NLM CYU 1

Physics Solution Q1 – Corrected

Solution to Question 1

Stability Analysis Method:

To check stability, imagine displacing the ball slightly from equilibrium. We compare the work done by gravity versus the work done by the string tension.

  • Let $y_1$ be the vertical height the ball rises.
  • Let $y_2$ be the length the string slackens (distance the counter-weight drops).

If $y_1 > y_2$, the system gains Potential Energy and wants to return (Stable).
If $y_1 < y_2$, the system loses Potential Energy and accelerates away (Unstable).

A: Valley STABLE Ball must rise to move. Any displacement raises COM. B: Steep Hill STABLE Slope is steeper than string. y₁ (rise) > y₂ (slack) C: Gentle Hill UNSTABLE Slope is flatter than string. y₁ (rise) < y₂ (slack)

Explanation:

  • Arrangement A (Valley): Any movement away from the bottom forces the ball to rise against gravity. The string length change is minimal compared to the rise. This is inherently Stable.
  • Arrangement B (Steep Hill): The hill is “steep”. If the ball moves down laterally, the geometry of the steep slope forces it to rise vertically ($y_1$) more than the string slackens ($y_2$). The net Potential Energy increases ($U$ increases), so the restoring force pushes it back. Stable.
  • Arrangement C (Gentle Hill): The hill is “flat”. Moving the ball laterally causes very little vertical rise ($y_1$ is small). However, moving it closer to the center slackens the string significantly ($y_2$ is large). The counterweight drops more than the ball rises. The net Potential Energy decreases ($U$ decreases), so it accelerates away. Unstable.
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