Proof 01: Tension in Vertical Circular Motion
For a particle of mass m moving at angular speed ω on a circle of radius r, prove that the required string tension is:
Angle convention: θ is measured from the downward vertical, so θ = 0° at the bottom and θ = 180° at the top. The result assumes the string is taut and the particle follows the prescribed circular path.
1. Apply Newton's second law radially
The acceleration of a particle following a circular path has an inward centripetal component of magnitude:
Choose the direction toward the centre as positive. Newton's second law gives:
2. Resolve the gravitational force
The tension T acts inward along the string. Weight mg always acts vertically downward. Since θ is measured from the downward vertical, the inward component of weight is −mg cos θ.
At the bottom, weight acts outward relative to the centre (−mg); at the top, it acts inward (+mg). This sign convention works for every angle.
3. Form the radial force equation
Rearranging:
4. Check three important positions
| Position | θ | Required tension |
|---|---|---|
| Bottom | 0° | T = mrω² + mg |
| Side | 90° | T = mrω² |
| Top | 180° | T = mrω² − mg |
5. What about centrifugal force?
In the inertial (laboratory) frame, centrifugal force is not an additional real force; the inward resultant produces the centripetal acceleration. In a frame rotating with the particle at constant ω, we may introduce an outward centrifugal pseudo-force of magnitude mrω². The radial balance in that rotating frame is:
This is algebraically the same result, not a different physical force law.
6. When does the string become slack?
A string can pull but cannot push, so the required tension must satisfy T ≥ 0. At the top, this means:
Important: This is the minimum angular speed at the top for a taut string. The simulator assumes constant ω, which generally requires an external drive. For a freely moving particle, ω changes with height; an energy-conservation model is needed.