Question 15.SP.14: Two adjacent identical wheels of a train can be modeled as r...
Two adjacent identical wheels of a train can be modeled as rolling cylinders connected by a horizontal link. The distance between A and D is 10 in. Assume the wheels roll without sliding on the tracks. Knowing that the train is traveling at a constant 30 mph, determine the acceleration of the center of mass of DE.
STRATEGY: The connecting bar DE is undergoing curvilinear translation, so the acceleration of every point is identical; that is, a_G = a_D. Therefore, all you need to do is determine the acceleration of D using the kinematic relationship between A and D.
MODELING and ANALYSIS: Model the wheels and bar DE as rigid bodies. The speed of A is v_A = 30 mph = 44 ft/s. Since the wheel does not slip, the point of contact with the ground, C (Fig. 1), has a velocity of zero, so
\omega=\frac{v_A}{r_{A / C}}=\frac{44 \mathrm{ft} / \mathrm{s}}{(20 / 12) \mathrm{ft} .}=26.4 \mathrm{rad} / \mathrm{s}
Acceleration of D. The acceleration of D is
\mathbf{a}_D=\mathbf{a}_A + \mathbf{a}_{D / A}=\mathbf{a}_A + \boldsymbol{\alpha} \times \mathbf{r}_{D / A} – \omega^2 \mathbf{r}_{D / A} (1)
The train is traveling at a constant speed, so a_A and α are both zero. Substituting known quantities into Eq. (1) gives
\begin{aligned}\mathrm{a}_D &=0 + 0 – (26.4 \mathrm{rad} / \mathrm{s})^2\left[\left(\frac{10}{12} \cos 60^{\circ} \mathrm{ft}\right) \mathbf{i} + \left(\frac{10}{12} \sin 60^{\circ} \mathrm{ft}\right) \mathbf{j}\right] \\&=-\left(290.4 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{i} – \left(503.0 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{j}\end{aligned}
\mathbf{a}_G=\mathbf{a}_D=-\left(290 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{i} – \left(503 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{j}
REFLECT and THINK: Instead of using vector algebra, you could have recognized that the direction of -\omega^2 \mathbf{r}_{D / A} is directed from D to A. So the final acceleration of D is simply -\omega^2 \mathbf{r}_{D / A} \text{⦫} 60^{\circ} .
