Question

Refer to the inverted slider crank mechanism shown in figure, at the instant the input crank (PS) is horizontal and perpendicular to fixed link and has an angular velocity of 1.25 rad/sec. The magnitude of the Coriolis acceleration is

• A. $360mm/se{c}^2$
• B. $480mm/se{c}^2$
• C. $960mm/se{c}^2$
• D. $625mm/se{c}^2$

Answer 1

The correct option is B $480mm/se{c}^2$


Magnitude of Coriolis acceleration = $2\omega V$ where $V_s$ is the sliding velocity alog link OQ.

$V_s=V\sin \theta =\omega r\sin \theta$ ($V_s$ is the velocity of point S on slider)

$\sin \theta =\frac{300}{500}=\frac{3}{5}$

$\cos \theta =\frac{4}{5}$

$V_s=\omega r\sin \theta$

$=1.25\times 400\times \frac{300}{500}=300mm/se{c}^2$

${\omega }_{OQ}ofthelinkOQ=\frac{V_s\cos \theta }{OS}=\frac{400mm/s}{500}$

${\omega }_{OS}=\frac{400}{500}=\frac{4}{5}$

$\text{Coriolis acceleration}=2\omega V_s$

$=2\times \frac{4}{5}\times 300=480mm/se{c}^2$