PhysicsMediumClass 11

Vertical Circular Motion — Minimum speed conditions

Work, Energy & Power

JEE Qs

Hard

min

Master the application of Newton's Laws and Conservation of Mechanical Energy simultaneously, paying close attention to the critical point (usually the top) where tension or normal force might become zero.

🧮 Key Formulas

E_initial = E_final => (1/2)mv_1^2 + mgh_1 = (1/2)mv_2^2 + mgh_2

F_centripetal = mv^2/R

At general point (angle θ from vertical downwards, T = tension/normal force): T - mg cos(θ) = mv^2/R

At the top point (θ=180°): T_top + mg = mv_top^2/R

At the bottom point (θ=0°): T_bottom - mg = mv_bottom^2/R

Minimum speed at the top point for completing circle (string/particle): v_top_min = sqrt(gR)

Minimum speed at the bottom point for completing circle (string/particle): v_bottom_min = sqrt(5gR)

Minimum speed at horizontal level for completing circle (string/particle): v_horizontal_min = sqrt(3gR)

✅ Key Points for JEE

1For an object attached to a string or moving along a track, the critical condition for completing a vertical circle is that the tension (or normal force) must remain non-negative (T ≥ 0 or N ≥ 0) throughout the motion, particularly at the topmost point.
2At the minimum speed required for a particle on a string to complete the circle, the tension in the string becomes zero precisely at the topmost point (T_top = 0).
3Energy conservation is the most powerful tool to relate the speed and potential energy of the object at different points in the circular path.
4Be careful to distinguish between situations involving a string (can only pull) and a rigid rod or moving inside a track (can both push and pull); the minimum speed at the top can be zero for a rod/track if merely completing the loop is required without specific tension criteria.

⚠️ Common Mistakes

✕Incorrectly applying the conservation of mechanical energy, often due to sign errors in potential energy calculations (e.g., taking 'h' from the wrong reference level).
✕Confusing the condition for a string (T_top = 0 at minimum speed) with that for a rigid rod or track (where v_top can be 0 without falling if the rod/track can support it).
✕Failing to correctly identify the components of gravitational force acting along or perpendicular to the radius at a general point, leading to errors in the centripetal force equation.
✕Assuming tension or normal force is constant throughout the motion instead of varying with speed and position.

NCERT Chapters

Class 11 Physics Ch 5: Laws of Motion
Class 11 Physics Ch 6: Work, Energy and Power