Centripetal Acceleration and Force

Centripetal Acceleration and Centripetal Force

Textbook Subtopics

• Derivation of centripetal acceleration
• 4.2.1 Centripetal force
• Possible sources of centripetal force
• Friction as centripetal force
• Normal force on a banked road
• Tension as centripetal force
• Gravitational force as centripetal force
• Centrifuge and its applications

1. Learning Outcomes

After studying this topic, students should be able to:

• Explain uniform circular motion.
• Explain why an object moving with constant speed can still accelerate.
• Define centripetal acceleration.
• Derive the centripetal-acceleration formula.
• Define centripetal force.
• Calculate centripetal force using linear and angular speed.
• Identify the physical force providing centripetal force.
• Analyse motion on level and banked curves.
• Explain the role of tension and gravity in circular motion.
• Describe the working principle of a centrifuge.

2. Uniform Circular Motion

Uniform circular motion is the motion of an object along a circular path with constant speed.

Although speed remains constant, velocity changes because its direction changes continuously.

Since acceleration is the rate of change of velocity, an object moving with constant speed in a circle possesses acceleration.

This acceleration points toward the centre.

3. Definition of Centripetal Acceleration

It is represented by:

a_c

4. Direction of Velocity and Acceleration

At any point on a circular path:

• Velocity acts tangent to the circle.
• Centripetal acceleration acts toward the centre.
• Velocity and centripetal acceleration are perpendicular.

Therefore:

v⃗ ⊥ a⃗_c

Centripetal acceleration changes the direction of velocity rather than its magnitude during uniform circular motion.

5. Diagram Placement

Centripetal Acceleration and Force

Keep these formulas available as searchable HTML:

a_c = v²r = rω² F_c = mv² / r = mrω²

6. Derivation of Centripetal Acceleration

Consider a particle moving at constant speed v around a circle of radius r.

During a short time interval Δt:

• Its position vector changes by Δr.
• Its velocity changes by Δv.
• The position and velocity triangles are similar.

Therefore:

Δv / v = Δrr

Rearranging:

Δv = v Δrr

Dividing by Δt:

ΔvΔt = (vr) (ΔrΔt)

As:

Δt → 0

we have:

ΔvΔt → a_c

and:

ΔrΔt → v

Therefore:

a_c = (vr)(v) a_c = v²r

7. Angular-Speed Form

The relation between linear and angular speed is:

v = rω

Substituting into:

a_c = v²r

we get:

a_c = (rω)² / r a_c = r²ω²r

Therefore:

a_c = rω²

8. Vector Form

The radial unit vector r̂ normally points outward.

Centripetal acceleration points inward.

Therefore:

a⃗_c = −(v²r) r̂

or:

a⃗_c = −rω² r̂

The negative sign represents the inward direction.

9. Units and Dimensions

SI Unit

m s⁻²

Dimensional Formula

[a_c] = [LT⁻²]

This is the same dimensional formula as every other form of acceleration.

10. Important Relationships

a_c = v²r

Therefore:

• If speed doubles, centripetal acceleration becomes four times greater.
• If speed triples, it becomes nine times greater.
• At constant speed, increasing radius decreases centripetal acceleration.

Using:

a_c = rω²

we find that at constant angular speed, centripetal acceleration increases with radius.

11. Solved Example: Centripetal Acceleration

Question: A particle moves in a circle of radius 2 m with a speed of 6 m s⁻¹. Find its centripetal acceleration.

Given

r = 2 m v = 6 m s⁻¹

Formula

a_c = v²r

Solution

a_c = (6)² / 2 a_c = 362 a_c = 18 m s⁻²

Final Answer

a_c = 18 m s⁻² toward the centre

4.2.1 Centripetal Force

12. Definition of Centripetal Force

13. Important Clarification

Centripetal force is not a separate or new type of force.

The term describes the role performed by the resultant inward force.

Depending on the situation, it may be provided by:

• Friction
• Tension
• Gravity
• Normal contact force
• Electric force
• Magnetic force
• A combination of forces

14. Derivation of Centripetal Force

According to Newton’s second law:

F = ma

For circular motion:

a = a_c

Therefore:

F_c = ma_c

Using:

a_c = v²r

we get:

F_c = mv² / r

Using:

a_c = rω²

we get:

F_c = mrω²

15. Vector Form of Centripetal Force

Since the force is directed inward:

F⃗_c = −(mv² / r) r̂

or:

F⃗_c = −mrω² r̂

16. Unit and Dimensions

SI Unit

N

Dimensional Formula

[F_c] = [MLT⁻²]

17. Effect of Different Quantities

From:

F_c = mv² / r

Centripetal force is:

• Directly proportional to mass.
• Directly proportional to the square of speed.
• Inversely proportional to radius.

Therefore:

F_c ∝ m F_c ∝ v² F_c ∝ 1r

A small increase in speed causes a much larger increase in the required inward force.

18. Solved Example: Book-Based Car Problem

Question: The centripetal force on a car of mass 856 kg moving along a curve is 7250 N. If its speed is 12.0 m s⁻¹, find the curve’s radius.

Given

m = 856 kg F_c = 7250 N v = 12.0 m s⁻¹

Formula

F_c = mv² / r

Rearrangement

F_c r = mv² r = mv² / F_c

Solution

r = (856)(12.0)² / 7250 r = (856)1447250 r = 1232647250 r = 17.0 m

Final Answer

r ≈ 17.0 m

32. Centripetal and Tangential Quantities

Centripetal and Tangential Quantities

33. Centripetal and Centrifugal Terms

In an inertial frame, the real force required for circular motion is inward and is called centripetal force.

An apparent outward effect may be described in a rotating frame using the term centrifugal force.

For standard Class 11 inertial-frame problems:

Σ F_inward = mv² / r

Do not add an extra outward “centripetal” force.

34. Key Logic

An object naturally tends to continue along a straight-line tangent because of inertia.

A net inward force continuously changes the direction of its velocity.

Therefore:

• No inward force means no circular path.
• Greater speed requires much greater inward force.
• A smaller radius requires greater inward force.
• Centripetal force is supplied by an existing physical force.

35. Important Facts

• Constant speed does not mean constant velocity.
• Centripetal acceleration points inward.
• Tangential velocity is perpendicular to the radius.
• Centripetal force does not increase speed in uniform circular motion.
• Centripetal force is a role, not a new force.
• Friction may provide centripetal force on a level road.
• A normal-force component provides it on an ideal banked road.
• Tension provides it for some string-constrained motions.
• Gravity provides it in orbital motion.
• Mass cancels from the ideal banking-angle formula.

36. Tips and Tricks

• Draw or identify the inward radial direction first.
• Write Σ F_inward = mv²/r.
• Do not automatically write T = mv²/r if other radial forces exist.
• Square the speed carefully.
• Convert all distances into metres.
• Use radius, not diameter.
• Keep angular velocity in rad s⁻¹.
• State “toward the centre” with acceleration and force answers.
• Separate vertical and horizontal equations on a banked curve.

37. Common Mistakes

• Saying acceleration is zero because speed is constant.
• Directing centripetal acceleration outward.
• Directing velocity toward the centre.
• Treating centripetal force as an additional force.
• Using diameter instead of radius.
• Forgetting the square of velocity.
• Assuming friction always acts opposite to the vehicle’s forward motion.
• Using N = mg on every banked surface.
• Ignoring other radial forces.
• Confusing centripetal and tangential acceleration.

38. Remember Box

39. Exam Facts

• Derivation of a_c = v²r is important.
• Centripetal force is the resultant radial force.
• For a level turn, static friction can supply centripetal force.
• For an ideal banked curve: tan θ = v²r g
• For string-constrained horizontal circular motion: T = mv²/r when tension is the only radial force.
• Gravity provides centripetal force for an orbiting satellite.
• A centrifuge separates substances through rapid circular motion.

40. Short-Question Preparation

Short-Question Preparation

41. Long-Question Preparation

Derive centripetal acceleration and force.

Consider an object moving with constant speed v around a circle. Its velocity is tangent to the path. During a short interval, its velocity changes by Δv. The position-vector and velocity-vector triangles are similar. Therefore: Δv / v = Δrr Rearranging: Δv = v Δrr Divide by Δt: ΔvΔt = (vr) (ΔrΔt) As Δt approaches zero: ΔvΔt → ac Also: ΔrΔt → v Therefore: ac = v²r Using v = rω: ac = rω² Newton’s second law gives: Fc = mac Therefore: Fc = mv² / r In angular form: F_c = mrω² Both centripetal acceleration and force point toward the centre.

42. MCQs

43. Sticky Notes

Sticky Notes

• Constant speed can still mean changing velocity.
• Velocity is tangent to the circle.
• Centripetal acceleration points inward.
• a_c = v²r = rω²
• F_c = mv²/r = mrω²
• Centripetal force is the net inward force.
• Level curve: friction may provide F_c.
• Banked curve: normal-force component provides F_c.
• Orbit: gravity provides F_c.
• Use radius, not diameter.

44. 60-Second Revision

ac = v²r ac = rω² Fc = mv² / r Fc = mrω² For an ideal banked curve: tan θ = v²r g Main idea: Circular motion requires a resultant force directed toward the centre.

45. Topic Summary

Centripetal acceleration changes the direction of an object’s velocity and always points toward the centre of its circular path. The net force producing this acceleration is called centripetal force. It may be supplied by friction, tension, gravity, a normal force or another real force.

Next verified book topic: 4.3 Moment of Inertia