Question

For an inline slider-crank mechanism, the length of the crank ans connecting rod are 3 m and 4m, respectively. At the instant when the connecting rod is perpendicular to the crank, if the velocity of the slider is 1 m/s, the magnitude of angular velocity (upto 3 decimal points accuracy) of the crank is radian/s.

Answer 1

Here $V_B=1m/s,{\omega }_{crank}=?$
Let us draw the I-centres

Applying angular velocity theorem at $I_{24}$
$\therefore {\omega }_2\left(I_{24}{I}_{12}\right)=V_4=V_B=1$
${\omega }_2\left(I_{24}{I}_{12}\right)=1$

$\theta =ta{n}^{-1}\left(4/3\right)={53.13010}^o$
$\alpha ={90}^o-\theta ={36.869}^o$
$cos\alpha =\frac{3}{I_{12}{I}_{24}}$
$I_{12}{I}_{24}=\frac{3}{cos{36.869}^o}=3.7499meter$
Now, ${\omega }_2\left(3.7499\right)=1$
${\omega }_2=\frac{1}{3.7499}=0.266$ rad/s