Centripetal force — OCR A-Level Physics
Test yourself on Centripetal force with OCR A-Level practice questions.
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Centripetal force explained
When the net force on a moving object is constant in magnitude and always perpendicular to its velocity, the force changes the direction of motion but not the speed.
Read the full explanation
The object therefore travels in a circular path at constant speed. The force is called the centripetal force and points towards the centre of the circle. For example, a ball on a string whirled in a horizontal circle experiences tension towards the centre; the tension is perpendicular to the velocity at every instant, so the ball moves in a circle. If the force were removed, the ball would continue in a straight line tangent to the circle, as Newton's first law predicts. The centripetal force is not a new kind of force; it is the resultant of real forces such as tension, gravity, friction or electrostatic attraction.
(b) constant speed in a circle; v = ωr
An object moving in a circle at constant speed has a velocity that changes direction continuously, so its velocity is not constant even though its speed is. The linear speed v is related to the angular velocity ω and the radius r by v = ωr, where v is in metres per second (m s⁻¹), ω is in radians per second (rad s⁻¹) and r is in metres (m). For example, a point 0.30 m from the axis of a wheel rotating at ω = 4.0 rad s⁻¹ moves at v = 4.0 × 0.30 = 1.2 m s⁻¹. Because ω is the same for all points on a rigid body, v is proportional to r, so points further from the axis move faster in a straight-line sense. The direction of v is always tangent to the circle.
(c) centripetal acceleration; a = v²/r; a = ω²r
Centripetal acceleration is the acceleration of a body moving in a circle at constant speed. Because velocity is a vector, its direction changes continuously, so the body accelerates even though its speed is constant. This acceleration acts towards the centre of the circle, perpendicular to the instantaneous velocity. For a body moving at speed v on a circle of radius r, the magnitude is a = v²/r. Since v = ωr, substituting gives a = (ωr)²/r = ω²r, so a = ω²r. For example, a ball whirled on a 0.50 m string at 4.0 m s⁻¹ has a = (4.0)²/0.50 = 32 m s⁻², directed towards the centre. Angular speed ω is measured in rad s⁻¹, so ω²r gives m s⁻².
(d)
This row is a specification sub-heading, not an assessed statement. It introduces the practical work on circular motion that follows in 5.2.2. When reading the specification, treat (d) as a signpost telling you that the next statements describe techniques and procedures you should be able to carry out or describe, not equations to memorise. Your task is to read on to the whirling-bung investigation, identify the variables involved, and connect the practical method to the centripetal force equations F = mv²/r and F = mω²r. Use the sub-heading to organise your notes: separate the theory statements (i) from the practical statement (ii), and check that you can explain how each measurement in the method feeds into the force calculation.
(i) centripetal force; F = mv²/r; F = mω²r
Centripetal force is the resultant force that keeps a body moving in a circular path. It acts towards the centre of the circle, perpendicular to the body's velocity, and changes the direction of motion without changing speed. By Newton's second law, F = ma, and substituting the centripetal acceleration gives F = mv²/r. Since v = ωr, substituting gives F = m(ωr)²/r = mω²r, so F = mω²r. For example, a 0.20 kg mass moving at 3.0 m s⁻¹ on a 0.60 m radius needs F = 0.20 × (3.0)²/0.60 = 3.0 N towards the centre. The force is provided by a real interaction such as tension, friction or gravity, not by a new outward force.
(ii) techniques and procedures used to investigate circular motion using a whirling bung.
The whirling-bung experiment investigates the relationship between centripetal force, mass, radius and speed. A rubber bung is attached to a string threaded through a glass tube; masses hang from the lower end while the bung is whirled in a horizontal circle. The tension in the string provides the centripetal force and equals the weight of the hanging masses, so F = Mg. The radius r is measured from the tube to the centre of the bung. The period T is timed over several revolutions and averaged, giving speed v = 2πr/T. Substituting into F = mv²/r allows the relationship between force, mass, radius and speed to be tested. Safety: wear goggles, check the string, and keep clear of the whirling bung.
Your focus
- Explain why a constant net force perpendicular to velocity produces circular motion at constant speed.
- Identify the real force that acts as the centripetal force in a given situation.
- Predict the path of an object when the centripetal force is removed.
Show all 18 objectives
- Use v = ωr to calculate the linear speed of an object in circular motion.
- Explain why the speed is constant but the velocity is not in uniform circular motion.
- Relate the linear speeds of points at different radii on a rigid rotating body.
- State that centripetal acceleration is directed towards the centre of the circular path.
- Apply a = v²/r to calculate centripetal acceleration from speed and radius.
- Apply a = ω²r to calculate centripetal acceleration from angular speed and radius.
- Recognise that sub-heading (d) signposts practical content rather than defining an equation.
- Identify the practical statements that follow and the skills they require.
- Organise revision notes so theory and practical content in 5.2.2 are clearly separated.
- State that centripetal force is the resultant force directed towards the centre of the circular path.
- Apply F = mv²/r to calculate the centripetal force on a body moving in a circle.
- Apply F = mω²r and identify the real force that provides the centripetal force in a given situation.
- Describe the apparatus and procedure used to investigate circular motion with a whirling bung.
- Explain how measurements of radius, period and hanging mass are used to test F = mv²/r.
- Evaluate sources of uncertainty and describe safety precautions in the investigation.
Centripetal force exam tips
Marking Points
- A constant net force perpendicular to the velocity changes direction but not speed.
- The resulting path is circular, with the force directed towards the centre of the circle.
- The centripetal force is the resultant of real forces such as tension, gravity, friction or electrostatic attraction.
- If the net force is removed, the object moves in a straight line tangent to the circle.
- The speed remains constant because the force does no work on the object.
- In uniform circular motion the speed is constant but the velocity changes direction continuously.
- The relationship between linear speed and angular velocity is v = ωr.
- v is measured in m s⁻¹, ω in rad s⁻¹ and r in m.
- For a rigid rotating body, ω is the same at all points, so v is proportional to r.
- The velocity vector is always tangent to the circular path.
- States that centripetal acceleration is directed towards the centre of the circular path, perpendicular to the instantaneous velocity.
- Uses a = v²/r correctly, substituting speed in m s⁻¹ and radius in m, to obtain acceleration in m s⁻².
- Uses a = ω²r correctly, with ω in rad s⁻¹ and r in m, to obtain acceleration in m s⁻².
- Explains that speed is constant but velocity changes direction, so the body accelerates.
- Links the two forms through v = ωr, showing a = v²/r and a = ω²r are equivalent.
- States that centripetal force is the resultant force directed towards the centre of the circular path.
- Applies F = mv²/r correctly, using mass in kg, speed in m s⁻¹ and radius in m to obtain force in N.
- Applies F = mω²r correctly, using ω in rad s⁻¹ and r in m to obtain force in N.
- Explains that the centripetal force is provided by a real force such as tension, friction or gravitational attraction.
- Links the two forms through v = ωr, showing F = mv²/r and F = mω²r are equivalent.
- Describes the apparatus: rubber bung on a string passing through a glass or plastic tube, with masses attached to the lower end.
- Explains that the tension in the string provides the centripetal force and is equal to the weight of the hanging masses, F = Mg.
- Measures the radius from the tube to the centre of the bung, and times several revolutions to find an average period T.
- Calculates speed using v = 2πr/T and substitutes into F = mv²/r to test the relationship between force, mass, radius and speed.
- Describes a control of variables, such as keeping the bung mass or radius constant while varying the hanging mass.
- Includes safety precautions such as wearing goggles, checking the string for wear and keeping observers clear of the rotating bung.
Examiner Tips
- 💡Draw the force arrow pointing towards the centre and the velocity arrow along the tangent to show the perpendicular relationship.
- 💡Name the real force providing the centripetal force, such as tension, friction or gravitational attraction, rather than writing 'centripetal force' alone.
- 💡Use Newton's first law to justify the straight-line path when the force is removed.
- 💡Convert any diameter to a radius before substituting into v = ωr.
- 💡Check that ω is in rad s⁻¹; if you are given a frequency, first use ω = 2πf.
- 💡State that the velocity direction is tangential when explaining why the velocity is not constant.
- 💡Check whether the question gives v or ω, then choose a = v²/r or a = ω²r to avoid an unnecessary conversion step.
- 💡Convert any angular speed in rev min⁻¹ to rad s⁻¹ by multiplying by 2π and dividing by 60 before substituting.
- 💡State the direction of the acceleration as towards the centre whenever the question asks for a vector quantity.
- 💡Use specification sub-headings to structure revision notes, grouping theory statements separately from practical statements.
- 💡When a sub-heading introduces practical work, list the apparatus, measurements and safety points you would need to describe.
- 💡Link each practical measurement back to the relevant equation so you can explain why it is taken.
- 💡Identify which real force supplies the centripetal force before substituting numbers, as questions often ask for this explanation.
- 💡Choose F = mv²/r when speed is given and F = mω²r when angular speed or period is given.
- 💡Check that the final force is quoted with a direction towards the centre when the question asks for a vector.
- 💡State clearly that the hanging weight provides the tension, and therefore the centripetal force, before using F = Mg.
- 💡Describe timing several revolutions and dividing by the number of revolutions to reduce percentage uncertainty in the period.
- 💡Identify at least one source of uncertainty, such as friction at the tube or judging the radius, and suggest how to reduce it.
Common Mistakes
- Thinking a force is needed to keep the object moving: the error is believing circular motion needs an outward force; the correction is that the net force is directed inward, towards the centre.
- Confusing centripetal force with a separate type of force: the error is treating it as an extra force; the correction is that it is the resultant of real forces such as tension or gravity.
- Believing the object would fly radially outward if the force stopped: the error is expecting motion along the radius; the correction is that it moves along the tangent, in a straight line.
- Assuming the force does work and changes the speed: the error is linking a perpendicular force to a speed change; the correction is that a force perpendicular to velocity does no work, so speed stays constant.
- Saying the velocity is constant: the error is treating speed and velocity as the same; the correction is that the speed is constant but the velocity changes direction, so it is not constant.
- Using degrees in v = ωr: the error is substituting an angle in degrees for ω; the correction is that ω must be in rad s⁻¹.
- Mixing up r and diameter: the error is using the full diameter as r; the correction is that r is the radius, half the diameter.
- Assuming all points on a rotating body have the same linear speed: the error is ignoring the r dependence; the correction is that v = ωr, so points further from the axis move faster.
- Thinking a body moving at constant speed in a circle has zero acceleration; correct this by noting velocity changes direction, so acceleration is non-zero and directed towards the centre.
- Treating centripetal acceleration as directed outwards or along the tangent; correct this by stating it always points towards the centre, perpendicular to velocity.
- Using diameter in place of radius in a = v²/r; correct this by halving any diameter before substituting.
- Mixing units by using ω in degrees per second or revolutions per minute; correct this by converting to rad s⁻¹ before using a = ω²r.
- Treating sub-heading (d) as a stand-alone fact to memorise; correct this by reading it as a signpost to the practical statements that follow.
- Skipping the practical statement because it contains no equation; correct this by noting that techniques and procedures are assessable content in this section.
- Assuming the sub-heading defines a required practical with a fixed mark allocation; correct this by focusing on the skills and measurements described in the following statement.
- Inventing an outward centrifugal force as the reaction to the centripetal force; correct this by stating that the centripetal force is the resultant inward force and any outward sensation arises from inertia.
- Using diameter instead of radius in F = mv²/r; correct this by halving the diameter before substituting.
- Forgetting to square the speed or angular speed; correct this by checking the exponent in the equation before calculating.
- Using ω in rev s⁻¹ rather than rad s⁻¹; correct this by multiplying revolutions per second by 2π.
- Measuring the radius to the outer edge of the bung rather than to its centre; correct this by measuring to the centre of mass of the bung.
- Timing a single revolution, which gives a large timing uncertainty; correct this by timing many revolutions and dividing to find the average period.
- Assuming the tension equals the hanging weight without checking that the bung moves in a horizontal circle at steady speed; correct this by ensuring the motion is steady and the string is horizontal at the bung.
- Neglecting the mass of the string or friction at the tube; correct this by keeping the string light and the tube smooth, and by acknowledging these as sources of uncertainty.