Work, Energy and Power · Lesson 10 of 16
Energy and Forces in a Vertical Circle
“Combine energy conservation with radial dynamics to find the minimum speed for a taut-string loop.”
• Calculate speed at different heights. • Derive the top contact condition. • Distinguish a taut string, a rigid rod and a cut string.
Energy determines speed; force determines contact
A bob attached to a light string can complete a vertical circle only if the string remains taut. Energy alone says whether it can reach a height, but not whether the string can enforce the circular path there. A string can pull, never push. Therefore solve an energy equation for speed and a radial force equation for tension, then check tension is nonnegative. Treat the pivot as fixed and ignore air resistance.
Set the energy zero at the bottom
At the side, the height is L; at the top it is 2L. Tension is radial and does no work along the circle. The kinetic energy lost while rising equals the gravitational potential gained. At the same height on descent the speed is the same if no energy has been lost, even though the velocity direction differs.
Write the radial equation carefully
Reaching the top with zero speed would require only v0 = √(4gL) in an energy calculation. But at the top zero speed cannot supply the downward curvature needed for a taut-string circular path: the force equation would demand negative tension. That energy-only result is therefore not the string-loop threshold. A constrained bead on a suitable rigid track or a rigid rod has a different contact constraint; do not import the string condition without checking the apparatus.
Problem
For a string pendulum in the limiting complete loop, find speeds and the side-to-top kinetic-energy ratio.
- 1.The threshold bottom speed is √(5gL). At the side, vside² = 5gL − 2gL = 3gL.
- 2.At the top, vtop² = 5gL − 4gL = gL.
- 3.The ratio Kside/Ktop equals vside²/vtop² = 3. At the bottom, tension is mv0²/L + mg = 6mg.
- 4.At the side, tension is 3mg; at the top it reaches zero only instantaneously in this limiting solution.
What happens after the top?
At the exact limiting top condition, tension is zero at that instant, but it need not stay zero. With the string uncut, the bob can continue its circular revolution and tension becomes positive again on descent. If the string is cut at the top, its constraint is removed: the bob has horizontal tangential velocity and downward acceleration due to gravity, so it follows projectile motion. If a slower launch makes the string slack before the top, the circular formulas stop applying from that point and the subsequent motion must be analysed separately.
Problem
A bob is attached to a string of length 1 m. Take g = 10 m/s². Find limiting bottom, side and top speeds.
- 1.Bottom speed is √50 = 7.07 m/s.
- 2.Side speed is √30 = 5.48 m/s. Top speed is √10 = 3.16 m/s.
- 3.The bob does not stop at the top. Its nonzero speed is required by the radial force condition.
Check your understanding
Quiz
For a taut string, tension must satisfy
Explanation: A string can pull but cannot push.
The limiting top speed squared is
Explanation: With Ttop=0, gravity supplies mv²/L.
Minimum bottom speed for the string loop is
Explanation: Energy rise 2mgL plus top K=½mgL gives bottom K=2.5mgL.
Tension does zero work here because
Explanation: A radial force is perpendicular to the local tangent.
Cutting the string at the top produces
Explanation: The existing tangential velocity remains when the constraint is removed.
Key Takeaways
• Use both energy and radial force equations. • The threshold bottom speed is √(5gL) for a taut-string loop. • A momentarily zero tension does not by itself mean the string remains slack.