Question

In the figure, link 2 rotates with constant angular velocity ${\omega }_2$. A slider link 3 moves outwards with a constant relative velocity $V_{Q/P}$, where Q is a point on slider 3 and P is a point on link 2. The magnitude and direction of Coriolis component of acceleration is given by

• A. $2{\omega }_2{V}_{Q/P}$; direction of $V_{Q/P}$ rotated by ${90}^o$ in the direction ${\omega }_2$
• B. ${\omega }_2{V}_{Q/P}$; direction of $V_{Q/P}$ rotated by ${90}^o$ in the direction ${\omega }_2$
• C. $2{\omega }_2{V}_{Q/P}$; direction of $V_{Q/P}$ rotated by ${90}^o$ opposite to the direction of ${\omega }_2$
• D. ${\omega }_2{V}_{Q/P}$; direction of $V_{Q/P}$ rotated by ${90}^o$ opposite to the direction ${\omega }_2$

Answer 1

The correct option is A $2{\omega }_2{V}_{Q/P}$; direction of $V_{Q/P}$ rotated by ${90}^o$ in the direction ${\omega }_2$


Coriolis acceleration
$=2\times$ velocity of slide $\times$ angular
velocity of link 2.
$a_c=2\times V_{a/p}\times {\omega }_2$
To determine the direction of the coriolis component, rotate the slider velocity vector by ${90}^o$ in the direction of angular velocity of the link.