A rod AB has an angular velocity of 2 rad/sec,counter clock wise as shown.End C of rod BC is free to move on a horizontal surface.Determine:

(1)Angular velocity of rod BC

(2)Velocity of C

**Given** : ωAB=2 rad/sec (anti clockwise)

**To find** : ωBC

VC

#### Solution

**BY GEOMETRY :**

Assume I to be the ICR of rod BC

In △IAB,

∠BIC=40°

∠IBC=90°

`tan 40 = (BC)/(IB) = (0.5)/(IB)`

`sin 40 = (BC)/(IC) = (0.5)/(IC)`**IB = 0.5959m and IC = 0.7779m**

v_{B} = rω

= ABxωAB

= 0.3 x 2

= 0.6 m/s

ω_{BC} = `(v_B)/r`

`= (v_B)/(IB)`

`= (0.6)/(0.5959)`

=1.0069 rad/sec

The direction is anti-clockwise

vC = rω

= IC x ωBC

= 0.7779 x 1.0069

= 0.7832 m/s

The direction of v_{C} is towards right

**Angular velocity of BC=1.0069 rad/sec(anto clockwise)v _{C}=0.7832 m/s(Towards right) **