Circular force balance

A car can go round a bend without changing its speed, yet its passengers certainly notice something. The direction of velocity is changing. Our job is to identify the real interaction that supplies that change, rather than add a mysterious new force to the diagram.

We will derive the required acceleration and then ask which interactions provide it. Level turns, banked roads, conical pendulums and vertical circles share a radial force balance, but their geometry and contact limits differ.

What you will learn to do

Derive v²/r from changing velocity direction.
Separate radial and tangential acceleration.
Identify real sources of the inward resultant.
Resolve banked-road and conical-pendulum forces with the correct radius.
Check local tension, friction and contact limits.

Useful before this chapterConnected bodies, sliding friction and accelerating frames →Vectors and their components →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Can acceleration be nonzero when speed stays constant?

Compare your reasoning

Yes. Velocity includes direction. In circular motion at constant speed, velocity keeps turning, so acceleration points toward the centre.

Revisit the explanation →

01 / Forces that turn a moving object

A constant speed can still mean a changing velocity

Imagine taking two photographs of a particle moving around a circle. Its velocity is tangent to the circle in each photograph. Even if the speed is unchanged, the two velocity arrows point in different directions. Their difference is nonzero, so the particle has an acceleration.

For a small angular turn Δθ, the velocity triangle has change magnitude approximately vΔθ. The travelled arc is rΔθ = vΔt, so dividing the velocity change by Δt gives v²/r in the limit. The change points toward the centre. This is the radial or centripetal acceleration.

If the speed also changes, there is a tangential acceleration dv/dt in addition to the inward radial acceleration. These components are perpendicular. The total acceleration is their vector sum, not simply v²/r and not the arithmetic sum of their magnitudes.

a_r=\frac{v^2}{r}=\omega^2r,\qquad a_t=\frac{dv}{dt},\qquad |\mathbf a|=\sqrt{a_r^2+a_t^2}

Here aᵣ denotes the inward radial magnitude for a circular path of fixed radius. The angular speed ω is v/r and uses radians per second.

WORKED EXAMPLE

Separate turning from speeding up

At one instant a particle moves at 6 m/s around a circle of radius 3 m while its speed increases at 5 m/s each second. Find the radial, tangential and total acceleration magnitudes.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: A constant speed can still mean a changing velocity
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
a_r=\frac{v^2}{r}=\omega^2r,\qquad a_t=\frac{dv}{dt},\qquad |\mathbf a|=\sqrt{a_r^2+a_t^2}
Meaning: Here aᵣ denotes the inward radial magnitude for a circular path of fixed radius. The angular speed ω is v/r and uses radians per second.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / Forces that turn a moving object

Centripetal describes the resultant, not a new interaction

Newton’s second law requires the inward component of the net real force to equal mv²/r. The force can be supplied by tension, gravity, normal reaction, friction or a combination. There is no additional force named “centripetal” to add after drawing those interactions.

Choose inward as the positive radial direction at the point you are analysing. Resolve each real force along that direction and add with signs. If outward is chosen instead, the acceleration component is −v²/r. Either convention works, but mixing them does not.

For uniform circular motion, the net tangential force is zero even though the net radial force is nonzero. For changing speed, the tangential force equals m dv/dt. Drawing radial and tangential axes at the particle keeps these two equations distinct.

\sum F_{\mathrm{inward}}=m\frac{v^2}{r},\qquad\sum F_{\mathrm{tangent}}=m\frac{dv}{dt}

These are components of one vector law. List actual interactions before resolving them.

Explain this equation

Read it in words

The inward components of real forces provide the acceleration that bends the path. Tangential components change the speed.

Symbols and units

$\sum F_{\mathrm{inward}}$: Net force component toward the instantaneous centre of curvature.Newtons (N).
$\sum F_{\mathrm{tangent}}$: Net force component along the direction of motion.Newtons (N).
$m$: Mass of the object.Kilograms (kg).
$v$: Instantaneous speed, a nonnegative scalar.m/s.
$r$: Radius of curvature, positive.Metres (m).
$\frac{dv}{dt}$: Rate at which speed changes.m/s².

Signs and operations

Σ adds signed components of forces already on the diagram. “Centripetal force” is their inward resultant, not an extra force. The v² term stays positive even when the orbit direction is reversed.

When it applies

Use an inertial frame and instantaneous radial/tangential directions. At constant speed the tangential sum is zero but the inward sum generally is not.

Make it concrete

A 0.50 kg object moving at 4 m/s on a 2 m radius needs 4 N net inward force. Doubling speed at fixed radius requires 16 N inward, while zero tangential force can still hold speed constant.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \sum F_{\mathrm{inward}}=m\frac{v^2}{r},\qquad\sum F_{\mathrm{tangent}}=m\frac{dv}{dt}
Read it as: The inward components of real forces provide the acceleration that bends the path. Tangential components change the speed.
Symbols:
\sum F_{\mathrm{inward}}: Net force component toward the instantaneous centre of curvature.; Newtons (N).
\sum F_{\mathrm{tangent}}: Net force component along the direction of motion.; Newtons (N).
m: Mass of the object.; Kilograms (kg).
v: Instantaneous speed, a nonnegative scalar.; m/s.
r: Radius of curvature, positive.; Metres (m).
\frac{dv}{dt}: Rate at which speed changes.; m/s².
Signs and operations: Σ adds signed components of forces already on the diagram. “Centripetal force” is their inward resultant, not an extra force. The v² term stays positive even when the orbit direction is reversed.
Conditions: Use an inertial frame and instantaneous radial/tangential directions. At constant speed the tangential sum is zero but the inward sum generally is not.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Static friction supplies a turn on a level road

A small vehicle follows a level circular path of radius 20 m. The tyres roll without slipping, and the available static-friction coefficient is 0.50. Neglect aerodynamic forces and take g = 10 m/s². Find the largest speed allowed by this elementary grip model.

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Compare required turning force with available grip

Inspect a plotted point: Speed (m/s) = 0

Required: 0
Static limit: 10

For a 2 kg model vehicle on the 20 m radius path, required force grows as v². The static limit is 10 N. Points above that line cannot maintain the stated radius under this grip model. Lines connect the plotted samples.

STOP & THINK

A ball moves in a horizontal circle because a string pulls inward. What forces belong on its diagram?

The string force and another inward centripetal force
The actual string force and any other real interactions
An inward string force balanced by an outward force in the inertial frame

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: Centripetal describes the resultant, not a new interaction
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
\sum F_{\mathrm{inward}}=m\frac{v^2}{r},\qquad\sum F_{\mathrm{tangent}}=m\frac{dv}{dt}
Meaning: These are components of one vector law. List actual interactions before resolving them.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

03 / Forces that turn a moving object

A tilted normal can provide horizontal acceleration

On a banked road, the normal is tilted away from vertical. Even without friction it can have an inward horizontal component. The vertical component balances weight if the car stays at constant height, while the inward component supplies radial acceleration.

Let the road bank angle be θ above the horizontal, rising outward. The normal is tilted inward by the same angle from vertical. Then N cos θ = mg and N sin θ = mv²/r. Dividing eliminates N and m, giving tan θ = v²/(rg).

This frictionless relation specifies one speed for a given radius and bank angle. At other speeds, friction may help maintain the path, with its direction determined by the tendency to slide on the bank. Do not assume it always points inward or always down the slope. The frictionless result itself is not a maximum-speed formula.

WORKED EXAMPLE

Find the speed requiring no friction on a bank

A circular road has radius 40 m and is banked so that tan θ = 0.25. Use g = 10 m/s². Find the speed at which no friction is needed.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: A tilted normal can provide horizontal acceleration
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / Forces that turn a moving object

The circle radius comes from the geometry

A conical pendulum has a mass moving in a horizontal circle while a taut string makes a fixed angle θ from vertical. The vertical acceleration is zero, so the vertical tension component balances weight. The horizontal tension component supplies radial acceleration.

If the string length is L, the radius of the circular path is L sin θ, not L. The distinction matters because the centripetal relation uses distance to the rotation axis. Write the geometry before substituting into the force equation.

Combining T cos θ = mg with T sin θ = mω²L sin θ gives ω² = g/(L cos θ) for a nonzero cone angle. The period is 2π/ω. At the exactly vertical stationary configuration, dividing by sin θ would not be valid; the formula describes the conical-motion family and its small-angle limit.

WORKED EXAMPLE

Use the horizontal radius, not the string length

A conical pendulum has string length 1.25 m, with sin θ = 0.6 and cos θ = 0.8. Take g = 10 m/s². Find the circle radius and speed.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: The circle radius comes from the geometry
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

05 / Forces that turn a moving object

The inward direction changes around a vertical circle

At the bottom of a vertical circle the centre lies above the particle. For a mass on a string, tension points inward upward while weight points outward downward, so T − mg = mv²/r. At the top, the centre lies below the particle and both tension and weight point inward, so T + mg = mv²/r.

The speeds at the top and bottom are not generally equal. Gravity changes kinetic energy as the mass rises and falls. A local radial equation uses the local speed. To relate speeds at different positions, add an energy calculation or another description of the tangential motion.

A string can pull but cannot push, so its tension magnitude must remain nonnegative. At the top, T = m(v²/r − g), requiring v² ≥ gr for the assumed taut circular motion. This is a local condition at the top. It does not alone tell you the required launch speed at the bottom; the height gain and energy transfer must also be considered.

WORKED EXAMPLE

Check whether the assumed taut circle is possible

A 0.50 kg mass passes the top of a vertical circle of radius 2 m at speed 6 m/s on a light string. Take g = 10 m/s². Find the tension there and decide whether the taut-string condition is consistent.

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STOP & THINK

A 1 kg mass passes the bottom of a vertical circle of radius 2 m at 8 m/s on a taut string. With g = 10 m/s², what is the tension?

Answer in N.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: The inward direction changes around a vertical circle
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

06 / Forces that turn a moving object

A road can push, but it cannot hold a vehicle down

At the top of a convex circular hump, inward points downward. Weight supplies the downward contribution and the normal force opposes it, so mg − N = mv²/r. As speed rises, the normal force falls. Ordinary contact is lost when it reaches zero; a negative value would require the road to pull downward.

At the bottom of a circular dip, inward points upward. The equation becomes N − mg = mv²/r, so the normal exceeds the weight. This difference explains why riders feel light over a crest and heavy in a dip, even though gravity itself has not changed.

For a general smooth curved path, the same local reasoning uses the instantaneous radius of curvature rather than a fixed global circle. The complete motion may also have a tangential acceleration. Always identify the local centre of curvature before assigning signs to force components.

WORKED EXAMPLE

A zero normal force marks loss of contact

A small vehicle crosses the top of a convex road hump with radius of curvature 10 m. Neglect downforce and take g = 10 m/s². Find the threshold speed for losing contact.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: A road can push, but it cannot hold a vehicle down
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Forces that turn a moving object

Choose the frame before naming an outward term

An inertial observer describes circular motion with an inward net force and an inward acceleration component. There is no outward force to add merely because a particle is moving in a circle. If an observer rotates with the particle, the coordinate description changes and requires inertial terms, including a centrifugal term under the relevant conditions.

These are alternative descriptions of the same motion, not two forces that must be inserted into every diagram. The translating-frame term −mA developed earlier does not by itself cover a rotating coordinate system. Until the rotating-frame terms are explicitly derived, use the inertial-frame radial and tangential equations.

A useful final check is to release the constraint mentally. If a string breaks, the particle’s immediate velocity is tangent to the former circle. It does not instantly acquire an outward radial velocity. Its subsequent path depends on the forces that remain, such as gravity.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Forces that turn a moving object
Focus: Choose the frame before naming an outward term
Learning goals:
- Derive v²/r from changing velocity direction.
- Separate radial and tangential acceleration.
- Identify real sources of the inward resultant.
- Resolve banked-road and conical-pendulum forces with the correct radius.
- Check local tension, friction and contact limits.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

Centripetal is a direction and role of the net force, not an extra interaction.
Use the instantaneous local speed in mv²/r.
Speed changes require a tangential force component too.
Geometry determines the circle radius and force projections.
A negative required normal or tension signals failure of the assumed contact or constraint.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A particle moves at 10 m/s on a circle of radius 25 m. What is the inward acceleration?

Answer in m/s².

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 2

At a fixed radius, doubling speed changes the required inward resultant by what factor?

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SELF CHECK 3

A string breaks during circular motion. What direction does the particle’s velocity have immediately after the break?

Radially outward
Tangent to the former circle
Radially inward

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 4

At the top of a vertical circle, a taut-string calculation gives T < 0. What does this imply?

The string pushes outward
The assumed taut circular motion is not possible with those conditions
Gravity points upward

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

How much inward force is required?Intermediate · Free sample · Hints and guided reasoning What speed keeps the string taut at the top?Advanced · Hints and guided reasoning Grip sets a turning speedIntermediate · Hints and guided reasoning Bank the road for a chosen speedIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

Where should you draw “centripetal force” on a force diagram?

Compare your reasoning

Draw the actual forces: tension, friction, gravity or contact forces as appropriate. Their inward resultant supplies mv²/r. Centripetal describes the role of that resultant, not an additional interaction.

Revisit the explanation →

A few days later: change the situation

At the top of a convex hill, why can a car’s normal force be less than its weight?

Compare your reasoning

The required acceleration points downward toward the centre of curvature. Gravity minus the upward normal force provides that net downward force. Losing contact begins when the required normal would otherwise become negative.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑