Question 15.10: In the Geneva mechanism of Sample Prob. 15.9, disk D rotates......
In the Geneva mechanism of Sample Prob. 15.9, disk D rotates with a constant counterclockwise angular velocity \text V_D of 10 rad/s. At the instant when f = 150°, determine the angular acceleration of disk S.
Referring to Sample Prob. 15.9, we obtain the angular velocity of the frame S attached to disk S and the velocity of the pin relative to S:
{a}_{P}={a}_{P^{\prime}}+{a}_{P / S}+{a}_{c} (1)
Each term of this vector equation is investigated separately.
Absolute Acceleration a_P. Since disk D rotates with a constant angular velocity, the absolute acceleration a_P is directed toward B. We have
\begin{aligned}a_{P} & =R \mathrm{v}_{D}^{2}=(500 \mathrm{~mm})(10 \mathrm{rad} / \mathrm{s})^{2}=5000 \mathrm{~mm} / \mathrm{s}^{2} \\{a}_{P} & =5000 \mathrm{~mm} / \mathrm{s}^{2}~ \mathrm{C} ~30^{\circ}\end{aligned}Acceleration a_{P^{\prime}} of the Coinciding Point {P^{\prime}}. The acceleration a_{P^{\prime}} of the point {P^{\prime}} of the frame S which coincides with P at the instant considered is resolved into normal and tangential components. (We recall from Sample Prob. 15.9 that r = 37.1 mm.)
\begin{aligned}& \left(a_{P^{\prime}}\right)_{n}=r \mathrm{~V}_{S}^{2}=(37.1 \mathrm{~mm})(4.08 ~\mathrm{rad} / \mathrm{s})^{2}=618 \mathrm{~mm} / \mathrm{s}^{2} \\& \left({a}_{P^{\prime}}\right)_{n}=618 \mathrm{~mm} / \mathrm{s}^{2} \mathrm{~d} \quad 42.4^{\circ} \\& \left(a_{P^{\prime}}\right)_{t}=r \mathrm{a}_{S}=37.1 \mathrm{a}_{S} \quad\quad\left({a}_{P^{\prime}}\right)_{t}=37.1 \mathrm{a}_{S} \mathrm{x~f}~ 42.4^{\circ}\end{aligned}Relative Acceleration a_{P/S}. Since the pin P moves in a straight slot cut in disk S, the relative acceleration a_{P/S} must be parallel to the slot; i.e., its direction must be a 42.4°.
Coriolis Acceleration a_c. Rotating the relative velocity \text v_{P/S} through 90° in the sense of \text V_S , we obtain the direction of the Coriolis component of the acceleration: h 42.4°. We write
\begin{gathered}a_{c}=2 \mathrm{v}_{S} v_{P / S}=2(4.08~ \mathrm{rad} / \mathrm{s})(477 \mathrm{~mm} / \mathrm{s})=3890 \mathrm{~mm} / \mathrm{s}^{2} \\{a}_{c}=3890 \mathrm{~mm} / \mathrm{s}^{2} \mathrm{~h}~ 42.4^{\circ}\end{gathered}We rewrite Eq. (1) and substitute the accelerations found above:
\begin{aligned}& {a}_{P}=\left({a}_{P^{\prime}}\right)_{n}+\left({a}_{P^{\prime}}\right)_{t}+{a}_{P / S}+{a}_{c} \\& {\left[5000 ~~\mathrm{c} \quad 30^{\circ}\right]=\left[\begin{array}{lll}618 \mathrm{~d} & 42.4^{\circ}\end{array}\right]+\left[37.1 \mathrm{a}_{S} \text { x f} ~42.4^{\circ}\right]} \\& +\left[\begin{array}{ll}a_{P / S} ~\text {z a}& 42.4^{\circ}\end{array}\right]+\left[\begin{array}{ll}3890 \mathrm{~h}~ 42.4^{\circ}\end{array}\right]\end{aligned}Equating components in a direction perpendicular to the slot,
\begin{aligned}5000~ \cos 17.6^{\circ}=37.1 \mathrm{a}_{S}-3890 & \\& \mathrm{~A}_{S}=\mathrm{A}_{S}=233~ \mathrm{rad} / \mathrm{s}^{2} ~\mathrm{i}\end{aligned}
