$\text { Take radius of disk 0.24m. }$

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : MAY 2014

Consider above diagram in which BX & AY are velocity vectors

AB is diameter of circle & C is ICR of disk.

Now, triangles XBI & IAY are similar

Therefore, $BI/IA = BX/AY = 3/1.5 =2 \\ BI = 2 IA ------------(i) $

Also, BI + IA = BA(diameter) = 0.48____(ii)

From (i) & (ii),

$BI = 0.32 m $ & $ IA = 0.16 m $

Now, as $BI = 0.32 $ & $BC = 0.24, $

IC(distance between centre of circle & ICR) = 0.08

Now, w of disc is $VB/IB = 3/0.32 = 9.375 rad/s \\ Vc = w * IC = 9.375*0.08 = 0.75 m/s $

Now, for calculation of VD , we need ID

Applying cosine rule,

$ID =\sqrt{(0.24)2+(0.08)2 - 2*(0.24)(0.08)\cos135} = 0.315 m $

Now, $VD = w * ID = 9.375 * 0.315 = 2.953 m/s $

Velocity of C is 0.75 m/s & velocity of D is 2.953 m/s