Vertical Circular Motion Revision Notes for AQA A-Level Further Maths

Vertical Circular Motion

Introduction

When an object moves in a circular path in a vertical plane, we call this vertical circular motion. Unlike horizontal circular motion where speed remains constant, in vertical circular motion both the angular velocity (ω) and tangential velocity (v) vary continuously as the object moves around the circle. This variation occurs because gravity acts on the object throughout its motion, causing it to speed up when moving downwards and slow down when moving upwards.

The varying velocity means that the acceleration also changes at different points around the circle, making vertical circular motion more complex than its horizontal counterpart. This type of motion appears in many real situations, such as a ball on a string being swung in a vertical circle, or vehicles travelling around vertical loops.

infoNote

The key difference between horizontal and vertical circular motion is that in vertical motion, gravity continuously affects the particle's speed, causing both velocity and acceleration to vary throughout the motion.

Acceleration components in vertical circles

When a particle moves in a vertical circle, its acceleration has two distinct components that describe different aspects of the motion.

Radial component

The radial component of acceleration always points towards the centre of the circle. This component is responsible for changing the direction of the velocity vector, keeping the particle moving in a circular path. For a circle of radius $r$ with the particle moving at speed $v$ , the magnitude of the radial acceleration is:

$a_{\text{radial}} = r\omega^2 = r\dot{\theta}^2 = \frac{v^2}{r}$

where $\omega$ represents the angular velocity and $\dot{\theta}$ is used as shorthand notation for $\frac{d\theta}{dt}$ .

This radial acceleration is essential for circular motion and must be produced by a resultant force directed towards the centre. In vertical circular motion, this resultant force comes from a combination of tension (or normal reaction) and the component of weight acting along the radius.

Tangential component

The tangential component of acceleration acts along the tangent to the circle. This component is responsible for changing the speed of the particle. The magnitude of tangential acceleration is:

$a_{\text{tangential}} = r\frac{dv}{dt} = r\frac{d\omega}{dt} = r\ddot{\theta}$

where $\ddot{\theta}$ is shorthand for $\frac{d^2\theta}{dt^2}$ .

chatImportant

In most problems involving vertical circular motion, you won't need to use the tangential component explicitly. The energy equation and Newton's second law applied to the radial direction are sufficient to solve the problems you'll encounter.

Energy considerations in vertical circular motion

As a particle moves around a vertical circle under gravity, there is a continuous exchange between gravitational potential energy (GPE) and kinetic energy (KE). Understanding this energy transformation is crucial for solving vertical circular motion problems.

Energy transformations

Consider a particle moving smoothly in a vertical circle. As it travels from the bottom of the circle towards the top:

It gains GPE as it rises higher
It loses KE as it slows down

Conversely, as it moves from the top back down towards the bottom:

It loses GPE as it descends
It gains KE as it speeds up

At the lowest point of the circle, the particle has minimum GPE and maximum KE. At the highest point, it has maximum GPE and minimum KE.

infoNote

Energy Cycle in Vertical Circles:

Think of the energy transformation as a continuous cycle: At the bottom, the particle has maximum speed (maximum KE) and minimum height (minimum GPE). As it rises, it trades kinetic energy for potential energy. At the top, it has minimum speed (minimum KE) and maximum height (maximum GPE). Then the cycle reverses as it descends.

The energy equation method

The energy equation is a powerful tool for analysing vertical circular motion. This equation can be stated in two equivalent ways:

The total mechanical energy remains constant (if there are no external forces doing work)
The gain in one form of energy equals the loss in the other form

Mathematically, the energy equation can be written as:

$\text{KE lost} = \text{GPE gained}$

$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mg(h_2 - h_1)$

where $u$ is the initial velocity, $v$ is the final velocity, and $(h_2 - h_1)$ is the change in height.

chatImportant

When to Use the Energy Equation:

The energy equation is particularly useful because it involves velocities and heights but not forces. This makes it the preferred method for finding velocities at different points in the circle. Use energy equations to find speeds, and Newton's second law to find forces like tension or normal reaction.

Newton's second law approach

While the energy equation helps us find velocities, we need Newton's second law when we want to determine forces such as tension in a string or normal reaction from a surface.

For a particle moving in a vertical circle of radius $r$ with velocity $v$ at any point, the radial component of acceleration is $\frac{v^2}{r}$ . Applying Newton's second law in the direction towards the centre of the circle:

$\text{Resultant force towards centre} = m \times \frac{v^2}{r}$

The resultant force is the combination of:

The tension $T$ (or normal reaction $R$ ) acting along the radius
The component of weight $mg$ acting along the radius

The sign and magnitude of these forces depend on the position of the particle in the circle.

infoNote

At different positions around the circle, you must carefully identify which forces contribute to the centripetal force. Draw a clear diagram showing all forces and their components to avoid errors.

Complete circles vs incomplete motion

A crucial question in vertical circular motion problems is whether the particle will complete a full circle or only oscillate in the lower part of the circle. The answer depends on whether the particle has sufficient energy.

High-energy particles

Particles with sufficient initial energy can complete full rotations around the circle. To do this, they must reach the highest point with enough velocity to maintain the required centripetal acceleration.

Low-energy particles

Particles with insufficient energy will only oscillate in the lower part of the circle. They rise to a certain maximum height, then fall back down, never completing the full circle.

The critical factor determining which behaviour occurs is whether the particle can satisfy the conditions for remaining in circular motion when it reaches the upper part of the circle. These conditions differ depending on whether the particle is attached to a light rod, a string, or is moving on a surface.

Different types of constraints

The type of connection between the particle and the circular path significantly affects the conditions for motion. Understanding these differences is essential for solving problems correctly.

chatImportant

Key Difference Between Constraints:

Light rod: Can PUSH and PULL
String: Can only PULL (never push)
Surface: Can only PUSH perpendicular to surface (via normal reaction)

This fundamental difference means each constraint requires different conditions for complete circular motion.

Light rod

When a particle is attached to a fixed point by a light rod:

The rod can push (provide thrust) or pull (provide tension)
Circular motion breaks when the particle's velocity reaches zero
At this point, the rod comes to instantaneous rest
The particle will then oscillate on a circular arc but won't complete the full circle

For a particle on a light rod to complete a full circle, it must reach the topmost point with a speed $v \geq 0$ . Using energy conservation from the lowest point (speed $u$ ) to the highest point (speed $v$ ):

$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mg(2r)$

For a complete circle, $v \geq 0$ , which gives:

$u^2 \geq 4gr$

Therefore, the minimum speed for a complete circle with a light rod is:

$u_{\text{min}} = \sqrt{4gr} = 2\sqrt{gr}$

String

When a particle is attached to a fixed point by a string:

The string can only pull (provide tension), never push
Circular motion breaks when the tension in the string becomes zero and the string goes slack
The particle then falls under gravity alone

For circular motion to continue, the tension must remain positive. At the topmost point $B$ , applying Newton's second law towards the centre:

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

For the string to remain taut, $T \geq 0$ , which means:

$\frac{mv^2}{r} \geq mg$

$v^2 \geq gr$

Using energy conservation from the lowest point to the highest point:

$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = 2mgr$

Combining these conditions: $u^2 - v^2 = 4gr$ and $v^2 \geq gr$ gives:

$u^2 \geq 5gr$

Therefore, the minimum speed for a complete circle with a string is:

$u_{\text{min}} = \sqrt{5gr}$

infoNote

Why Strings Need Higher Speed:

A string requires a higher minimum speed than a light rod ( $\sqrt{5gr}$ vs $\sqrt{4gr}$ ) because the particle must maintain sufficient speed at the top to keep the string taut. Since the string cannot push, the tension alone must provide the required centripetal force, which demands a higher velocity at the top of the circle.

Surface (inside or outside)

When a particle moves on a circular surface:

The surface provides a normal reaction force
The normal reaction can only push perpendicular to the surface, not pull
Circular motion breaks when the normal reaction becomes zero
The particle then leaves the surface and follows a parabolic path under gravity

For a particle on the inside of a circular surface, contact is lost when the normal reaction $R = 0$ , which typically occurs in the upper half of the circle. For a particle on the outside of a surface, contact is lost when $R = 0$ in a similar manner.

Problem-solving strategy

To solve problems involving vertical circular motion effectively, follow this systematic approach:

Step 1: Draw a clear diagram

Start by drawing a clear diagram showing:

The circular path with centre marked
The position of the particle
All forces acting on the particle (weight, tension/reaction)
The angle or position clearly labelled
The radius of the circle

infoNote

A well-drawn diagram is crucial for identifying all forces correctly and setting up the right equations. Take time to label all components clearly, including angles, heights, and force directions.

Step 2: State the condition

Identify what condition applies for the situation described:

For completing a circle with a rod: velocity at top $v \geq 0$
For completing a circle with a string: tension at top $T \geq 0$ (or $v^2 \geq gr$ )
For losing contact with a surface: normal reaction $R = 0$

Step 3: Write appropriate equations

Depending on what you need to find, use:

Energy equation to find velocities at different heights
Newton's 2nd law ( $F = \frac{mv^2}{r}$ towards centre) to find forces

Often you'll need to use both equations together: first find the velocity using energy, then substitute into the force equation to find the tension or reaction.

Worked examples

lightbulbExample

Example 1: Finding velocity and tension in a string

A 2 kg mass $P$ is attached to a string $OP$ of length 1.2 m where $O$ is a fixed point. $P$ is given a horizontal velocity $u = 4 \text{ m s}^{-1}$ when hanging vertically below $O$ at point $Q$ . When $\angle QOP = 60°$ , find: a) The velocity $v$ of $P$ b) The tension $T$ in the string

Solution:

Part a) Let $P$ have velocity $v$ when $\angle QOP = 60°$ .

First, find the vertical height of $P$ above $Q$ :

$h = OQ - OP\cos 60° = 1.2 - 1.2\cos 60°$

$h = 1.2(1 - \cos 60°) = 1.2(1 - 0.5) = 0.6 \text{ m}$

Using the energy equation from $Q$ to $P$ :

$\text{KE lost} = \text{GPE gained}$

$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgh$

$\frac{1}{2} \times 2 \times 4^2 - \frac{1}{2} \times 2 \times v^2 = 2 \times 9.8 \times 0.6$

$16 - v^2 = 11.76$

$v^2 = 4.24$

$v = 2.06 \text{ m s}^{-1}$

Part b) Apply the equation of motion along the radius towards $O$ :

$T - mg\cos 60° = m \times \frac{v^2}{r}$

$T = m\frac{v^2}{r} + mg\cos 60°$

$T = 2 \times \frac{4.24}{1.2} + 2 \times 9.8 \times \cos 60°$

$T = 2 \times \frac{4.24}{1.2} + 2 \times 9.8 \times 0.5$

$T = 16.9 \text{ N}$

Key points:

Finding vertical height as $r(1 - \cos\theta)$ is very common in these problems
Energy is conserved since there's no friction or external forces
The resultant force equals mass times acceleration

lightbulbExample

Example 2: Minimum speed for complete circle with a light rod

A particle $P$ of mass $m$ at point $A$ hangs vertically from a fixed point $O$ where $OA$ is a light rod of length $r = 1.5$ m. $P$ is given a horizontal speed $u$ at $A$ . Find the least value of $u$ for $P$ to perform a complete circle about $O$ .

Solution:

For $P$ to complete a circle with a light rod, it must reach the topmost point $B$ with a speed $v \geq 0$ .

The rod can push or pull, so the critical condition is simply that the velocity doesn't become zero before reaching the top.

Gain in GPE from $A$ to $B$ is $m \times g \times 2r = 2mgr$

Loss of KE from $A$ to $B$ is $\frac{1}{2}mu^2 - \frac{1}{2}mv^2$

Using the energy equation:

$2mgr = \frac{1}{2}mu^2 - \frac{1}{2}mv^2$

Simplifying:

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

For a complete circle, $v \geq 0$ , so $v^2 \geq 0$ :

$u^2 - 4gr \geq 0$

$u^2 \geq 4gr$

Minimum speed: $u = \sqrt{4 \times 9.8 \times 1.5} = \sqrt{58.8} = 7.67 \text{ m s}^{-1}$

lightbulbExample

Example 3: Minimum speed for complete circle with a string

The light rod $OA$ in Example 2 is replaced by a string of the same length. Find the new least value of $u$ for $P$ to perform a complete circle about $O$ .

Solution:

With a string, $P$ completes a circle if the string is still taut at $B$ ; that is, if the tension $T \geq 0$ at $B$ .

The energy equation is unchanged: $v^2 = u^2 - 4gr$ ... [1]

At point $B$ (the top), applying Newton's 2nd law towards $O$ :

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

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

For the string to remain taut, $T \geq 0$ :

$\frac{mv^2}{r} \geq mg$

$v^2 \geq gr$ ... [2]

From equations [1] and [2]:

$u^2 - 4gr \geq gr$

$u^2 \geq 5gr$

Minimum speed: $u = \sqrt{5 \times 9.8 \times 1.5} = \sqrt{73.5} = 8.57 \text{ m s}^{-1}$

Important: The minimum speed for a string (8.57 m s⁻¹) is greater than for a light rod (7.67 m s⁻¹) because the string cannot push, only pull. The particle must have sufficient speed at the top to create the required centripetal force through tension alone.

lightbulbExample

Example 4: Particle losing contact with a sphere

A particle $P$ of mass $m$ is at the lowest point $A$ on the inside of a thin, smooth sphere of radius 1.2 m with centre $O$ . $P$ is projected along the surface from $A$ with a velocity $u = 6 \text{ m s}^{-1}$ . Find $\angle AOP$ when $P$ is about to lose contact with the sphere.

Solution:

Let the normal reaction be $R$ when the velocity is $v$ .

Let $\theta = \angle BOP$ where $B$ is the topmost point.

The particle loses contact with the surface when $R = 0$ , which occurs when $P$ is in the upper half of the circle.

Using the energy equation from $A$ to $P$ :

$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mg(r + r\cos\theta)$

$u^2 - v^2 = 2gr + 2gr\cos\theta$ ... [1]

Applying Newton's 2nd law towards $O$ :

$R + mg\cos\theta = m\frac{v^2}{r}$

The particle leaves the circle when $R = 0$ :

$v^2 = gr\cos\theta$ ... [2]

From [1] and [2]:

$u^2 - gr\cos\theta = 2gr + 2gr\cos\theta$

$u^2 = 2gr + 3gr\cos\theta$

$\cos\theta = \frac{u^2 - 2gr}{3gr} = \frac{36}{3 \times 9.8 \times 1.2} - \frac{2}{3}$

$\cos\theta = \frac{36}{35.28} - \frac{2}{3} = 0.3537$

$\theta = 111°$

Therefore, $\angle AOP = 180° - \theta = 180° - 111° = 69°$

The particle leaves the sphere when $\angle AOP = 69°$ (or equivalently when it has travelled $180° - 69° = 111°$ from the lowest point).

lightbulbExample

Example 5: Particle leaving a sphere and becoming a projectile

A particle $P$ of mass 4 kg rests at the topmost point $A$ of the surface of a smooth sphere of radius 1.5 m, fixed to a horizontal plane at point $B$ . $P$ is slightly disturbed from rest and moves down the surface of the sphere before leaving it at $C$ . If $P$ strikes the plane at $D$ , find the distance $BD$ .

Solution:

Let the normal reaction be $R$ and angle $\angle AOC$ be $\theta$ when particle $P$ is at $C$ .

The particle leaves the sphere when $R = 0$ .

Using the energy equation whilst $P$ is on the sphere:

$\text{KE gained} = \text{GPE lost}$

$\frac{1}{2}mv^2 = mg(r - r\cos\theta)$

$\frac{1}{2}mv^2 = mgr(1 - \cos\theta)$ ... [1]

Applying Newton's 2nd law to the centre $O$ :

$mg\cos\theta - R = m\frac{v^2}{r}$

The particle leaves the sphere when $R = 0$ :

$v^2 = gr\cos\theta$ ... [2]

From [1] and [2]:

$\frac{1}{2}mgr\cos\theta = mgr - mgr\cos\theta$

$\frac{1}{2}mgr\cos\theta = mgr(1 - \cos\theta)$

$\frac{1}{2}\cos\theta = 1 - \cos\theta$

$\frac{3}{2}\cos\theta = 1$

$\cos\theta = \frac{2}{3}$

So $\theta = 48.2°$

From equation [2]:

$v^2 = \frac{2gr}{3} = \frac{2 \times 9.8 \times 1.5}{3} = 9.8$

$v = 3.13 \text{ m s}^{-1}$

The particle leaves the sphere at a distance $r + r\cos\theta = \frac{5r}{3} = 2.5$ m above the plane.

At point $C$ , the velocity has components:

Vertical: $v\sin\theta = 3.13 \times \sin 48.2° = 2.33 \text{ m s}^{-1}$ (downwards)
Horizontal: $v\cos\theta = 3.13 \times \cos 48.2° = 2.09 \text{ m s}^{-1}$

Now $P$ becomes a projectile. Using the equation $s = ut + \frac{1}{2}at^2$ for vertical motion (taking downwards as positive):

$2.5 = 2.33t + \frac{1}{2} \times 9.8 \times t^2$

$4.9t^2 + 2.33t - 2.5 = 0$

Using the quadratic formula (taking the positive root):

$t = 0.515 \text{ s}$

Horizontal distance travelled in this time:

$v\cos\theta \times t = 2.09 \times 0.515 = 1.08 \text{ m}$

Distance $BD$ = distance $BC$ + horizontal distance $CD$

$BD = 1.5\sin 48.2° + 1.08 = 1.12 + 1.08 = 2.20 \text{ m}$

Key points:

State the correct condition for completing a full circle or breaking away from the circle
Use energy equations for finding velocities
Use Newton's 2nd law for finding forces
When the particle leaves the circular path, treat it as projectile motion

bookmarkSummary

Key Points to Remember:

In vertical circular motion, both angular velocity (ω) and tangential velocity (v) vary due to gravity, unlike horizontal circular motion where they remain constant.
The radial acceleration is $\frac{v^2}{r}$ and the resultant force towards the centre equals $m\frac{v^2}{r}$ by Newton's second law.
Use the energy equation (KE lost = GPE gained) to find velocities at different heights, and Newton's 2nd law to find forces such as tension or normal reaction.
For complete circles: a light rod requires minimum speed $u_{\min} = \sqrt{4gr}$ , while a string requires the higher minimum speed $u_{\min} = \sqrt{5gr}$ because strings can only pull, not push.
Always draw a clear diagram, state the condition for the motion (such as $R = 0$ for losing contact or $T \geq 0$ for a taut string), and write both energy and force equations where appropriate.
At the top of the circle: Maximum GPE, Minimum KE
At the bottom of the circle: Minimum GPE, Maximum KE

Vertical Circular Motion (AQA A-Level Further Maths): Revision Notes