Question

If link AB of the four-bar linkage has a constant counterclockwise angular velocity of 58 rad/s during an interval which includes the instant represented, determine the angular acceleration of AO (positive if counterclockwise, negative if clockwise) and the acceleration of point D.

Answer 1

Answer:Vb=-6i-(-0.1ωab+8)j m/s

Explanation:

Va=V0+Va0

Va=V0+(ra0 x ωao)

ω=Angular velocity of link A0

Using r0a=0.1m;

Va=V0+(0.1i x ω0a K)

Va=0

ixk=j

Va=0+0.1ω0aj

Calculating te velocity of using te equation below

Vb=Va+Vba

Vb=Va+ωab x rba

ωab=40rad/s

rab=-0.21i+0.15j

Va=0.1ω0aj

Vb=Va+ωabxrba

Vb=0.1ω0aj+40k x -(0.21i+0.15j)

Vb=0.1ω0aj-8j-6i

Vb=-6i-(-0.1ωab+8)j m/s

Answer 2

The angular acceleration of the bar $AO$ is 560.673 radians per square second and the acceleration of point $D$ is 3501595.696 milimeters per square second.

How to solve a general plane motion problem

Bars $AO$ and $CB$ are only rotating, whereas bar $AB$ experiments a general plane motion, that is, a combination of rotational and translational motion. In addition, the latter have a constant angular velocity ($\vec \omega_{AB}$), in radians per second.

To obtain the given information, we must find all angular speeds and accelerations from relative velocity and acceleration formulas:

Relative velocity

$\vec v_{B}-\vec v_{A} = \vec v_{B/A}$ (1)

$(-r_{CB}\cdot \omega_{B},0) - (0, r_{OA}\cdot \omega_{A}) = (-r_{AB}\cdot \omega_{AB}\cdot \cos \theta_{AB}, -r_{AB}\cdot \omega_{AB}\cdot \sin \theta_{AB})$

Whose solution is:

$\omega_{A} = \frac{r_{AB}\cdot \omega_{AB}\cdot \cos \theta_{AB}}{r_{CB}}$   (2)

$\omega_{B} = \frac{r_{AB}\cdot \omega_{AB}\cdot \cos \theta_{AB}}{r_{CB}}$   (3)

Relative acceleration

$\vec a_{B}^{n} + \vec a_{B}^{t}-\vec a_{A}^{n}-\vec a_{A}^{t} = \vec a_{B/A}^{n}+\vec a_{B/A}^{t}$ (4)

$(0, -\omega_{B}^{2}\cdot r_{CB}) + (-\alpha_{B}\cdot r_{CB}, 0)-(-\omega_{A}^{2}, 0)-(0, \alpha_{A}\cdot r_{OA}) = (r_{BA}\cdot \cos \theta_{AB}, -r_{BA}\cdot \sin \theta_{AB}) + (0,0)$

Whose solution is:

$\alpha_{A} = \frac{r_{BA}\cdot \omega_{AB}^{2}\cdot \sin \theta_{AB}-\omega_{B}^{2}\cdot r_{CB}}{r_{OA}}$   (5)

$\alpha_{B} = \frac{\omega_{A}^{2}\cdot r_{OA}-r_{BA}\cdot \omega_{BA}^{2}\cdot \cos \theta_{AB}}{r_{CB}}$   (6)

The angular acceleration of the bar $AO$ is given by (5) and the acceleration of point $D$ is given by (3) and (6) and the following formula:

$a_{D} = r_{CD}\cdot \sqrt{\omega_{B}^{4}+\alpha_{B}^{2}}$   (7)

If we know that $r_{BA} = 300\,mm$, $r_{CB} = 180\,mm$, $r_{OA} = 120\,mm$, $\theta_{AB} \approx 36.870^{\circ}$, $r_{CD} = 260\,mm$ and $\omega_{AB} = 58\,\frac{rad}{s}$, then the angular acceleration of AO and the acceleration of point D are, respectively:

By (2) and (3):

$\omega_{A} = 87\,\frac{rad}{s}$, $\omega_{B} = 116\,\frac{rad}{s}$

By (5) and (6):

$\alpha_{A} = -15137.988\,\frac{rad}{s^{2}}$, $\alpha_{B} = 560.673\,\frac{rad}{s^{2}}$

And by (7):

$a_{D} = 3501595.696\,\frac{mm}{s^{2}}$

The angular acceleration of the bar $AO$ is 560.673 radians per square second and the acceleration of point $D$ is 3501595.696 milimeters per square second. $\blacksquare$

Remark

Statement is incomplete and the figure of the four-bar system is missing. We proceed to include the entire statement below:

If link $AB$ of the four-bar linkage has a constant counterclockwise angular velocity of 58 radians per second during an interval which includes the instant represented, determine the angular acceleration of $AO$ (positive if counterclockwise, negative if clockwise) and the acceleration of point $D$.

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