The Crank OA Is Rotating In The Counter-Clockwise Direction At a Speed Of $180 rpm$. Increasing At The Rate Of $50 Rad/S^2$.

The Dimension Of The Various Links Are As Follows: OA=180 mm; CB=240 mm; AB = 360 mm; And BD=540 mm.

For The Given Configuration, Find -

1. Velocity Of Slider D And Angular Velocity Of BD, and

2. Acceleration Of Slider D and Angular Acceleration Of BD.

Subject: Kinematics of Machinery

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

Difficulty: Medium

Given: $N_{AO}$ = 180 RPM. OR $\omega_{AO}$ = $2\pi \times\frac{120}{60}$ = 18.85 rad/s, OA= 180 mm = 0.18m;

CB = 240 mm =0.24 m, AB= 360 mm =0.36 m, BD = 540 mm =0.54m.

We know that velocity if A with respect to O or velocity of A,

$v_{AO}$ = $v_{A}$ = $\omega_{AO}$ $\times$ OA

= 18.85 $\times$ 0.18 = 3.4 m/s

Vector oa= $v_{AO}$ = $v_{A}$ = 3.4 m/s.

By measurement, we find that velocity of B with respect to A,

$v_{AB}$ = vector ab = 0.9 m/s.

velocity of B with respect to C,

$v_{BC}$ = vector cb = 2.8 m/s.

velocity of D with respect to B,

$v_{DB}$ = vector bd = 2.4 m/s.

and velocity of slider D $v_{D}$ = vector cd = 2.05 m/s.

Angular velocity of BD

We know that the angular velocity of BD,

$\omega_{BD}$ = $v_{DB}$ / BD = 2.4/0.54 = 4.5 m/s.

2. Acceleration of slider D and angular acceleration of BD

Since the angular acceleration of OA increases at the rate of 50 $rad/s^{2}$ , i.e. $\alpha_{AO}$ = 50 $rad/s^{2}$,

Therefore,

Tangential component of the acceleration of A with respect to O,

$a_{AO}^t$ = $\alpha_{AO}\times$ OA = 50 $\times$ 0.18 = 9 $m/s^{2}$

Radial component of the acceleration of A with respect to O,

$a_{AO}^r$ = $v_{AO}^2$ / OA =$(3.4)^2$ /0.18 = 63.9 $m/s^{2}$

Radial component of the acceleration of B with respect to A,

$a_{BA}^r$ = $v_{BA}^2$ / AB =$(0.9)^2$ /0.36 = 2.25 $m/s^{2}$

Radial component of the acceleration of B with respect to C,

$a_{BC}^r$ = $v_{BC}^2$ / CB =$(2.8)^2$ / 0.24 = 32.5 $m/s^{2}$

Radial component of the acceleration of D with respect to B,

$a_{DB}^r$ = $v_{DB}^2$ / BD =$(2.4)^2$ /0.54 = 10.8 $m/s^{2}$

vector xa' = $a_{AO}^t$ = 9 $m/s^{2}$

vector a'y = $a_{BA}^r$ =2.25 $m/s^{2}$

vector c'z = $a_{BC}^r$ = 32.5 $m/s^{2}$

vector b's = $a_{DB}^r$ = 10.8 $m/s^{2}$

By measurement , we find that acceleration of slider D,

$a_{D}$ = vector c'd' = 13.3 $m/s^{2}$

Angular acceleration of BD

By measurement , we find that tangential component of the acceleration of D with respect to B,

$a_{DB}^t$= vector sd' = 38.5 $m/s^{2}$

we know that angular acceleration of BD,

$\alpha_{BD}$ = $a_{DB}^t$/BD = 38.5/0.54 = 71.3 $rad/s^{2}$ …(Clockwise)