Circular motion: practice problems with step-by-step solutions
Free Circular motion practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: centripetal force, friction on a curve, vertical circle.
Free practice problemEasy
A 1600 kg car drives around a flat curve of radius 100 m at a constant 29 m/s. Find the inward (centripetal) force Fc needed.
A car in a bend, a stone on a string and a satellite around the Earth all move on a circle. Even at constant speed, their direction keeps changing.
A change of direction is a change of velocity, so the body accelerates. This acceleration points to the centre of the circle. It is called the centripetal acceleration ac.
The velocity is always along the tangent, at right angles to the radius. If the inward pull suddenly stopped, the body would fly off in a straight line along the tangent, not outwards.
Self-check A car drives around a roundabout at constant speed. In which direction does it accelerate?
Forwards, along its velocity
It does not accelerate
Outwards, away from the centre
Towards the centre of the circle
Answer: Towards the centre of the circle. The speed is constant but the direction changes, so there is an acceleration. It points to the centre. The “outward” feeling is your body trying to keep going straight.
2Centripetal acceleration and force
The centripetal acceleration depends on the speed v (m/s) and the radius r (m):
ac = v² / r
Twice the speed gives four times the acceleration. A tighter curve (smaller r) gives a larger acceleration.
By Newton’s second law, something must provide an inward force of that size:
Fc = m · v² / r
Fc is not a new force. It is the net inward force made by real forces: friction on a flat curve, tension in a string, gravity for a satellite, or the normal force on a banked road.
Spinning objects are often described by the angular speed ω in rad/s. One revolution is 2π rad, so 60 rpm is 2π rad/s.
ω = v / r = 2π / T ac = ω² · r
Self-check A cyclist rides at 10 m/s around a curve of radius 50 m. What is the centripetal acceleration?
0.2 m/s²
2 m/s²
5 m/s²
5000 m/s²
Answer: 2 m/s². ac = v²/r = 100/50 = 2 m/s². Forgetting to square the speed gives 10/50 = 0.2 m/s².
3The method step by step
Convert speeds to m/s, radii to m and rpm to rad/s.
Draw the body at the point in question and mark the direction of the centre.
Draw the real forces: weight, normal force, friction, tension.
Take “towards the centre” as positive and add the force components in that direction.
Set this net inward force equal to m·v²/r and solve.
On a flat curve, friction alone gives the inward force. The highest speed before skidding follows from μ·m·g = m·v²/r. The mass cancels:
vmax = √(μ · g · r)
4A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A car takes a flat curve of radius 40 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
1
Which force points inwards?
On a flat road, gravity and the normal force are vertical. Only friction acts sideways, towards the centre of the curve.
2
The largest friction force
On a level road N = m·g, so the friction can be at most:
Ff,max = μ · m · g
3
Set it equal to the centripetal force
μ · m · g = m · v² / r
The mass m is on both sides and cancels, so the answer is the same for a small car and a lorry.
4
Solve for the speed
v² = μ · g · r = 0.5 · 9.81 · 40 = 196.2 m²/s²
vmax = √196.2 ≈ 14.0 m/s
5
Check the answer
14.0 m/s × 3.6 ≈ 50 km/h. That is a sensible top speed for a tight bend of 40 m. A slippery road (smaller μ) would give a lower speed.
Answer
vmax ≈ 14.0 m/s, about 50 km/h.
5Vertical circles
In a vertical circle the weight sometimes points towards the centre and sometimes away from it. Look at the top and the bottom separately.
At the top, tension (or the normal force) and weight both point down, towards the centre.
At the bottom, the tension points up to the centre but the weight points away from it. The string must hold the weight and give the centripetal force.
At the top, the string just stays tight when T = 0. Then the weight alone gives the inward force: m·g = m·v²/r, so vmin = √(g·r).
If you know the speed at one point and need it at another, use energy: going up by 2r costs 2·g·r of v²/2 per kg, so vtop² = vbottom² − 4·g·r.
6Common mistakes
Drawing the centripetal force as an extra force next to friction or tension. It is the sum of the real forces.
Forgetting to square the speed in v²/r.
Using km/h or rpm directly. Convert to m/s or rad/s first.
Getting the sign of the weight wrong in a vertical circle: it helps at the top and works against the string at the bottom.
Thinking there is an outward force on the body. In the ground frame there is none.
Self-check A 0.5 kg ball on a 1 m string moves at 4 m/s at the lowest point of a vertical circle. With g = 9.81 m/s², what is the tension there?
12.9 N
3.1 N
4.9 N
8 N
Answer: 12.9 N. At the bottom T − m·g = m·v²/r, so T = 0.5 · 9.81 + 0.5 · 16/1 = 4.9 + 8 = 12.9 N. 8 N forgets the weight. 3.1 N subtracts the weight instead of adding it.
7How exam questions look
Easy: centripetal force. Find ac or Fc for a car in a curve, a stone on a string or a point on a spinning disc.
Medium: friction on a curve and related cases: the highest speed on a flat curve, a banked curve, a hump in the road, a conical pendulum or a fairground rotor.
Hard: the vertical circle. Tension at the top or bottom, the smallest speed at the top, or a loop-the-loop with energy conservation.
In every case, ask first: which real force, or which combination, points to the centre?
A 1700 kg car drives around a flat curve of radius 110 m at a constant 14 m/s. Find the inward (centripetal) force Fc needed.
m = 1700 kgv = 14 m/sr = 110 m
Find Fc(N)
Show hints
Moving in a circle at constant speed still means accelerating, towards the centre.
a = v²/r and F = m·a.
v² = 196 m²/s².
Show the answer and the step-by-step solution
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Centripetal acceleration
ac = v²/r = 14²/110 = 1.782 m/s²
Centripetal force
Fc = m·a = 1700·1.782 = 3029 N
Answer
Fc = 3029 N
Example 2Medium
A car takes a flat curve of radius 50 m. The coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
r = 50 mμs = 0.3g = 9.81 m/s²
Find vmax(m/s)
Show hints
Friction is the only inward force, and it can be at most μ·m·g.
Set m·v²/r = μ·m·g. The mass cancels.
v = √(μ·g·r).
Show the answer and the step-by-step solution
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Friction provides the centripetal force
m·v²/r = μ·m·g
v = √(μ·g·r) = √(0.3·9.81·50) = 12.13 m/s
Answer
vmax = 12.13 m/s (43.67 km/h)
Example 3Hard
A 1 kg cart is released from rest at a height of 11.9 m on a frictionless track and runs through a vertical loop of radius 3.5 m. Use g = 9.81 m/s². Find the normal force Nbottom of the track on the cart at the bottom of the loop.
m = 1 kgh = 11.9 mR = 3.5 mg = 9.81 m/s²
Find Nbottom(N)
Show hints
Energy: at the bottom v² = 2g·h.
At the bottom: N − m·g = m·v²/R.
v² = 233.5 m²/s².
Show the answer and the step-by-step solution
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Speed from energy conservation
v² = 2·9.81·11.9 = 233.5 m²/s²
Forces
Towards the centre is positive.
N = m·v²/R + m·g = 1·233.5/3.5 + 9.81 = 76.52 N
Answer
Nbottom = 76.52 N
Questions and answers
What is Circular motion?
A car in a bend, a stone on a string and a satellite around the Earth all move on a circle. Even at constant speed, their direction keeps changing.
A change of direction is a change of velocity, so the body accelerates. This acceleration points to the centre of the circle. It is called the centripetal acceleration ac.
How do I solve Circular motion problems step by step?
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
What are common mistakes in Circular motion?
Square the speed: a = v²/r.
That is v². Take the square root: v = √(μ·g·r).
Give the speed in m/s, not km/h.
What do the Circular motion exercises cover?
Easy: centripetal force
Medium: friction on a curve
Hard: vertical circle
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.