Animated Solution for Physics - Rotational Motion: The centre of a wheel rolling on a plane surface moves with a speed $v_0$. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt{x}{v}_0$. Then, the value of $x$ is ……… .

Imagine a wheel rolling smoothly along a flat surface without slipping. This motion might seem simple, but it is actually a beautiful superposition of two independent motions: pure translation and pure rotation.

In pure translation, every single point on the wheel moves forward with the exact same velocity as the center of mass, ${\vec{v}}_C=v_0\hat{i}$.

In pure rotation, the wheel spins around its center of mass with an angular velocity $\omega$. For a point on the rim at a distance $R$ from the center, this creates a tangential velocity of magnitude $v_{rot}=\omega R$. Because the wheel is rolling without slipping, the point touching the ground must be momentarily at rest. This gives us the crucial constraint equation: $v_0=\omega R$.

The Master Key

To find the actual velocity of any point $P$ on the wheel, we use the principle of superposition. The net velocity ${\vec{v}}_P$ is the vector sum of the translational velocity and the rotational velocity:

Let's apply this to the specific point mentioned in the problem: a particle on the rim at the same level as the centre.

Let's place the center of the wheel at the origin of our relative coordinate system. The particle $P$ is at the rightmost edge, so its position vector relative to the center is ${\vec{r}}_{P,C}=R\hat{i}$.

1. Translational Component: The center moves forward, so ${\vec{v}}_C=v_0\hat{i}$. 2. Rotational Component: The wheel rotates clockwise (assuming it moves right). The velocity of $P$ with respect to the center is tangential, pointing straight down. Mathematically, ${\vec{v}}_{P,C}=\vec{\omega }\times {\vec{r}}_{P,C}=(-\omega \hat{k})\times (R\hat{i})=-\omega R\hat{j}$. Since $\omega R=v_0$, we get ${\vec{v}}_{P,C}=-v_0\hat{j}$.

Now, we add these two perpendicular vectors:

The speed is the magnitude of this velocity vector:

The problem states the speed is $\sqrt{x}{v}_0$. Comparing the two expressions, we find:

Top vs. Side

A Note on a Common Error: Some textbooks and reference materials mistakenly calculate the speed at the top of the wheel and conclude that $x=4$. Let's see why that happens.

At the highest point of the wheel, the rotational velocity points in the exact same direction as the translational velocity (forward).

The speed at the top is $2v_0$, which can be written as $\sqrt{4}{v}_0$. However, the question explicitly asks for the speed at the same level as the centre, making $x=2$ the only physically correct answer.