Circular Motion Lab

Explore Uniform Circular Motion and vertical circular paths with slack tension projectile simulation.

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Lesson

The theory — Circular Motion Lab

Something moving in a circle at constant speed is still accelerating, because acceleration is a change in velocity and velocity includes direction. That acceleration always points at the centre — which is what the word centripetal means, and it is why a circle needs a force to maintain it at all.

What each symbol means

r: the radius of the circle, 1.50 m here. It sits in both v = ωr and a_c = ω²r, so widening the circle at a fixed ω makes the object travel faster and accelerate harder — the two are not independent.
ω: the angular velocity in radians per second — how fast the angle sweeps, 2.0 rad/s. It says nothing about size on its own.
v: the tangential speed, along the circle: v = ωr = 3.00 m/s.
a_c: the centripetal acceleration, ω²r = 6.00 m/s². Notice what is not in it.
T: the period, one full turn: 2π/ω = 3.14 s, or 0.32 turns per second.

Where the formula comes from

Over a small time the velocity vector does not change length, only direction — it rotates through the same angle Δθ = ω·Δt that the object does.
Rotating a vector of length v through a small angle changes it by v·Δθ, aimed perpendicular to the velocity, which is towards the centre. So the acceleration has size v·Δθ/Δt = v·ω.
Substituting v = ωr gives the two forms the page uses: a_c = ω²r, and equivalently a_c = v²/r. At the defaults, 2.00² × 1.50 = 6.00 m/s².

Assumes: Uniform circular motion: constant speed, constant radius, and a rigid constraint holding the object on its path. Speed up or let the radius vary and a second, tangential acceleration appears that none of these formulas describes.
Breaks when: Two things worth noticing, and the second is a genuine misconception. First, a_c = ω²r contains no mass at all — change the mass box and the acceleration does not move; mass only enters when you ask for the force, F = m·a_c. Second, there is no outward force. The outward push you feel on a roundabout is your own inertia carrying you straight while the seat turns; the only real force is inward, and if it vanishes you continue along the tangent — not outwards.

Constant speed, and still accelerating the whole way round 🖖

Nothing about the ball's speed changes on this tool's first preset — it holds 3.00 m/s from start to finish — and yet the readout also shows an acceleration of 6.00 m/s² that never switches off. Both are true, because acceleration is a change in velocity, and velocity carries a direction. The speed is fixed; the direction is turning every instant, so something must be pulling. That is why "uniform" circular motion is not unaccelerated motion, and it is the single most common thing students get wrong here. The pull also grows viciously fast: a_c = ω²r, so raising the spin from 2.0 to 4.5 rad/s — only 2.25 times — multiplies the acceleration by 5.06, to 30.38 m/s². That is about 3.1g on a hammer-thrower's wire.

Centripetal force is a job, not a force 🖖

Nothing labelled centripetal ever pushes an object; the word just names whichever real force happens to point toward the centre — tension in a string, gravity for a satellite, friction for a turning car. It always aims inward, along the radius. The concrete takeaway: cut the string and the ball does not fly straight outward — it leaves along the tangent in a straight line, because no force is left to keep bending its path.

Roller-coaster loops are teardrops, not circles 🖖

A truly circular vertical loop needs a high entry speed, and whipping through the bottom at that speed would crush riders with brutal g-forces. Because the minimum speed at the very top is only √(gr) — independent of mass — engineers tighten the radius near the top and widen it at the bottom, giving the teardrop clothoid (Euler spiral) shape pioneered by Werner Stengel in 1976. Less speed needed up high, gentler g-forces down low.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

The vertical loop preset swings a 1.00 m string at v₀ = 7.3 m/s and reports success. The slack string preset uses a 1.50 m string at 6.0 m/s and fails. Type 7.3 into the slack preset’s bottom speed: does the speed that worked on the shorter string rescue the longer one?

Show answer

No — it still goes slack, top tension 0.0 N. The condition is v₀ ≥ √(5gr), which grows with the square root of the radius: 7.00 m/s at r = 1.00 m, but 8.57 m/s at r = 1.50 m. Walk the speed up on the 1.5 m loop and the verdict flips between 8.5 m/s (top tension −0.8 N, slack) and 8.6 m/s (0.3 N, taut), exactly where the formula says it should. The 5 is not a fudge factor: energy conservation gives v_top² = v₀² − 4gr, and staying taut all the way over needs gravity alone to supply the centripetal force at the top, v_top² = gr. Add those two and the 5 appears.
The centripetal force preset spins 2.0 kg on a 1.80 m arm at ω = 2.2 rad/s and reports a꜀ = 8.7 m/s² — just short of one g. What ω makes the centripetal acceleration exactly g, and what force does the panel then report?

Show answer

ω = √(g/r) = √(9.8/1.8) = 2.33 rad/s. Enter it and the panel reads a꜀ = 9.8 m/s² and F = 19.5 N, which is what the 2.0 kg mass weighs. This is the crossing where a conical pendulum hangs at 45°: the sideways pull it needs has grown to match the downward pull it already had. Note that you have to bring g yourself — the panel prints the acceleration and the force, never the weight, so the comparison is yours to make.

Problem solved in full

1 kg ball on a 1 m string swung in a vertical circle 5 steps
A 1 kg ball on a 1 m string, swung in a vertical circle and passing the bottom at 7.3 m/s. Find whether the string stays taut over the top, and what it has to survive at the bottom. This is Vertical Loop with r = 1.0 m, m = 1.0 kg, g = 9.8 m/s² and v₀ = 7.3 m/s.
1. At the top of the circle the centre is directly below, so gravity and the string pull the same way. A string can pull but never push, so the limiting case is the one where it contributes nothing and gravity alone has to supply the whole centripetal requirement.
  
  $T_{top} + m g = \frac{m v _{top}^{2}}{r}, T_{top} \geq 0 ⟹ v_{top, min} = g r = 9.8 \times 1.0 = 3.13 m/s$
2. The speed at the top comes from energy, not from forces. Climbing 2r is paid for out of kinetic energy, and combining that with the condition from step 1 gives a minimum bottom speed of √5 times the minimum top speed.
  
  $\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v_{top}^{2} + m g (2 r) ⟹ v_{0}^{2} = v_{top}^{2} + 4 g r, v_{0, m i n} = 5 g r = 49.0 = 7.00 m/s$
3. So 7.3 m/s clears the 7.00 m/s minimum by 4.3%. In v² that surplus is 4.29, and the climb subtracts a fixed 4gr rather than a fraction, so the whole 4.29 survives to the top — where it sits against a requirement of only 9.8, and is therefore worth 44% there.
  
  $v_{0}^{2} - v_{0, m i n}^{2} = 53.29 - 49.00 = 4.29; v_{top}^{2} = 53.29 - 39.20 = 14.09 ⟹ v_{top} = 3.75 m/s$
4. Now the two tensions. At the top gravity points inward and pays part of the bill, so the string covers the shortfall — and that shortfall is the 4.29 from step 3, because m/r is 1 here. At the bottom gravity points outward, so the string carries the requirement plus the full weight.
  
  $T_{top} = \frac{m v _{top}^{2}}{r} - m g = 14.09 - 9.80 = 4.29 N, T_{bot} = \frac{m v _{0}^{2}}{r} + m g = 53.29 + 9.80 = 63.09 N$
5. Subtract them. The two v² terms differ by exactly 4gr, the two weight terms add instead of cancelling, and every trace of the speed disappears.
  
  $T_{bot} - T_{top} = \frac{m ( v _{0}^{2} - v _{top}^{2} )}{r} + 2 m g = 4 m g + 2 m g = 6 m g = 58.8 N$
Answer

The tool prints 63.1 N at the bottom, 4.3 N at the top, and a top-of-loop minimum speed of 3.13 m/s. Step 5 is the one worth keeping: the gap between the two tensions is 6mg — 58.8 N here — and it does not depend on the speed at all, so swinging faster raises both together and never narrows it. Run at the bare minimum, the top tension is zero and the bottom tension is therefore exactly 6mg, which for a rider in a circular rollercoaster loop is 6 g of apparent weight. That is past what a person can take sitting upright, and it is why no modern loop is circular: a clothoid tightens the radius at the top, where a small radius means a small required speed, and opens it out at the bottom, where v²/r is what presses on the passenger.

$T_{bot} - T_{top} = 6 m g \forall v_{0}$