Question 15.SP.12: Collars A and B are pin-connected to bar ABD and can slide a...
Collars A and B are pin-connected to bar ABD and can slide along fixed rods. Knowing that, at the instant shown, the velocity of A is a constant 0.9 m/s to the right, determine the angular acceleration of AB and the acceleration of B.
STRATEGY: Use the kinematic equation that relates the acceleration of two points on the same rigid body. Because you know that the directions of the accelerations of A and B must be along the fixed rods, choose these two points to relate.
MODELING and ANALYSIS: Model bar ABD as a rigid body. From Sample Prob. 15.5, you know \omega=3.00 \mathrm{rad} / \mathrm{s} \circlearrowleft . The accelerations of A and B are related by
\mathbf{a}_B=\mathbf{a}_A + \mathbf{a}_{B / A}=\mathbf{a}_A + \boldsymbol{\alpha} \times \mathbf{r}_{B / A} – \omega^2 \mathbf{r}_{B / A}
Substituting in known values (Fig. 1) and assuming α = αk gives
\begin{aligned}&a_B \cos 60^{\circ} \mathbf{i} + a_B \sin 60^{\circ} \mathbf{j}=0 \mathbf{i} + \alpha \mathbf{k} \times\left[\left(0.3 \cos 30^{\circ}\right) \mathbf{i} + \left(0.3 \sin 30^{\circ}\right) \mathbf{j}\right] \\&\quad-3^2\left[\left(0.3 \cos 30^{\circ}\right) \mathbf{i} + \left(0.3 \sin 30^{\circ}\right) \mathbf{j}\right] \\ &\quad 0.500 a_B \mathbf{i} + 0.866 a_B \mathbf{j}=(0-0.15 \alpha – 2.338) \mathbf{i} + (0.260 \alpha – 1.350) \mathbf{j}\end{aligned}
Equating components, you have
\begin{array}{ll}\text { i: } & 0.500 a_B=-0.15 \alpha – 2.338 \\\text { j: } & 0.866 a_B=0.260 \alpha-1.350\end{array}Solving these equations gives a_B=-3.12 \mathrm{~m} / \mathrm{s}^2 \text { and } \alpha=-5.20 \mathrm{rad} / \mathrm{s}^2
\begin{aligned}&\boldsymbol{\alpha}=5.20 \mathrm{rad} / \mathrm{s}^2\circlearrowright \\&\mathbf{a}_B=3.12 \mathrm{~m} / \mathrm{s}^2 \text { ⦫ } 60^{\circ} \\&\end{aligned}
REFLECT and THINK: Even though A is traveling at a constant speed, bar AB still has an angular acceleration, and B has a linear acceleration. Just because one point on a body is moving at a constant speed doesn’t mean the rest of the points on the body also have a constant speed.
