A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be :

For a wheel that rolls without slipping, the centre of mass (CM) moves with linear speed $$v_{CM}$$ and the wheel simultaneously rotates with angular speed $$\omega$$ such that $$v_{CM}=R\omega$$, where $$R$$ is the radius.

The linear velocity of any point on the rim is the vector sum of the translational velocity $$v_{CM}$$ (same for every point, directed horizontally) and the tangential velocity due to rotation $$\omega R$$ (perpendicular to the radius through that point).

The highest point on the rim has its rotational velocity in the same horizontal direction as $$v_{CM}$$, so the speeds add: $$v_{\text{top}} = v_{CM}+R\omega = v_{CM}+v_{CM}=2v_{CM}$$.

Given $$v_{\text{top}} = 8 \text{ m/s}$$, we get
$$2v_{CM}=8 \implies v_{CM}=4 \text{ m/s}$$.

Consider a point on the rim that lies at the same vertical level as the centre (rightmost point if the wheel rolls to the right).
 • Translational velocity: $$4 \text{ m/s}$$ to the right.
 • Rotational (tangential) velocity: $$\omega R = v_{CM}=4 \text{ m/s}$$ vertically downward.

These two perpendicular components give the resultant speed
$$v_{\text{side}}=\sqrt{(4)^2+(4)^2}=4\sqrt{2} \text{ m/s}$$.

Therefore, the speed of the particle on the rim at the same level as the centre is $$4\sqrt{2} \text{ m/s}$$. Hence, Option A is correct.