Problem 2: Velocity of Particle A
1. Geometry and Coordinate System
Let the angle between the two rods be $\theta$. Since the triangle formed by the rods and the floor is isosceles (equal rod lengths $l$), the angle each rod makes with the vertical is $\alpha = \theta/2$.
Coordinates:
Let the origin be on the floor directly below the initial position of A. Let $y_A$ be the height of A and $x_A$ be its horizontal displacement.
- Height of A: $y_A = l \cos \alpha = l \cos(\theta/2)$
- Position of B: $x_B = x_A – l \sin \alpha$
- Position of C: $x_C = x_A + l \sin \alpha$
Velocities (Differentiating with respect to time):
Let $\dot{\alpha} = \omega$. Then:
- $\dot{y}_A = -l \sin \alpha \cdot \dot{\alpha}$
- $\dot{x}_B = \dot{x}_A – l \cos \alpha \cdot \dot{\alpha}$
- $\dot{x}_C = \dot{x}_A + l \cos \alpha \cdot \dot{\alpha}$
2. Conservation of Horizontal Momentum
There are no external horizontal forces on the system (frictionless floor). Initially, the system is at rest, so the total horizontal momentum is zero.
$$m \dot{x}_A + m \dot{x}_B + 2m \dot{x}_C = 0$$Substitute the velocity expressions:
$$m \dot{x}_A + m(\dot{x}_A – l \cos \alpha \dot{\alpha}) + 2m(\dot{x}_A + l \cos \alpha \dot{\alpha}) = 0$$ $$4m \dot{x}_A + ml \cos \alpha \dot{\alpha} = 0$$ $$\dot{x}_A = -\frac{1}{4} l \cos \alpha \cdot \dot{\alpha}$$3. Conservation of Energy
The only work done is by gravity on mass A. Masses B and C move horizontally on the floor, so their potential energy is constant.
- Initial Energy (Rods vertical, $\alpha=0$): $U_i = mgl$.
- Final Energy (Angle $\theta$): $U_f = mg(l \cos \alpha)$.
- Kinetic Energy: $K = \frac{1}{2}m v_A^2 + \frac{1}{2}m v_B^2 + \frac{1}{2}(2m) v_C^2$.
Loss in PE = Gain in KE:
$$mg l (1 – \cos \alpha) = \frac{1}{2}m v_A^2 + \frac{1}{2}m v_B^2 + m v_C^2$$We need to express the velocities in terms of a common parameter, say $K = l \dot{\alpha}$.
Velocity of A:
$$v_{Ax} = -\frac{1}{4} K \cos \alpha$$ $$v_{Ay} = -K \sin \alpha$$ $$v_A^2 = v_{Ax}^2 + v_{Ay}^2 = K^2 \left( \frac{1}{16}\cos^2 \alpha + \sin^2 \alpha \right)$$Velocity of B and C:
$$v_B = \dot{x}_A – K \cos \alpha = -\frac{1}{4}K \cos \alpha – K \cos \alpha = -\frac{5}{4} K \cos \alpha$$ $$v_C = \dot{x}_A + K \cos \alpha = -\frac{1}{4}K \cos \alpha + K \cos \alpha = \frac{3}{4} K \cos \alpha$$4. Solving for Velocity
Substitute velocities into the energy equation:
$$2gl(1 – \cos \alpha) = v_A^2 + v_B^2 + 2v_C^2$$Substituting expressions in terms of $K$:
$$2gl(1 – \cos \alpha) = K^2 \left[ \left(\frac{\cos^2 \alpha}{16} + \sin^2 \alpha\right) + \frac{25 \cos^2 \alpha}{16} + 2\left(\frac{9 \cos^2 \alpha}{16}\right) \right]$$Grouping terms:
$$[…] = \sin^2 \alpha + \frac{\cos^2 \alpha}{16} (1 + 25 + 18) = \sin^2 \alpha + \frac{44 \cos^2 \alpha}{16} = \sin^2 \alpha + \frac{11}{4} \cos^2 \alpha$$ $$[…] = \frac{4 \sin^2 \alpha + 11 \cos^2 \alpha}{4}$$Now, we relate $K$ back to $v_A$. From the $v_A^2$ equation:
$$K^2 = \frac{v_A^2}{\frac{1}{16}\cos^2 \alpha + \sin^2 \alpha} = \frac{16 v_A^2}{\cos^2 \alpha + 16 \sin^2 \alpha}$$Substitute $K^2$ into the main energy balance:
$$2gl(1 – \cos \alpha) = \left( \frac{16 v_A^2}{\cos^2 \alpha + 16 \sin^2 \alpha} \right) \left( \frac{11 \cos^2 \alpha + 4 \sin^2 \alpha}{4} \right)$$ $$2gl(1 – \cos \alpha) = v_A^2 \frac{4(11 \cos^2 \alpha + 4 \sin^2 \alpha)}{\cos^2 \alpha + 16 \sin^2 \alpha}$$Solving for $v_A$ and substituting $\alpha = \theta/2$:
$$v_A = \sqrt{\frac{gl(1 – \cos(\theta/2))}{2} \cdot \frac{\cos^2(\theta/2) + 16\sin^2(\theta/2)}{11\cos^2(\theta/2) + 4\sin^2(\theta/2)}}$$