Accelerations of points on rolling wheel

Given: $r = 0.5$ m, $v_O = 6$ m/s, $a_O = 2$ m/s² (both forward, +x); rolling without slipping.

Angular motion

The wheel rolls forward, so the rotation is clockwise.

$$\omega = \frac{v_O}{r} = \frac{6}{0.5} = 12.0\ \text{rad/s} \qquad \alpha = \frac{a_O}{r} = \frac{2}{0.5} = 4.0\ \text{rad/s}^2$$

Relative acceleration equation

For any point P: $\mathbf{a}_P = \mathbf{a}_O + \mathbf{a}_{P/O}$, where $\mathbf{a}_{P/O}$ has a tangential part $\alpha r$ (perpendicular to OP, in the sense of $\alpha$) and a normal part $\omega^2 r$ directed from P toward O.

$$\omega^2 r = 12^2 \times 0.5 = 72.0\ \text{m/s}^2 \qquad \alpha r = 4 \times 0.5 = 2.0\ \text{m/s}^2$$

For a clockwise rotation, the tangential part points backward at the bottom point, forward at the top point and downward at the front point.

(a) Contact point C (directly below O)

• $a_O = 2$ m/s² forward
• Tangential: $\alpha r = 2$ m/s² backward
• Normal: $\omega^2 r = 72$ m/s² upward (toward O)

$$a_{Cx} = 2 - 2 = 0 \qquad a_{Cy} = 72.0\ \text{m/s}^2 \Rightarrow a_C = 72.0\ \text{m/s}^2\ \text{upward}$$

(b) Top point T

• Tangential: 2 m/s² forward; normal: 72 m/s² downward

$$a_{Tx} = 2 + 2 = 4.0,\quad a_{Ty} = -72.0 \Rightarrow a_T = 72.11\ \text{m/s}^2$$

directed $\tan^{-1}(72/4) = 86.8^\circ$ below the horizontal (forward).

(c) Front point F (level with O)

• Tangential: 2 m/s² downward; normal: 72 m/s² backward (toward O)

$$a_{Fx} = 2 - 72 = -70.0,\quad a_{Fy} = -2.0 \Rightarrow a_F = 70.03\ \text{m/s}^2$$

directed backward at $\tan^{-1}(2/70) = 1.6^\circ$ below the horizontal.

Answer: ω = 12 rad/s clockwise, α = 4 rad/s² clockwise; a_C = 72.0 m/s² (vertically up); a_T = 72.1 m/s² (86.8° below the horizontal); a_F = 70.0 m/s² (backward, 1.6° below the horizontal).