Q.The speed of the center of a wheel rolling on a horizontal surface is $v_0$. A point on the rim, in level with the center, will be moving at a speed of,

Step 1. Recall the two-part description of rolling.

Rolling without slipping is equivalent to a translation of the center of mass at speed $v_0$ combined with rotation about that same center of mass with angular speed $\omega$, where $v_0 = R\omega$ (the pure-rolling condition).

Step 2. Locate the point in question.

'A point on the rim, in level with the center' means the point on the wheel's edge that is at the same height as the axle — i.e., at the side of the wheel (the 3 o'clock or 9 o'clock position relative to the direction of travel), at horizontal distance $R$ from the center.

Step 3. Find the two velocity components at that point.

• Translational component: every point on the wheel shares the center's translational velocity, $v_0$, directed horizontally (the direction of travel).
• Rotational component: this point is at distance $R$ from the axis, so its speed due to rotation alone is $v_{rot} = R\omega = v_0$ (using $v_0=R\omega$). Because this point is at the side of the wheel (radius vector horizontal), its rotational velocity — always tangential, i.e. perpendicular to the radius — points vertically (straight up if the point is on the leading/appropriate side for the sense of rotation).

Step 4. Add the two perpendicular components as vectors.

The translational part ($v_0$, horizontal) and the rotational part ($v_0$, vertical) are perpendicular to each other, so their vector sum has magnitude

directed at 45° to the horizontal.

Step 5. Sanity-check against the other two well-known points on a rolling wheel. …

Vector addition of translational velocity v₀ and rotational velocity Rω = v₀ (perpendicular component …

• Automatically applying the bottom-point rule (components cancel, speed = 0) or the top-point rule (components add, speed = 2v₀) without noticing that at the side, the two components are perpendicular, not parallel — so they must be combined with Pythagoras, giving √2 v₀. …