Question

A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in the figure. The crank is rotating with a uniform angular speed of 10 rad/s, counter clockwise. For the given configuration, the speed (in m/s ) of the slider is

Answer 1

Method I:

L = 240 mm
r = 60 mm
$n=\frac{L}{r}=\frac{240}{60}=4$
$\omega =10$ rad/s
From the given configuration,
$cos\theta =\frac{sin\theta }{n}ortan\theta =n$
or $\theta =ta{n}^{-1}4={75.96}^o$
$\therefore V_s=\omega r[sin\theta +\frac{sin2\theta }{2n}]$
$=10\times 0.06[sin{75.96}^o+\frac{sin{151.92}^o}{2\times 4}]$
$=0.6[0.97+0.0588]$
$=0.6\times 1.0288=0.617m/s$
or $V_0=\omega r$
Since slider is $\perp$ar to the crank
so, $V_0=\omega r=10\times 0.06$
$=0.6$m/s

Method II:

$V_s=$ Velocity of slider
$A{C}^2={60}^2+{240}^2$
$AC=247.386$ mm
Velocity, $V=\left(BC\right)\times \omega$
$=0.06\times 10=0.6$ m/s
Since, crank and connecting rod is perpendicular.
$V_{c}cos\theta =V$
$V_s=\frac{V}{cos\theta }=\frac{0.6}{\left(\frac{240}{297.386}\right)}$
$=0.618$m/s