AB = 400 mm.

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : MAY 2016

$$V_c= ?,V_D= ?,W_AB=3rad/sec $$

$AB = 400 mm. \\ V_B=r.w \\ =AB.ω_{AB } \\ =100×3=300 mm/s $

$IC = 100 mm \\ IB=\sqrt{125^2-100^2}=75mm \\ ∠ IBC = \tan^{-1} (\dfrac {100}{75})=53.13^0 \\ ω_{CD} = \dfrac {V_B}{IB} = \dfrac {300}{75}=4 rad/s \\ V_C=I_C.ω_{CD}=100×4=400 mm/s$

In ΔDBI

$∠DBI=180-53.13=126.87^0$

By cosine Rule,

$I_D = \sqrt{ IB^2+BD^2-2IBBD× \cos ∠DBI} \\ =\sqrt{75^2+ 125^2-2×75×125×\cos⁡126.87 } \\ = 180.3 m \\ V_D=ID.ω_{CD}=180.3×400=721.2 mm/s$

In Δ DMI

$∠ DMI=180-90-33.69=56.31^0 $

Instantaneous velocity of $C = 400mm/s (→)$

Instantaneous velocity of $D = 721.2mm/s (56.31^0 )$