Question

Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

Answer 1

The linear acceleration of collar D when θ = 60° is - 693.867 inches per square second.

How to determine the angular velocity of a collar

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 ($a_{D}$), in inches per square second, we need to determine first all linear and angular velocities ($v_{D}$, $\omega_{BD}$), in inches per second and radians per second, respectively, and later all linear and angular accelerations ($a_{D}$, $\alpha_{BD}$), the latter in radians per square second.

By definitions of relative velocity and relative acceleration we build the following two systems of linear equations:

Velocities

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$   (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$   (2)

Accelerations

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$   (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$   (4)

If we know that $\theta = 60^{\circ}$, $\gamma = 19.889^{\circ}$, $r_{BD} = 10\,in$, $\omega_{AB} = 16\,\frac{rad}{s}$, $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$, then the solution of the systems of linear equations are, respectively:

Velocities

$v_{D}+3.402\cdot \omega_{BD} = -41.569$   (1)

$9.404\cdot \omega_{BD} = -24$   (2)

$v_{D} = -32.887\,\frac{in}{s}$, $\omega_{BD} = -2.552\,\frac{rad}{s}$

Accelerations

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$   (3)

$-9.404\cdot \alpha_{BD} = -687.264$   (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$, $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The linear acceleration of collar D when θ = 60° is - 693.867 inches per square second. $\blacksquare$

Remark

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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