Determine the angular velocity of the wheel and velocity of points P, Q and R on the wheel.

We assume centre O has only linear velocity.

So the instantaneous centre of rotation (1) of the wheel will the point of contact with flat surface.

Diameter of wheel $= 2m$

$\therefore $ radius-OR-OQ-OI-1m

instantaneous velocity of centre $O =4m/s \rightarrow$

$\therefore $ Angular velocity of the wheel $\omega=\dfrac {v_0}r=\dfrac 4{IO}=\dfrac 4I 4rad/s $

Now $IQ = 2$

$\therefore $ Instantaneous velocity of $Q= rw= IQ \times w = 2 \times 4=8m/s$

$$\triangle IOP, OI =OR \space text {and } \space \angle ROI =90 $$

$ \therefore IOR \space \is \space 45^\circ -45^\circ -90^\circ \space triangle \\ IR=IO\sqrt2=1\times 1.4142=1.4142 $

$\therefore $ Instantaneous velocity of $R=r\omega =IR\times \omega =1.4142\times 4=5.6569 m/s \\ In \triangle IOP, \angle IOP=90^\circ + 30^\circ =120^\circ \\ \text { by cos inerule }, IP^2 =IO^2 +OP^2 -2\times IO\times OP \times \cos \angle IOP \\ =0.6^2+1^2 -2\times 0.6 \times 1\times \cos 120 \\ \therefore IP^2=1.96 \\ \therefore IP=1.4 m$

By sine rule,

$\dfrac {OP}{\sin \angle PIO}=\dfrac {IP}{\sin \angle IOP} \\ \therefore \dfrac {0.6}{\sin \angle PIO}=\dfrac {1.4}{\sin 120} \\ \therefore \sin \angle PIO =\dfrac {0.6\sin 120}{1.4}=0.3712 \\ \therefore PIO =\sin ^{-1}(0.3712)=21.7868^\circ $

Instantaneous velocity of $P = r\omega =Ip \times \omega =1.4 \times 4 = 5.6m/s$

Hence,

$\therefore $Instantaneous velocity of $R = 5.6569m/s \rightarrow 45^\circ $

$\therefore $ Instantaneous velocity of $Q= r\omega =IQ \times \omega =2 \times 4 = 8m/s \rightarrow$

$\therefore $ Instantaneous velocity of $P = 5.6m/s \rightarrow 21.7868^\circ $