Vertical Circular Motion

Overview

Vertical Circular Motion refers to circular motion in a vertical plane, where the object moves through changing heights. Unlike many horizontal circular-motion problems, the speed usually changes because gravitational potential energy and kinetic energy interchange.

This means two ideas are commonly used together:

1. Circular-motion force analysis
2. Conservation of mechanical energy

This page extends Circular Motion and Circular Motion Force Analysis.

Why It Matters

Gravity changes both the speed and the radial force balance at different points on the path. Vertical circle problems therefore often require both a radial force equation at a point and an energy equation between points.

Definition

Vertical circular motion occurs when an object moves in a circular path in a vertical plane.

Key Representations

$$\sum F_{\text{inward}}=\frac{m{v}^2}{r}$$ $$T+mg=\frac{m{v}_{\text{top}}^2}{r}$$ $$T-mg=\frac{m{v}_{\text{bottom}}^2}{r}$$ $$KE+PE=\text{constant}$$

Why Vertical Circular Motion Is Different

In a vertical circle:

• the direction of velocity changes continuously
• the height changes continuously
• gravity may speed up or slow down the object

Therefore:

• speed is often greatest at the bottom
• speed is often smallest at the top

Common Examples

• bucket of water swung in a circle
• roller coaster loop
• mass on string moving in a vertical circle
• toy car on loop-the-loop track

Forces Acting

Typical real forces:

• weight $mg$
• tension $T$ (string problems)
• normal reaction $N$ (track problems)

Do not draw an extra centripetal force arrow.

The inward resultant force is produced by the real forces.

Radial Equation

At any point:

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

Where:

• inward means toward centre of the circle
• $v$ is speed at that point

Because inward direction changes with position, draw a fresh diagram each time.

Top and Bottom of the Circle

These are the most common exam positions.

At the Top of the Circle

The centre lies below the object, so inward direction is downward.

Example: Mass on String

Real forces downward:

• tension $T$
• weight $mg$

Hence:

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

At the Bottom of the Circle

The centre lies above the object, so inward direction is upward.

Real forces:

• tension $T$ upward
• weight $mg$ downward

Hence:

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

So:

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

This is often the largest tension.

Minimum Speed at the Top

To just maintain contact / just keep string taut:

• tension becomes zero at limiting case

So at top:

$$T=0$$

Hence:

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

Therefore:

$$v=\sqrt{gr}$$

This is the minimum speed at the top.

If speed is lower than this:

• string may slacken, or
• object loses contact with track

Why Speed Changes

Use conservation of mechanical energy if friction is negligible.

$$KE+PE=\text{constant}$$

As object rises:

• $PE$ increases
• $KE$ decreases
• speed decreases

As object descends:

• $PE$ decreases
• $KE$ increases
• speed increases

Typical Speed Relationship

For many loop problems:

• speed maximum at bottom
• speed minimum at top

Worked Example 1: Minimum Top Speed

A mass moves in a vertical circle of radius $0.80\text{ m}$.

Find minimum speed at top to keep string taut.

Solution

$$v=\sqrt{gr}=\sqrt{9.81(0.80)}=\sqrt{7.848}=2.80{\text{ m s}}^{-1}$$

Worked Example 2: Tension at Bottom

A $0.50\text{ kg}$ mass moves at $6.0{\text{ m s}}^{-1}$ at the bottom of a circle of radius $1.2\text{ m}$.

Find tension.

Solution

At bottom:

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

So:

$$T=\frac{m{v}^2}{r}+mg$$ $$T=\frac{0.50(36)}{1.2}+0.50(9.81)$$ $$T=15.0+4.91=19.9\text{ N}$$

Worked Example 3: Speed at Top from Energy

A particle moves from bottom to top of a smooth vertical circle of radius $r$.

If speed at bottom is $u$, find speed $v$ at top.

Solution

Height gained:

$$2r$$

Use conservation of energy:

$$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mg(2r)$$

So:

$$u^2=v^2+4gr$$

Hence:

$$v^2=u^2-4gr$$

Track Contact Problems

For roller coaster or car on hump:

If Just Losing Contact

Normal reaction:

$$N=0$$

Then remaining force provides centripetal requirement.

Example at top of hump:

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

Exam Strategy

Step 1

Identify position:

• top?
• bottom?
• side?

Step 2

Mark centre and inward direction.

Step 3

Draw real forces only.

Step 4

Use:

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

Step 5

Use energy equation if speed changes between positions.

Common Exam Pitfalls

1. Wrong Inward Direction

At top inward is downward.
At bottom inward is upward.

2. Forgetting Speed Changes

Do not assume constant speed unless stated.

3. Adding Centripetal Force as Extra Force

Wrong.

4. Using $v=\sqrt{gr}$ Everywhere

This is only the limiting minimum speed at top in many standard cases.

5. Wrong Height Change

Bottom to top is usually:

$$2r$$

not $r$.

Summary

Vertical circular motion combines force analysis and energy methods.

Key equations:

Radial condition

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

Bottom of circle

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

Top of circle

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

Minimum top speed

$$v=\sqrt{gr}$$

Speed is commonly:

• highest at bottom
• lowest at top

Links

• Circular Motion
• Centripetal Acceleration and Force
• Circular Motion Force Analysis
• Orbital Motion in Gravity
• Dynamics
• Work, Energy and Power