Ball on a String — Vertical Circle
Whirl a ball in a vertical loop and find the speed the string needs to stay taut.
About this simulation
A ball whirled in a vertical circle needs enough speed at the top that gravity alone cannot supply more than the required centripetal force — below the critical speed the string goes slack and the circular path breaks. Students vary the launch speed v₀ and string length L, watching the tension around the loop and the make-or-break condition at the top (v_top² ≥ gL).
How to use it
- Set the Launch speed slider and watch whether the ball gets over the top or falls inside the circle.
- Read the text above the circle: it shows the speed needed to complete the loop, or the speed the ball keeps at the top once it does.
- Change the String length and notice that a longer string needs a bigger launch speed to loop.
- Switch Vectors to 'v and T' to see the tension arrow grow at the bottom and shrink near the top.
- Try the presets 'Just short of the loop' and 'Completes the loop' and watch the label at the bottom left change between 'string taut' and 'string slack'.
What to notice
- The ball needs a launch speed of at least √(5gL) at the bottom to go all the way round; while the ball falls short, the text above the circle gives this value for your string as 'needs … m/s'.
- When the launch speed is too small to loop but more than √(2gL), the string goes slack above the pivot level and the ball falls inward along a projectile path.
- With a launch speed of √(2gL) or less, the ball never rises above the pivot and simply swings like a pendulum.
- The tension arrow is longest at the bottom, where the string must support the weight as well as supply the centripetal force.
The formulas
- v₀ ≥ √(5gL)
- v₀ is the speed at the bottom, g is the acceleration due to gravity and L is the string length; this is the condition to complete the loop.
- v_top² = v₀² − 4gL
- Energy conservation between the bottom and the top, which is 2L higher; v_top is the speed at the top.
- T = m(v²/L + g cos θ)
- Tension in the string when the ball is at angle θ from the lowest point, moving at speed v; m is the mass of the ball.
Questions
What is the minimum speed at the top of a vertical circle?
At the top, both gravity and tension point towards the centre, so T + mg = mv²/L. The string can only pull, so the smallest possible tension is zero. Setting T = 0 gives v_top = √(gL). Any slower and gravity alone is more than the centripetal force needed, so the ball falls inside the circle. In the simulation, move the launch speed just below the critical value and watch the string go slack near the top.
Why is the minimum speed at the bottom of a vertical circle √(5gL)?
Going from the bottom to the top, the ball rises by 2L, so ½mv₀² = ½mv_top² + mg(2L). Putting in the smallest allowed top speed, v_top² = gL, gives v₀² = gL + 4gL = 5gL. So the launch speed at the bottom must be at least √(5gL). In the simulation, set a launch speed too low to loop and the text above the circle shows this number, as 'needs … m/s', for the string length you set.
Where is the tension maximum in vertical circular motion?
Tension is greatest at the lowest point. There the string must hold up the weight and also provide the centripetal force, so T = mv²/L + mg. At the top, gravity helps, so T = mv²/L − mg. For a ball that just completes the loop, the difference between the two is 6mg. In the simulation, turn on the tension arrow and watch it shrink as the ball climbs.
What happens if the speed is less than the critical speed in a vertical circle?
If the ball has enough speed to rise above the level of the pivot but not enough to loop, the tension drops to zero somewhere in the upper half. From that point the string is slack and the ball moves as a projectile under gravity alone, falling inside the circle until the string becomes tight again. In the simulation, try the 'Just short of the loop' preset and watch the path leave the circle.
Is the condition different for a ball on a rod instead of a string?
Yes. A rigid rod can push as well as pull, so the ball may reach the top with almost zero speed and still be held on the circle. Energy conservation then gives a minimum bottom speed of √(4gL) = 2√(gL). A string can only pull, which is why it needs the larger value √(5gL). This simulation models a string, so it shows the slack-string case.
