At Point C On AB ,150 mm From A, The Rod CE 350 mm Long Is Attached. This Rod CE Slides In a Slot In A Trunnion At D. The End E Is connected By a Link EF, 300mm Long To The Horizontally Moving Slider F.

For The Mechanism In The Position Shown, Find

1.Velocity Of F.

2.Velocity Of Sliding Of CE In The Trunnion , And

3.Angular Velocity Of CE.

Subject: Kinematics of Machinery

Topic: Velocity Analysis of Mechanism (Mechanism up-to 6 links)

Difficulty: High

Given: $v_{AO}$ =120 rpm., or $\omega_{AO}$ = $2\pi \times\frac{120}{60} = 4\pi$ rad/s.

Since the length of crank OA =100 mm =0.1m, therefore velocity of A with respect to O or velocity of A (because O is a fixed point),

$v_{AO}$ = $v_{A}$ = $\omega_{AO}$ $\times$ OA = $2\pi \times$ 0.1 = 1.26 m/s …(perpendicular to AO)

1.Velocity of F.

First of all draw the space diagram, to some suitable scale, as shown in fig.(a) , now the velocity diagram, as shown in fig.(b) is drawn as discussed below:

Vector oa= $v_{AO}$ = $v_{A}$ = 1.26 m/s.

By measurement, we find that velocity of F,

$v_{F}$ = vector of = 0.53 m/s

2.Velocity of sliding of CE in the trunnion

Since velocity of skidding of CE in the trunnion is the velocity of D, therefore velocity of sliding Of CE in the trunnion.

= vector od= 1.08 m/s.

3.Angular velocity of C

By measurement, we find that linear velocity of C with respect to E,

$v_{CE}$ =vector ec= 0.44 m/s

Since the length CE = 350 mm = 0.35 m, therefore angular velocity of CE,

$\omega_{CE}$ = $v_{CE}$ /CE = 0.44/0.35 = 1.26 rad/s. (Clockwise about E)