Velocities in plane motion by instantaneous centre

(a) Ladder

The velocity of A is horizontal and the velocity of B is vertical (along the wall). Draw perpendiculars to these velocities at A and B (a vertical line through A and a horizontal line through B). They meet at the instantaneous centre $I$, which is the corner of the rectangle formed by the floor, the wall and the two perpendiculars, opposite the wall-floor corner O.

   wall
   |
   B -------------- I    <- horizontal line through B
   | \              |
   |    \  ladder   |    <- vertical line through A
   |       \        |
   O------------- A ---> 2 m/s     floor

Distances from $I$: $IA = L\sin 60^\circ = 3.464$ m (equal to the height of B), $IB = L\cos 60^\circ = 2.0$ m (equal to the distance of A from the wall).

$$\omega_{AB} = \frac{v_A}{IA} = \frac{2}{3.464} = 0.577\ \text{rad/s (anticlockwise)}$$ $$v_B = \omega_{AB}\cdot IB = 0.577 \times 2.0 = 1.155\ \text{m/s (downwards)}$$

The midpoint M of the ladder is the centre of the rectangle (the diagonals OI and AB bisect each other), so $IM = L/2 = 2$ m.

$$v_M = \omega_{AB}\cdot IM = 0.577 \times 2 = 1.155\ \text{m/s}$$

directed perpendicular to $IM$.

(b) Rolling wheel

For pure rolling, the point of contact with the road is the instantaneous centre. Angular velocity:

$$\omega = \frac{v_C}{r} = \frac{6}{0.3} = 20.0\ \text{rad/s}$$

• Topmost point (distance $2r = 0.6$ m from the IC): $v_{top} = 20 \times 0.6 = 12.0$ m/s, horizontal, in the direction of travel.
• Rim point at the level of the centre (distance $r\sqrt{2}$ from the IC): $v = 20 \times 0.3\sqrt{2} = 8.485$ m/s, at $45^\circ$ to the horizontal.

Answer: (a) $\omega = 0.577$ rad/s, $v_B = 1.155$ m/s, $v_M = 1.155$ m/s; (b) top $= 12.0$ m/s, side point $= 8.485$ m/s.