Question

Rod AB of length 3 m is kept on a smooth plane as shown in the given figure.The velocity of end A is 5 m/s along the inclined plane.Locate the ICR and find velocity of end B.

Given : Length of rod AB = 3m
                 va = 5 m/s
To find: ICR
              Velocity of end B

Answer 1

Solution :

Given : AB   =    3m
               v_A =5m /s 
To find : ICR
v_B 
Solution:
ICR is shown in the diagram denoted by point I

Assume ω to be the angular velocity of rod AB
BY GEOMETRY:

∠CAD=30^o, ∠ABD=20^o

∠CAB= ∠ABD=20^o

∠CAI=90^o-30^o 
        =60^o

∠BAI = ∠CAI+ ∠CAB=60^o+20^o 
        =80^o

In △IAB. ∠AIB = 180^o-80^o-70^o
                       =30^o

BY SINE RULE :

`(AB)/(sin I) =(IB)/(sin A )=(IA)/(sin B)`

`3/(sin 30) = (IB)/(sin 80 )  = (IA)/(sin 70)`

IB=5.9088 m
IA=5.6382 m

`ω = (v_a)/r = (v_a)/(IA) = 5/(5.6382)` = 0.8868 rad/s(anti clockwise)

v_B = r ω

= IB x ω 
= 5.9088 x 0.8868

=5.2401 m/s (Towards right)

Velocity of end B=5.2401 m/s (towards right)

Answer 2

Given : Length of rod AB = 3m
                 va = 5 m/s
To find: ICR
              Velocity of end B
Solution :

Given : AB   =    3m
               v_A =5m /s 
To find : ICR
v_B 
Solution:
ICR is shown in the diagram denoted by point I

Assume ω to be the angular velocity of rod AB
BY GEOMETRY:

∠CAD=30^o, ∠ABD=20^o

∠CAB= ∠ABD=20^o

∠CAI=90^o-30^o 
        =60^o

∠BAI = ∠CAI+ ∠CAB=60^o+20^o 
        =80^o

In △IAB. ∠AIB = 180^o-80^o-70^o
                       =30^o

BY SINE RULE :

`(AB)/(sin I) =(IB)/(sin A )=(IA)/(sin B)`

`3/(sin 30) = (IB)/(sin 80 )  = (IA)/(sin 70)`

IB=5.9088 m
IA=5.6382 m

`ω = (v_a)/r = (v_a)/(IA) = 5/(5.6382)` = 0.8868 rad/s(anti clockwise)

v_B = r ω

= IB x ω 
= 5.9088 x 0.8868

=5.2401 m/s (Towards right)

Velocity of end B=5.2401 m/s (towards right)