Question 15.6: The center of the double gear of Sample Prob. 15.2 has a vel......

The center of the double gear of Sample Prob. 15.2 has a velocity of 1.2 m/s to the right and an acceleration of 3 m/s² to the right. Recalling that the lower rack is stationary, determine (a) the angular acceleration of the gear, (b) the acceleration of points B, C, and D of the gear.

15.6.1
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a. Angular Acceleration of the Gear. In Sample Prob. 15.2, we found that x_{A}=-r_{1} \text {u}\text{ and }v_{A}=-r_{1} \text {v}. Differentiating the latter with respect to time, we obtain a_{\mathrm{A}}=-r_{1} \mathrm{a}.

v_A=-r_1 \text v \quad \quad 1.2 ~m / s =-(0.150 ~m ) \text v \quad \quad \text v =-8~ rad / s \\ \\ a_{ A }=-r_1 a \quad\quad 3 ~m / s ^2=-(0.150~ m ) a \quad\quad a =-20~ rad / s ^2

A = ak = -(20 rad/s²)k

b. Accelerations. The rolling motion of the gear is resolved into a translation with A and a rotation about A.

Acceleration of Point B. Adding vectorially the accelerations corresponding to the translation and to the rotation, we obtain

\begin{aligned}{a}_{B} & ={a}_{A}+{a}_{B / A}={a}_{A}+\left({a}_{B / A}\right)_{t}+\left({a}_{B / A}\right)_{n} \\& ={a}_{A}+{a} {k} \times {r}_{B / A}-\mathrm{v}^{2} {r}_{B / A} \\& =\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}-\left(20~ \mathrm{rad} / \mathrm{s}^{2}\right) {k} \times(0.100 \mathrm{~m}) {j}-(8 ~\mathrm{rad} / \mathrm{s})^{2}(0.100 \mathrm{~m}){j} \\& =\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}+\left(2 \mathrm{~m} / \mathrm{s}^{2}\right) {i}-\left(6.40 \mathrm{~m} / \mathrm{s}^{2}\right) {j} \\& \quad {a}_{B}=8.12~ \mathrm{~m} / \mathrm{s}^{2} ~\mathrm{c}~ 52.0^{\circ}\end{aligned}

Acceleration of Point C

\begin{aligned}{a}_{C} & ={a}_{A}+{a}_{C / A}={a}_{A}+\mathrm{a} {k} \times {r}_{C / A}-\text {v}^{2} {r}_{C / A} \\& =\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}-\left(20~ \mathrm{rad} / \mathrm{s}^{2}\right){k} \times(-0.150 \mathrm{~m}) {j}-(8 ~\mathrm{rad} / \mathrm{s})^{2}(-0.150 \mathrm{~m}) {j} \\& =\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}-\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}+\left(9.60 \mathrm{~m} / \mathrm{s}^{2}\right) {j} \\& {a}_{C}=9.60 \mathrm{~m} / \mathrm{s}^{2} ~\text x\end{aligned}

Acceleration of Point D

\begin{aligned}{a}_{D} & ={a}_{A}+{a}_{D / A}={a}_{A}+a {k} \times {r}_{D / A}-\text{v}^{2} {r}_{D / A} \\& =\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}-\left(20 ~\mathrm{rad} / \mathrm{s}^{2}\right) {k} \times(-0.150 \mathrm{~m}) {i}-(8~ \mathrm{rad} / \mathrm{s})^{2}(-0.150 \mathrm{~m}) {i} \\& =\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {i}+\left(3 \mathrm{~m} / \mathrm{s}^{2}\right) {j}+\left(9.60 \mathrm{~m} / \mathrm{s}^{2}\right) {i} \\& \quad {a}_{D}=12.95 \mathrm{~m} / \mathrm{s}^{2} \text { a } 13.4^{\circ}\end{aligned}
15.6.2
15.6.3
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