ = 129}{\text{.99 rad/s}}^{\text{2}}\end{array}$

Conclusion:

Angular acceleration of gear C = 129.99 rad/s²

To determine

ii.

The tangential force that gear B exerts on gear C.

Answer to Problem 16.37P

Force exerted by gear B on gear C = 9.33 lb

Explanation of Solution

Given:

weight of Gear A, w_a = 1 lb radius of gyration of gear A, $\overline{k_a}$ = 1.3 in. = 1.3/12 ft

radius of gear A = 2 in.

weight of gear B, w_b = 6 lb

radius of gyration of gear B, $\overline{k_b}$ = 3 in. = 0.25 ft

radius of gear B, r₁ = 4 in.; r₂ = 2 in.

weight of Gear C, w_c = 9 lb

radius of gyration of gear C, $\overline{k_c}$ = 4.3 in. = 4.3/12 ft

radius of gear C = 6 in.

Magnitude of Couple applied on gear A, M = 40lb-in. = 40/12 lb-ft = 3.333 lb-ft

Angular acceleration of gear C = 129.99 rad/s²

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:

Mass of gear A = $\frac{\text{1lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.03106 lb-s}}^{\text{2}}\text{/ft}$

Mass of gear B = $\frac{\text{6lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.18634 lb-s}}^{\text{2}}\text{/ft}$

Mass of gear C = $\frac{\text{9lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.2795 lb-s}}^{\text{2}}\text{/ft}$

Mass moment of inertia of gear A = ${\text{ m}}_{\text{A}}{\text{k}}_{\text{A}}^{\text{2}}\text{ = (0}{\text{.03106 lb-s}}^{\text{2}}\text{/ft)×(}\frac{\text{1}\text{.3}}{\text{12}}{\text{)}}^{\text{2}}\text{ = 3}{\text{.6448×10}}^{\text{-4}}{\text{lb×s}}^{\text{2}}\text{ft}$

Mass moment of inertia of gear B = ${\text{ m}}_B{\text{k}}_B^{\text{2}}\text{ = (0}{\text{.18634 lb-s}}^{\text{2}}\text{/ft)×(}{\text{.25)}}^{\text{2}}\text{ = 0}{\text{.01165 lb×s}}^{\text{2}}\text{ft}$

Mass moment of inertia of gear A = ${\text{ m}}_C{\text{k}}_C^{\text{2}}\text{ = (0}{\text{.2795 lb-s}}^{\text{2}}\text{/ft)×(}\frac{\text{4}\text{.3}}{\text{12}}{\text{)}}^{\text{2}}\text{ = 0}{\text{.3589 lb×s}}^{\text{2}}\text{ft}$

At the point of contact between A and B

$\begin{array}{l}\text{tangential acceleration of gear A is equal to the tangential acceleration of gear B}\\ \text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ {\text{Therefore, a}}_{\text{t}}{\text{ = r}}_{\text{A}}{\text{α}}_{\text{A}}{\text{ = r}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{2α}}_{\text{A}}{\text{ = 4α}}_{\text{B}}\\ {\text{α}}_{\text{A}}{\text{ = 2α}}_{\text{B}}\end{array}$

At the point of contact between B and C

$\begin{array}{l}\text{tangential acceleration of gear B is equal to the tangential acceleration of gear C}\\ \text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ {\text{Therefore, a}}_{\text{t}}{\text{ = r}}_{\text{2}}{\text{α}}_{\text{B}}{\text{ = r}}_{\text{C}}{\text{α}}_{\text{C}}\\ {\text{2α}}_{\text{B}}{\text{ = 6α}}_{\text{C}}\\ {\text{α}}_{\text{B}}{\text{ = 3α}}_{\text{C}}\\ {\text{ Therefore, α}}_{\text{A}}{\text{ = 2α}}_{\text{B}}{\text{ = 6α}}_{\text{C}}\end{array}$

For gear A,

$\begin{array}{l}{\text{ΣM}}_{\text{B}}{\text{ = I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{M-F}}_{\text{AB}}{\text{×r}}_{\text{A}}{\text{= I}}_{\text{A}}{\text{α}}_{\text{A}}\\ {\text{F}}_{\text{AB}}\text{×(2/12 ft) = 3}\text{.333lb×ft - (3}{\text{.6448×10}}^{\text{-4}}{\text{) lb×ft×s}}^{\text{2}}{\text{ × (6α}}_{\text{C}}\text{)}\\ {\text{F}}_{\text{AB}}\text{= 20lb - 0}{\text{.01312α}}_{\text{C}}\text{ lb}\end{array}$

For gear B,

$\begin{array}{l}{\text{ΣM}}_{\text{B}}{\text{ = I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{AB}}{\text{×r}}_{\text{1}}{\text{ - F}}_{\text{BC}}{\text{×r}}_{\text{2}}{\text{= I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{BC}}{\text{×(2/12 ft) = F}}_{\text{AB}}\text{×(4/12 ft) - (0}{\text{.01165) lb×ft×s}}^{\text{2}}{\text{ × (3α}}_{\text{C}}\text{)}\\ {\text{F}}_{\text{BC}}\text{ = ((20lb - 0}{\text{.01312α}}_{\text{C}}\text{ lb)×(4/12 ft) - (0}{\text{.01165) lb×ft×s}}^{\text{2}}{\text{ × (3α}}_{\text{C}}\text{))×}\frac{\text{1}}{\text{2/12}}\\ {\text{F}}_{\text{BC}}\text{ = 40lb - 0}{\text{.23594α}}_{\text{C}}\end{array}$

As found in part I,

Angular acceleration of gear C = 129.99 rad/s²

Tangential force of gear B, on gear C-

$\begin{array}{l}{\text{F}}_{\text{BC}}\text{ = 40lb - 0}{\text{.23594α}}_{\text{C}}\\ {\text{F}}_{\text{BC}}\text{ = 40lb - 0}\text{.23594(129}{\text{.9rad/s}}^2\text{)}\\ {\text{F}}_{AC}\text{= 9}\text{.33 lb}\end{array}$

Conclusion:

Angular acceleration of gear C = 129.99 rad/s²

Force exerted by gear B on gear C = 9.33 lb