Two slender rods released, angular velocity of AB

Central and eccentric impact

Motion of the two rods

Rods $AB$ and $BD$ are identical ($L=0.75$ m, $m=6$ kg), joined by a pin at $B$. $A$ is a fixed pin and $D$ slides on a smooth horizontal surface (rollers). Released from rest at $\beta=50^\circ$, the system moves under gravity only, so use the work-energy principle.

          B
         / \
        /   \
       /     \
    A /_______\ D   (rollers)
       beta

Kinematics

Let $\omega_{AB}$ be the angular velocity of $AB$. $\vec v_B \perp AB$ and $\vec v_D$ is horizontal. The ICR $I$ of $BD$ lies on the line $AB$ extended and on the vertical through $D$. By symmetry of the triangle $ABD$, $IB = BD = L$ and $ID = 2L\sin\beta$, so

$$\omega_{BD} = \frac{v_B}{IB} = \frac{\omega_{AB}L}{L} = \omega_{AB} = \omega$$

(the two rods turn in opposite senses). The centre $G_2$ of $BD$ is at $(1.5L\cos\beta,\ 0.5L\sin\beta)$ and $I$ is at $(2L\cos\beta,\ 2L\sin\beta)$, so

$$\left(\frac{IG_2}{L}\right)^2 = 0.25\cos^2\beta + 2.25\sin^2\beta$$

At $\beta = 20^\circ$: $0.25\cos^2 20^\circ + 2.25\sin^2 20^\circ = 0.4840$, so $IG_2 = 0.5218$ m.

Kinetic energy at $\beta = 20^\circ$

$$T_{AB} = \tfrac12\left(\tfrac13mL^2\right)\omega^2 = \tfrac12(1.1250)\,\omega^2$$ $$T_{BD} = \tfrac12\left[\tfrac1{12}mL^2 + m(IG_2)^2\right]\omega^2 = \tfrac12(1.9146)\,\omega^2$$ $$T_2 = \tfrac12(1.1250 + 1.9146)\,\omega^2 = 1.5198\,\omega^2$$

Work of gravity

Heights of the mass centres: $\tfrac L2\sin\beta$ for each rod, so the total potential energy is $V = mgL\sin\beta$.

$$U_{1\to2} = mgL(\sin50^\circ - \sin20^\circ) = 6(9.81)(0.75)(0.7660 - 0.3420) = 18.72\ \text{J}$$

(a) Angular velocity of AB

$$T_1 + U_{1\to2} = T_2:\quad 0 + 18.72 = 1.5198\,\omega^2 \;\Rightarrow\; \omega_{AB} = 3.51\ \text{rad/s}$$

AB turns clockwise (B moving down and to the right).

(b) Velocity of D

$D$ is $ID = 2L\sin20^\circ$ from the ICR:

$$v_D = \omega_{BD}\,(ID) = 3.51\times2(0.75)\sin20^\circ = 3.51\times0.5130 = 1.80\ \text{m/s}\ (\to)$$

Answer: $\omega_{AB} = 3.51$ rad/s clockwise; $v_D = 1.80$ m/s horizontally to the right.