Question 18: Constrained Motion & Energy
Step 1: Kinematics Relationship
Let $y$ be the drop of mass C and $x$ be the distance of sleeves from the center.
Relationship: $v_C = u_A \cot \theta$.
Step 2: Optimization
Using conservation of energy, we derived:
$$ v_C^2 = gL \sin \theta \cos^2 \theta $$
Maximizing this gives $\tan \theta = 1/\sqrt{2}$.
This corresponds to:
- Height: $h = \frac{L}{2} \sin \theta = 1.5$ m.
- Max Speed: $v_C = 2\sqrt{5}$ m/s.
- Net Force: Zero at max speed (acceleration is zero).
Answer: (b), (c), (d)