Acceleration analysis of slider-crank

Given: $r = OA = 0.15$ m, $l = AB = 0.6$ m, $\omega_{OA} = 20$ rad/s (constant, counter-clockwise), $\theta = 45^\circ$.

Geometry

$\sin\phi = \dfrac{0.15\sin45^\circ}{0.6} = 0.1768$, $\phi = 10.18^\circ$. Horizontal projection of AB: $d = l\cos\phi = 0.5906$ m. Vertical drop from A to B: $0.15\sin45^\circ = 0.1061$ m.

Velocity analysis

$v_A = \omega r = 20 \times 0.15 = 3$ m/s, perpendicular to OA: $\mathbf{v}_A = (-2.121,\ 2.121)$ m/s. B moves horizontally, so the vertical components give

$$\omega_{AB} = \frac{2.121}{0.5906} = 3.592\ \text{rad/s (clockwise)}$$

Acceleration analysis

Since $\omega_{OA}$ is constant, A has only a normal acceleration toward O:

$$a_A = \omega^2 r = 20^2 \times 0.15 = 60\ \text{m/s}^2,\quad \mathbf{a}_A = (-42.43,\ -42.43)\ \text{m/s}^2$$

Relative acceleration of B with respect to A has:

• normal part $\omega_{AB}^2\,l = (3.592)^2(0.6) = 7.74$ m/s² directed from B toward A,
• tangential part $\alpha_{AB}\,l$ perpendicular to AB.

Resolve $\mathbf{a}_B = \mathbf{a}_A + \mathbf{a}^n_{B/A} + \mathbf{a}^t_{B/A}$ with B having no vertical acceleration. The vector from A to B is $(d,\ -0.1061)$ m. Vertical components:

$$0 = a_{Ay} + \alpha_{AB}\,d + \omega_{AB}^2(0.1061) \text{ with } a_{Ay} = -42.43 \;\Rightarrow\; \alpha_{AB} = \frac{42.43 - \omega_{AB}^2(0.1061)}{d}$$ $$\alpha_{AB} = \frac{42.43 - (12.903)(0.1061)}{0.5906} = 69.52\ \text{rad/s}^2\ \text{(counter-clockwise)}$$

Horizontal direction:

$$a_B = a_{Ax} + \alpha_{AB}(0.1061) - \omega_{AB}^2\,d$$ $$a_B = -42.43 + (69.52)(0.1061) - (12.903)(0.5906) = -42.67\ \text{m/s}^2$$

The slider's acceleration is directed toward the crank centre O.

Check (approximate formula): $a_B \approx -\omega^2 r\left(\cos\theta + \dfrac{\cos2\theta}{n}\right)$ with $n = l/r = 4$ gives $-60(0.7071 + 0) = -42.4$ m/s², which agrees closely.

Answer: a_B = 42.67 m/s² toward O; α_AB = 69.52 rad/s² counter-clockwise.