NBPhO 2026

Difficulty medium

Prerequisites

• Inextensible-string constraints and related rates
• Differentiating geometric relations (chain rule, implicit differentiation)
• Newton's second law in 2D with a normal-force constraint
• Unilateral contact: floor can push but not pull, so $N\ge 0$

Learning objectives

• Project a constrained body's velocity onto a rope direction to translate $\dot L$ into linear speed
• Differentiate a holonomic constraint twice; recognise that constant $\dot L$ implies $\ddot L = 0$
• Derive lift-off from the normal-force loss condition $N\to 0$, using both components of Newton's law simultaneously
• Identify the dimensionless group $v^{2}/(gH)$ as the parameter that organises the lift-off behaviour
• Use limiting-case checks ($\alpha\to 0$, $\alpha\to\pi/2$, $v\to 0$, $v\to\infty$) to validate kinematic and dynamic results

Watch out for

• The weight does not move at speed $v$. The free end of the rope moves at $v$, but the weight's velocity must be projected onto the rope, giving $u=v/\sin\alpha$ — divergent as $\alpha\to 0$.
• Even though $\dot L\ne 0$, the second derivative $\ddot L$ vanishes because $v$ is held constant. Forgetting this leaves an extra $L\ddot L$ term and produces a wrong acceleration.
• A frictionless floor still exerts a vertical normal force. Until $N$ reaches zero, the weight is constrained to $y=0$ and both Newton equations (horizontal and vertical) must be invoked at lift-off — using $T=Mg/\cos\alpha$ alone gives the wrong answer.
• Lift-off is not simply when the rope is vertical or when $T=Mg$; only the vertical component $T\cos\alpha$ supports the weight, and $T$ itself is fixed by the horizontal equation. The correct condition is $\tan\alpha_{0}=\lvert a\rvert/g$.