Question 15.SP.20: In the Geneva mechanism of Sample Prob. 15.19, disk D rotate...
In the Geneva mechanism of Sample Prob. 15.19, disk D rotates with a constant counterclockwise angular velocity \boldsymbol{\omega}_D of 10 rad/s. At the instant when \phi = 150°, determine the angular acceleration of disk S.
STRATEGY: You have two rigid bodies whose motions are related; therefore use rigid-body kinematics. Since point P is moving in a slot, use a rotating reference frame.
MODELING and ANALYSIS: Since you are computing accelerations instead of velocities, you need to use Eq. (15.36), which includes the Coriolis acceleration. You found the angular velocity of the frame \mathcal{S} attached to disk S and the velocity of the pin relative to \mathcal{S} in Sample Prob. 15.19:
\mathbf{a}_P=\mathbf{a}_{P^{\prime}} + \mathbf{a}_{P / \mathscr{F}} + \mathbf{a}_C (15.36)
\begin{gathered}\omega_{\mathcal{S}}=4.08 \mathrm{rad} / \mathrm{s} \circlearrowright \\ \beta=42.4^{\circ} \quad \mathbf{v}_{P/ \mathcal{S}}=477 \mathrm{~mm} / \mathrm{s} \text{⦫} 42.4^{\circ}\end{gathered}
Since pin P moves with respect to the rotating frame \mathcal{S}, you have
\mathbf{a}_P=\mathbf{a}_{P^{\prime}} + \mathbf{a}_{P /\mathcal{S}} + \mathbf{a}_c (1)
Investigate each term of this vector equation separately.
Absolute Acceleration a_P. Since disk D rotates with a constant angular velocity, the absolute acceleration a_P is directed toward B. This gives
\begin{aligned}&a_P=R \omega_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 \text { ⦪ }^{\circ} 0^{\circ}\end{aligned}
Acceleration a_{P^{\prime}} of the Coinciding Point P^{\prime}. Resolve into normal and tangential components the acceleration a_{P^{\prime}} of the point P^{\prime} of the frame \mathcal{S} that coincides with P at the given instant. (Recall from Sample Prob. 15.19 that r = 37.1 mm.)
\begin{aligned}&\left(a_{P^{\prime}}\right)_n=r \omega_{\mathcal{S}}^2=(37.1 \mathrm{~mm})(4.08 \mathrm{rad} / \mathrm{s})^2=618 \mathrm{~mm} / \mathrm{s}^2 \\&\left(\mathbf{a}_{P^{\prime}}\right)_n=618 \mathrm{~mm} / \mathrm{s}^2 \text{⦫} 42.4^{\circ} \\&\left(a_{P^{\prime}}\right)_t=r \alpha_{\mathcal{S}}=37.1 \alpha_{\mathcal{S}} \quad\left(\mathbf{a}_{P^{\prime}}\right)_t=37.1 \alpha_{\mathcal{S}} \text{ ⦮ } 42.4^{\circ}\end{aligned}
Relative Acceleration a_{P/\mathcal{S}}. Since the pin P moves in a straight slot cut in disk S, the relative acceleration a_{P/\mathcal{S}} must be parallel to the slot; i.e., its direction must be ⦨ 42.4°.
Coriolis Acceleration a_C. Rotating the relative velocity \mathrm{v}_{P/\mathcal{S}} through 90° in the sense of \omega_{\mathcal{S}}, you obtain the direction of the Coriolis component of the acceleration: \text{ ⦭ } 42.4°. You have
\begin{gathered}a_C=2 \omega_{\mathcal{S}} \mathrm{v}_{P /\mathcal{S}}=2(4.08 \mathrm{rad} / \mathrm{s})(477 \mathrm{~mm} / \mathrm{s})=3890 \mathrm{~mm} / \mathrm{s}^2 \\\mathbf{a}_C=3890 \mathrm{~mm} / \mathrm{s}^2 \text{⦭} 42.4^{\circ}\end{gathered}
Rewrite Eq. (1) and substitute the accelerations found (Fig. 1):
\begin{aligned}&\mathbf{a}_p=\left(\mathbf{a}_p^\prime \right)_n + \left(\mathbf{a}_p^\prime\right)_t + \mathbf{a}_{P /\mathcal{S}} + \mathbf{a}_C \\&{\left[5000 \text { ⦪ } 30^{\circ}\right]=\left[618 \text{⦫} 42.4^{\circ}\right] + \left[37.1 \alpha_g \text{⦮} 42.4^{\circ}\right]} \\&+\left[a_{P /\mathcal{S}} \text{ ⦨} 42.4^{\circ}\right] + \left[3890 \text{ ⦭ } 42.4^{\circ}\right] \\& \end{aligned}
Equating components in a direction perpendicular to the slot,
\begin{array}{r}5000 \cos 17.6^{\circ}=37.1 \alpha_{\mathcal{S}} – 3890 \\\alpha_S=\alpha_{\mathcal{S}}=233 \mathrm{rad} / \mathrm{s}^2 \circlearrowright \end{array}
REFLECT and THINK: It seems reasonable that, since disk S starts and stops over the very short time intervals when pin P is engaged in the slots, the disk must have a very large angular acceleration. An alternative approach would have been to use the vector algebra.
