Question

The cross head velocity in the slider crank mechanism, for the position shown in the figure

• A. $V_{c}cos(90-\overline{\alpha +\beta })cos\beta$
• B. $V_{c}cos(90-\overline{\alpha +\beta })sec\beta$
• C. $V_{c}cos(90-\overline{\alpha -\beta })cos\beta$
• D. $V_{c}cos(90-\overline{\alpha -\beta })sec\beta$

Answer 1

The correct option is B $V_{c}cos(90-\overline{\alpha +\beta })sec\beta$

Method 1:

Cross head is 4th link
$V_4=?$

$V_4={\omega }_2\left(I_{24}{I}_{12}\right)$

$V_4={\omega }_2\left(I_{24}{I}_{12}\right)$

$\Rightarrow$ $V_4=\frac{V_c}{r}\left(I_{24}{I}_{12}\right)$
$\Rightarrow \frac{r}{sin({90}^o-\beta )}=\frac{x}{sin(\alpha +\beta )}[Let{I}_{24}{I}_{12}=x]$

$x=\frac{rsin(\alpha +\beta )}{cos\beta }$

$\Rightarrow V_4=\frac{V_c}{r}\times \frac{rcos[90-(\alpha +\beta \left)\right]}{cos\beta }$

$\Rightarrow V_4=V_{c}cos[{90}^o-(\alpha +\beta \left)\right]sec\beta$

Method II:
Considering $\Delta COM,$

$CM=lsin\beta =rsin\alpha$
$sin\beta =\frac{r}{l}sin\alpha$

$cos\beta =\sqrt{1-\frac{r^{2}si{n}^2\alpha }{l^2}}$

Velocity of cross-head,

$V=r\omega [sin\alpha +\frac{rsin\alpha cos\alpha }{l^2-r^{2}si{n}^2\alpha }]$

$=V_C[sin\alpha +\frac{rsin\alpha cos\alpha }{l^{2}co{s}^2\beta }]$

$=V_c[sin\alpha \frac{cos\beta }{cos\beta }+\frac{rsin\alpha cos\alpha }{lcos\beta }]$

$=V_c[\frac{sin\alpha cos\beta }{sin\beta }+\frac{sin\beta cos\alpha }{lcos\beta }]$

$=V_{C}sec\beta [sin\alpha cos\beta +cos\alpha sin\beta ]$

$=V_{C}sec\beta sin(\alpha +\beta )$

$=V_{C}sec\beta cos(90-\overline{\alpha +\beta })$

$=V_{C}cos(90-\overline{\alpha +\beta })sec\beta$