HC Verma Chapter 7 Problem 21 — Condition for circular motion in vertical plane with rod

Problem Statement

A ball is attached to a rigid rod and swung in a vertical circle. Unlike a string, the rod can push as well as pull. What is the minimum speed at the top?

Given Information

  • See problem statement for all given quantities.

Physical Concepts & Formulas

Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.

  • $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
  • $F_c = mv^2/R$ — net centripetal force needed
  • Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
  • Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\boxed{v_{min} = 0\text{ at top (rod can push inward)}}$$

$$\text{Numerical result} = \text{given expression substituted with values}$$

$$\boxed{\boxed{v_{min} = 0\text{ at top (rod can push inward)}}}$$

Answer

$$\boxed{v_{min} = 0\text{ at top (rod can push inward)}}$$

Physical Interpretation

The centripetal force is not a ‘new’ force but the net inward resultant of real forces. If that resultant falls below $mv^2/r$, the object cannot maintain circular motion and will fly outward — this is the critical condition for minimum speed problems.


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