Question

Figure shows a collar B which moves upwards with constant velocity of 1.5 m/s.At the instant when θ=50o.Determine :
(i)The angular velocity of rod pinned at B and freely resting at A against 25o sloping ground.
(ii)The velocity of end A of the rod.

Answer 1

ICR is shown in the given figure
BY USING GEOMETRY:
In △ABC
∠ABC = 50
∠ACB = 90
∠BAC = 40
∠CAV = 25
∠BAV = 40 – 25 = 15
IA ⊥ VA
∠IAB = 90 – 15 = 75
∠IBA = 90 – 50 = 40
In △IBA
∠BIA = 180 – 75 = 65
In △IBA
AB=1.2 m
APPLYING SINE RULE
`(AB)/(sin I)=(IB)/sinA=(IA)/(sinB)`

`1.2/sin65=(IB)/(sinA)=(IA)/sin40`
IB=1.2789 m
IA=0.8511 m
Assume ω_AB be the angular velocity of AB
ω_AB = `V_v/r=V_B/(IB)=1.5/1.2789` = 1.1728 rad/s
vA = r x AB = IA x ωAB = 0.8511 x 1.7288 = 0.99825 m/s
Angular velocity of rod AB= 1.1728 rads (Anti-clockwise)
Instantaneous velocity of A = 0.9982 m/s( 25° in first quadrant)