Question

In the mechanism given below, if the angular velocity of the eccentric circular disc is 1 rad/s, the angular velocity (rad/s) of the follower link for the instant shown in the figure is
[Note : All dimensions are in mm.]

• A. 0.05
• B. 0.1
• C. 5.0
• D. 10.0

Answer 1

The correct option is B 0.1

Method I:


$\Delta PQO\sim \Delta SRO$

$\therefore \frac{PQ}{SR}=\frac{PO}{SO}$ ....(1)

Applying Pythagoras theorem in $\Delta PQO.$

$(PO{)}^2=(PQ{)}^2+(OQ{)}^2$

$(PO{)}^2=\sqrt{(PQ{)}^2+(OQ{)}^2}$

$=\sqrt{(50{)}^2-(25{)}^2}=43.3mm$

Putting value of PQ in Eq. (1), we get

$\frac{43.3}{SR}=\frac{45+5}{5}$

$SR=\frac{43.3\times 5}{50}=4.33mm$

Velocity of Q is equal to velocity or R.

$V_Q=V_R=SR\times \omega$
$=4.33\times 1=4.33mm/s$

${\omega }_{PQ}=\frac{V_{PQ}}{PQ}=\frac{4.33}{43.3}=0.1rad/s$

Given, $\omega =1$ rad/s

and ${\omega }_{\text{link}}=?$

In $\Delta PSR$,

$cos\theta =\frac{25}{50}=\frac{1}{2}$

$\theta ={60}^o$

So, $\alpha ={30}^o$

Applying sine rule, $\Delta PQS$,

$\frac{5}{sin\beta }=\frac{x}{sin\theta }$

$x=\frac{5sin\theta }{sin\beta }=\frac{5sin{60}^o}{sin\beta }$

Velocity at point P,
$V=PQ\times \omega =x\times \omega =x\times 1=x$

$V=x=\frac{5sin{60}^o}{sin\beta }$

$Vsin\beta =5sin{60}^o$

Angular velocity of follower link;

${\omega }_{\text{link}}=\frac{Vsin\beta }{50cos\alpha }=\frac{5sin{60}^o}{50cos{30}^o}=\frac{1}{10}$

$=0.1rad/s$