Step 1

We are given a slider block $B$ which is sliding horizontally with a certain velocity and acceleration. The slider block is connected to a rod $AB$ which is rotating in a circular slot.

The linear velocity of the slider block is, ${v_B} = 5{\rm{ ft/s}}$.

The linear acceleration of the slider block is, ${a_B} = 3{\rm{ ft/}}{{\rm{s}}^2}$.

We are required to determine the angular acceleration of point $A$ at the instant shown.

 
Step 2

The velocity diagram of the link $AB$ is given below:

The expression for the linear acceleration of the slider block in the horizontal direction is given by,

\[{{\rm{a}}_B} = \left( {3{\bf{i}}} \right){\rm{ ft/}}{{\rm{s}}^2}\].

The expression for the length of the link $AB$ in terms of the polar coordinates is given by,

\[{r_{A/B}} = \left( { - 2\cos 30^\circ {\bf{i}}{\rm{ + }}2\sin 30^\circ {\bf{j}}} \right)\]

From the velocity diagram it is clear that thedirection of the velocity of point $A$ is along the tangent of thecircular slot. So, the location of the Instantaneous Centers for rod AB can be determined by showing the direction of the velocities of points $A$ and $B$ of the rod.

The distance of point $B$ relative to the $IC$ for the rod $AB$ is given by,

\[\begin{array}{c} {r_{B/C}} = 2{\rm{ ft}} \times {\rm{sin30}}^\circ \\ {r_{B/C}} = 1{\rm{ ft}} \end{array}\]

The distance of point $A$ relative to the $IC$ for the rod $AB$ is given by,

\[\begin{array}{c} {r_{A/IC}} = 2{\rm{ ft}} \times {\rm{cos30}}^\circ \\ {r_{A/IC}} = 1.732{\rm{ ft}} \end{array}\]

At the instant, the angular velocity of the rod $AB$ is given by,

\[\begin{array}{c} {\omega _{AB}} = \frac{{{v_B}}}{{{r_{B/IC}}}}\\ {\omega _{AB}} = \frac{{5{\rm{ ft/s}}}}{{1{\rm{ ft}}}}\\ {\omega _{AB}} = 5{\rm{ rad/s}} \end{array}\]

Then using the relative velocity formula, the linear velocity of the point $A$ is given by,

\[\begin{array}{c} {v_A} = {\omega _{AB}} \times {r_{A/IC}}\\ {v_A} = 5{\rm{ rad/s}} \times \left( {1.732{\rm{ ft}}} \right)\\ {v_A} = 8.660{\rm{ ft/s}} \end{array}\]
 
Step 3

The point $A$ of the rod is moving along a circular slot, so the expression for the normal component of the acceleration of the point $A$ is given by,

\[{\left( {{a_A}} \right)_n} = \frac{{{v_A}^2}}{r}\]

Here, $r = 1.5{\rm{ ft}}$ is the radius of the circular slot.

On substituting the values in the above expression, we get,

\[\begin{array}{c} {\left( {{a_A}} \right)_n} = \frac{{{{\left( {8.660{\rm{ ft/s}}} \right)}^2}}}{{1.5{\rm{ ft}}}}\\ {\left( {{a_A}} \right)_n} = 50{\rm{ ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 4

Now using the relative acceleration equation, the expression for the acceleration for the point $A$ relative to the point $B$ is given by,

\[\begin{array}{c} {{\rm{a}}_A} = {{\rm{a}}_B} + {\alpha _{AB}}{r_{A/B}} - {\omega _{AB}}^2{r_{A/B}}\\ {\left( {{a_A}} \right)_n}{\bf{i}} + {\left( {{a_A}} \right)_t}{\bf{j}} = {{\rm{a}}_B} + {\alpha _{AB}}{r_{A/B}} - {\omega _{AB}}^2{r_{A/B}} \end{array}\]

Here, ${\left( {{a_A}} \right)_t}$ represents the tangential component of the acceleration of the point $A$.

On substituting the values in the above expression, we get,

\[\begin{array}{c} 50{\bf{i}} - {\left( {{a_A}} \right)_t}{\bf{j}} = 3{\bf{i}} + \left( {{\alpha _{AB}}{\bf{k}}} \right) \times \left( { - 2\cos 30^\circ {\bf{i}}{\rm{ + }}2\sin 30^\circ {\bf{j}}} \right) - {\left( 5 \right)^2} \times \left( { - 2\cos 30^\circ {\bf{i}}{\rm{ + }}2\sin 30^\circ {\bf{j}}} \right)\\ 50{\bf{i}} - {\left( {{a_A}} \right)_t}{\bf{j}} = \left( {46.30 - {\alpha _{AB}}} \right){\bf{i}} + \left( {1.732{\alpha _{AB}} + 25} \right){\bf{j}} \end{array}\]

Comparing all the $i$ components in the above expression, we get,

\[\begin{array}{c} 50 = 46.30 - {\alpha _{AB}}\\ {\alpha _{AB}} = - 3.70{\rm{ rad/}}{{\rm{s}}^2} \end{array}\]

The negative sign indicates that the rod $AB$ is rotating in the clockwise direction.

Similarly, comparing all the $j$ components in the above expression, we get,

\[\begin{array}{l} {\left( {{a_A}} \right)_t} = 1.732{\alpha _{AB}} + 25\\ {\left( {{a_A}} \right)_t} = 1.732 \times \left( { - 3.70} \right) + 25\\ {\left( {{a_A}} \right)_t} = 18.59{\rm{ ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 5

Then using the expression for the total acceleration, the magnitude of acceleration of the point $A$ is given by,

\[\begin{array}{c} {a_A} = \sqrt {{{\left( {{a_A}} \right)}_t}^2 + {{\left( {{a_A}} \right)}_n}^2} \\ {a_A} = \sqrt {{{\left( {18.59{\rm{ ft/}}{{\rm{s}}^2}} \right)}^2} + {{\left( {50{\rm{ ft/}}{{\rm{s}}^2}} \right)}^2}} \\ {a_A} = 53.3{\rm{ ft/}}{{\rm{s}}^2} \end{array}\]

And, the direction of the acceleration of point $A$ is given by,

\[\begin{array}{c} \theta = {\tan ^{ - 1}}\left( {\frac{{{{\left( {{a_A}} \right)}_t}}}{{{{\left( {{a_A}} \right)}_n}}}} \right)\\ \theta = {\tan ^{ - 1}}\left( {\frac{{18.59\;{\rm{ft}}}}{{50\;{\rm{ft}}}}} \right)\\ \theta = 20.4^\circ \end{array}\]