Circular Motion Force Analysis

Overview

Circular Motion Force Analysis focuses on how to analyse forces in circular-motion problems. Many examination errors arise not from formulas, but from drawing the wrong forces or using the phrase centripetal force incorrectly.

Key principle:

• Circular motion requires a resultant inward force
• This inward resultant is sometimes called the centripetal force
• It is not an extra physical force added to the free-body diagram

This page builds on Circular Motion and Centripetal Acceleration and Force.

Why It Matters

Most circular motion errors begin with the force diagram. The radial equation only works after the real forces and the centre of the circular path are identified.

Definition

Circular motion force analysis means drawing all real forces, resolving them into radial and tangential components, and applying Newton’s second law toward the centre.

Key Representations

$$\sum F_{\text{radial}}=\frac{m{v}^2}{r}$$ $$\sum F_{\text{radial}}=m{\omega }^{2}r$$

Core Method

For any circular-motion question:

1. Identify the object being analysed
2. Draw only real forces acting on that object
3. Choose the radial direction (toward centre usually positive)
4. Resolve forces into:
   • radial components
   • tangential components (if needed)
5. Apply radial equation:

$$\sum F_{\text{radial}}=\frac{m{v}^2}{r}$$

or

$$\sum F_{\text{radial}}=m{\omega }^{2}r$$

Real Forces vs Resultant Force

Real Forces

Examples:

• weight $mg$
• tension $T$
• normal contact force $N$
• friction $f$
• lift
• electric force

Resultant Inward Force

The vector sum of radial components gives the inward resultant force.

Important Reminder

Do not draw:

• weight
• tension
• normal force
• centripetal force

as four separate forces.

“Centripetal force” is already the resultant of the real forces.

Radial and Tangential Directions

At any instant:

Radial Direction

• toward centre or away from centre

Tangential Direction

• tangent to path
• perpendicular to radius

For uniform circular motion:

• tangential resultant force = 0
• radial resultant force provides circular motion

Choosing Sign Convention

Most convenient choice:

• inward radial direction = positive

Then:

$$\sum F_{\text{inward}}=\frac{m{v}^2}{r}$$

Be consistent if another convention is chosen.

Standard Setup 1: Object on String in Horizontal Circle

A mass moves in a horizontal circle on a string.

Possible inward force:

• tension

Hence:

$$T=\frac{m{v}^2}{r}$$

If the string is inclined (conical pendulum), resolve tension.

Standard Setup 2: Conical Pendulum

A mass moves in a horizontal circle while hanging from a string at angle $\theta$ to the vertical.

Real forces:

• tension $T$
• weight $mg$

Vertical Equilibrium

$$T\cos \theta =mg$$

Horizontal Radial Direction

$$T\sin \theta =\frac{m{v}^2}{r}$$

Standard Setup 3: Car on Flat Road

A car turns on a horizontal road.

Real forces:

• weight
• normal reaction
• friction

Vertical forces balance:

$$N=mg$$

Horizontal inward force from friction:

$$f=\frac{m{v}^2}{r}$$

If friction is insufficient, skidding occurs.

Standard Setup 4: Banked Road (No Friction)

Normal force has components.

Real forces:

• weight $mg$
• normal reaction $N$

Resolve:

Vertical

$$N\cos \theta =mg$$

Horizontal Inward

$$N\sin \theta =\frac{m{v}^2}{r}$$

Hence:

$$\tan \theta =\frac{v^2}{rg}$$

Standard Setup 5: Rotating Drum Ride

Person pressed against vertical wall of rotating cylinder.

Real forces:

• weight $mg$
• normal reaction $N$
• friction $f$

Radial Inward

$$N=\frac{m{v}^2}{r}$$

Vertical Balance

$$f=mg$$

Friction prevents sliding.

Worked Example 1: Flat Curve

A $1200\text{ kg}$ car turns on a flat road of radius $60\text{ m}$ at $18{\text{ m s}}^{-1}$.

Find friction required.

Solution

Radial inward force is friction:

$$f=\frac{m{v}^2}{r}=\frac{1200(18^2)}{60}$$ $$f=\frac{1200(324)}{60}=6480\text{ N}$$

Worked Example 2: Conical Pendulum

A bob of mass $0.50\text{ kg}$ hangs on a string making angle $30^{\circ }$ to vertical.

Find tension.

Solution

Use vertical equilibrium:

$$T\cos 30^{\circ }=mg$$ $$T=\frac{0.50(9.81)}{\cos 30^{\circ }}\approx 5.66\text{ N}$$

Worked Example 3: Banked Track Angle

A car moves at $20{\text{ m s}}^{-1}$ around radius $50\text{ m}$ on frictionless banked road.

Find angle.

Solution

$$\tan \theta =\frac{v^2}{rg}=\frac{400}{50(9.81)}=\frac{400}{490.5}=0.815$$ $$\theta \approx 39^{\circ }$$

Common Exam Pitfalls

1. Adding Centripetal Force as Extra Force

Wrong. Use only real forces.

2. Using All Forces in Radial Equation

Only radial components contribute.

3. Forgetting Vertical Equilibrium

Many horizontal-circle questions have:

$$\sum F_y=0$$

4. Wrong Centre Direction

Always identify centre before writing equations.

5. Confusing Tangential and Radial Components

Keep perpendicular directions separate.

Summary

Circular-motion force analysis is mainly about good free-body diagrams.

Use:

$$\sum F_{\text{radial}}=\frac{m{v}^2}{r}$$

The inward resultant force may come from:

• tension
• friction
• normal reaction component
• gravity
• combinations of forces

“Centripetal force” is a name for the resultant inward force, not a separate force.

Links

• Circular Motion
• Centripetal Acceleration and Force
• Vertical Circular Motion
• Orbital Motion in Gravity
• Forces
• Dynamics
• Vectors