A wheel of radius R = 0.5 m rolls without slipping along a…

A wheel of radius $R = 0.5\ \mathrm{m}$ rolls without slipping along a horizontal road. The velocity of its centre is $c = 2\ \mathrm{m/s}$. Find the velocities of the points on the rim that are $\pi/2$ away from the point where the wheel touches the ground.

Explanation of Method

Velocities of rim points of a rolling wheel.

Step 1: Which points?

Going $\pi/2$ (90°) around the rim from the contact point P in either direction brings us to the two ends of the horizontal diameter: the front point F and the back point B, level with the centre. (The top point is $\pi$ away from P, not $\pi/2$.)

Step 2: Angular velocity

Rolling without slipping means the contact point is momentarily at rest, so $c = \omega R$: \[\omega = \frac{c}{R} = \frac{2}{0.5} = 4\ \mathrm{rad/s}.\]

Step 3: Velocity of F and B

Each rim point has the centre's velocity $c$ (forward) plus a rotational velocity $\omega R = 2\ \mathrm{m/s}$ tangent to the rim. At F and B the tangent is vertical: downward at F, upward at B (the wheel turns clockwise when rolling to the right). The two parts are perpendicular:

\[ v_F = v_B = \sqrt{c^2 + (\omega R)^2} = \sqrt{2^2 + 2^2} = 2\sqrt2\ \mathrm{m/s}. \]

\[v_F = v_B = c\sqrt2 \approx 2.83\ \mathrm{m/s}, \ \text{at } 45^\circ \text{ to the road}\]

F moves forward and down; B moves forward and up.

Check with the instantaneous axis

At each instant the wheel turns about the contact point P, so $ v = \omega \cdot |PF| = 4 \cdot 0.5\sqrt2 \approx 2.83\ \mathrm{m/s}\ \checkmark $

Top point: $ v = \omega \cdot 2R = 4\ \mathrm{m/s} = 2c $; centre: $ \omega R = 2\ \mathrm{m/s} = c\ \checkmark $

$c = 2\ \mathrm{m/s}$ $2c = 4\ \mathrm{m/s}\ (\text{top})$ $v_F = c\sqrt2 \approx 2.83\ \mathrm{m/s}$ $v_B \approx 2.83\ \mathrm{m/s}$ $P\ (v = 0)$ $F$ $B$ $\omega$ $R\sqrt2$

The wheel rolls to the right. F and B are the rim points 90° from the contact point P. At any instant the wheel turns about P, so each velocity is perpendicular to the dashed line to P and proportional to its length $R\sqrt2$: $v = \omega R\sqrt2 = c\sqrt2$.

Result

\[\boxed{\,v_F = v_B = c\sqrt2 \approx 2.83\ \mathrm{m/s}, \ \text{at } 45^\circ \text{ to the road}\,}\]

Remark

About the original solution. It took the "points $\pi/2$ from the contact point" to be the top and the front of the wheel. The top point is actually $\pi$ away (its speed, 4 m/s, is shown above only for comparison). It also said the rotational velocity at the front point points upward; for a wheel rolling forward it points downward. The magnitude 2.83 m/s was right, and it applies to both F and B.