The linear acceleration of collar D when θ = 60° is - 693.867 inches per square second.
In this question we have a system formed by three elements, the element AB experiments a pure rotation at constant velocity, the element BD has a general plane motion, which is a combination of rotation and traslation, and the ruff experiments a pure translation.
To determine the linear acceleration of the collar ([tex]a_{D}[/tex]), in inches per square second, we need to determine first all linear and angular velocities ([tex]v_{D}[/tex], [tex]\omega_{BD}[/tex]), in inches per second and radians per second, respectively, and later all linear and angular accelerations ([tex]a_{D}[/tex], [tex]\alpha_{BD}[/tex]), the latter in radians per square second.
By definitions of relative velocity and relative acceleration we build the following two systems of linear equations:
[tex]v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta[/tex] (1)
[tex]\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta[/tex] (2)
[tex]a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma[/tex] (3)
[tex]-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma[/tex] (4)
If we know that [tex]\theta = 60^{\circ}[/tex], [tex]\gamma = 19.889^{\circ}[/tex], [tex]r_{BD} = 10\,in[/tex], [tex]\omega_{AB} = 16\,\frac{rad}{s}[/tex], [tex]r_{AB} = 3\,in[/tex] and [tex]\alpha_{AB} = 0\,\frac{rad}{s^{2}}[/tex], then the solution of the systems of linear equations are, respectively:
[tex]v_{D}+3.402\cdot \omega_{BD} = -41.569[/tex] (1)
[tex]9.404\cdot \omega_{BD} = -24[/tex] (2)
[tex]v_{D} = -32.887\,\frac{in}{s}[/tex], [tex]\omega_{BD} = -2.552\,\frac{rad}{s}[/tex]
[tex]a_{D}+3.402\cdot \alpha_{BD} = -445.242[/tex] (3)
[tex]-9.404\cdot \alpha_{BD} = -687.264[/tex] (4)
[tex]a_{D} = -693.867\,\frac{in}{s^{2}}[/tex], [tex]\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}[/tex]
The linear acceleration of collar D when θ = 60° is - 693.867 inches per square second. [tex]\blacksquare[/tex]
The statement is incomplete and figure is missing, complete form is introduced below:
Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.
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