Vertical Circular Motion – Minimum Top Speed and Force Balances

Physical Setup

• Thought experiment: a stone tied to a light, inextensible string is set into vertical circular motion.
  • Low speed: stone follows a small arc, may lose tension and fall.
  • Progressively higher launch speeds: trajectory grows until a full vertical circle becomes possible.
• Central question: “Under what condition will the stone NOT fall?”
  • i.e., what is the minimum speed at the highest point so the stone keeps moving in a circle rather than the string slackening?

Forces on the Stone (Top of the Circle)

• Two real forces act when the stone is exactly at the topmost point:
  • Weight: $mg$ (always vertically downward).
  • Tension in the string: $T$ (directed toward the circle’s centre, i.e., downward at the top).
• In a rotating frame one might talk about a “centrifugal” $m\,v^{2}/r$ acting upward, but in Newtonian analysis
  • The required centripetal force is $m\,v^{2}/r$ and must point toward the centre (downward at the top).
  • Hence, $T + mg = m\,v^{2}/r$.

Numerical Illustration (Given in the Lecture)

• Example values introduced to build intuition:
  • Take $mg = 40\ \text{N}$.
  • Suppose the necessary centripetal force is $m\,v^{2}/r = 70\ \text{N}$.
  • Substituting, $T = 70 - 40 = 30\ \text{N}$
  • Positive result ⇒ string remains taut, stone stays on its circular path.
  • Students asked whether the stone would “fly away.” Conclusion: No, because the string supplies the extra 30 N inward.

Analogy: Trolley on a Vertical Loop

• Similar dynamics for a trolley (e.g., roller-coaster car) going over the top of a loop.
  • Real forces: weight $mg$ (down), normal reaction $N$ from the rails (down at the top).
  • Condition for staying on the rails: $N ≥ 0$.
  • If $m\,v^{2}/r$ just balances $mg$, then $N = 0$ (momentarily weightless state).

Reducing Speed – What Fails First?

• Decrease the speed until $m\,v^{2}/r = mg$.
  • For stone on string: $T = 0$ (string is just slack-free).
  • Any slower: $T$ would need to be negative (impossible) → string goes slack → stone falls.
• For trolley: $N = 0$ at critical speed; any slower and trolley loses contact with rail.

Minimum Speed Condition

• For a light string that cannot push:
  • Critical condition (top of circle): $T_{\text{min}} = 0$.
  • Apply centripetal requirement:
    $mg = m\,v<em>{\text{min}}^{2}/r$ ⇒ $v</em>{\text{min}} = \sqrt{r\,g}.$
• Length of string $\ell$ equals circle radius $r$. Instructor occasionally writes $\sqrt{r g}$ as $\sqrt{\ell g}$ interchangeably.

Energy-Based Perspective (sketched verbally)

• If the stone is projected upward from the bottom with speed $v<em>0$, mechanical energy conservation (neglecting air drag) gives $\tfrac{1}{2}m v</em>0^{2} = \tfrac{1}{2}m v_{\text{top}}^{2} + mg (2r).$
• Setting $v<em>{\text{top}} = v</em>{\text{min}} = \sqrt{r g}$ yields launch-speed requirement at bottom: $v_{0,\text{min}} = \sqrt{5\,r\,g}.$
  • Mentioned only implicitly; full derivation left for future problems.

Additional Instructor Remarks / Conceptual Nuggets

• “String is always perpendicular to motion” – i.e., the string (radial) and velocity (tangential) vectors are orthogonal in uniform circular motion.
• Emphasis on distinguishing real forces (weight, tension, normal) from the required net centripetal force.
• Misconception check: A larger outward (centrifugal) notion does NOT make the mass “fly off” while the string/rail still constrains it.
• Ethical/practical note (implied): Roller-coaster design must ensure $v ≥ \sqrt{r g}$ at loop tops to keep passengers safely seated.

Quick Recap – Key Equations to Memorise

• Critical (top) condition, stone on string: $T = 0.$
• Minimum speed at the highest point: $v_{\text{min}} = \sqrt{r g}.$
• General force balance at the top: $T + mg = m v^{2}/r.$
• For trolley/rail loop: $N + mg = m v^{2}/r,$ with $N ≥ 0.$

Common Exam Triggers

• Compute minimum launch speed from bottom given the loop radius.
• Qualitative: Explain why a stone falls when speed < $\sqrt{r g}$ at the top.
• Distinguish between tension becoming zero (string) vs. normal reaction vanishing (rail).