Given: $N_{AO}$ = 20 rpm. or $\omega_{AO}$ =$2\pi\times\frac{20}{60}$ = 2.1 rad/s:

OA = 300 mm=0.3m, AB= 1200 mm,=1.2 m, BC = CD = 450 mm =0.45m

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

$N_{AO}$ = $v_{A}$ = $\omega_{AO}$ × OA =2.1 × 0.3 =0.63 m/s

**1.Velocities of sliding at B and D**

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) and acceleration diagram, as shown in fig (c) is drawn as discussed below:

Vector oa =$v_{AO}$ = $v_{A}$ = 0.63 m/s

By measurement, we find that velocity of sliding at B,

$v_{B}$ = vector ob =0.4 m/s

$v_{D}$ = vector od = 0.24 m/s.

**2.Angular velocity of CD**

$v_{DC}$ = vector cd=0.37 m/s

Angular velocity of CD

$\omega_{CD}$ = $v_{DC}$/CD = 0.37/0.45 =0.82 rad/s (Anticlockwise)

**3. Linear Acceleration of D**

We know that the radial component of the acceleration of A with respect to O or acceleration of A.

$a_{AO}^r$ = $a_{A}$ = $\frac{v_{AO}^2}{OA}=\omega_{AO}^2\times OA=(2.1)^{2}\times 0.3$ = 1.323 $m/s^{2}$

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

$a_{BA}^r$ = $\frac{v_{BA}^2}{AB}=\frac{(0.54)^{2}}{1.2}$= 0.243 $m/s^{2}$ …….(By measurement , νBA =0.54 m/s)

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

$a_{DC}^r$ = $\frac{v_{DC}^2}{CD}=\frac{(0.37)^{2}}{0.45}$ = 0.304 $m/s^{2}$

Now the acceleration diagram, as shown in fig.(c) is drawn as discussed below:

Vector o'a'= $a_{AO}^r$ = $a_{A}$ =1.323 $m/s^{2}$

Vector a'x = $a_{BA}^r$ = 0.243 $m/s^{2}$

Vector c'y = $a_{DC}^r$ =0.304 $m/s^{2}$

$a_{D}$ = vector o'd' =0.16 $m/s^{2}$

**4.Angular acceleration of CD**

From the acceleration diagram, we finfd that the tangential component of the acceleration of D with respect to C,

$a_{DC}^t$ = vector yd'=1.28 $m/s^{2}$ …..(By measurement)

Angular acceleration of CD,

$a_{CD}$ = $a_{DC}^t$/CD =1.28/0.45 =2.84 $rad/s^{2}$ (Clockwise)