Further circular motion

Q: What are the definitions and formulas for angular speed, centripetal acceleration, and centripetal force?

Angular speed is \(\omega = \frac{v}{r}\), centripetal acceleration is directed towards the centre with magnitude \(a = \frac{v^2}{r} = r\omega^2\), and centripetal force is the resultant inward force \(F_{\text{net}} = \frac{mv^2}{r}\).

Q: How is Conservation of Mechanical Energy applied to find the speed of a particle at any point in vertical circular motion?

\(v^2 = u^2 - 2gr(1 - \cos\theta)\), where \(u\) is the speed at the lowest point and \(\theta\) is the angle measured from the downward vertical.

Q: How do you formulate the radial equation of motion for a particle attached to a string in vertical circular motion?

Resolve forces towards the centre: \(T - mg\cos\theta = \frac{mv^2}{r}\), which gives \(T = \frac{mv^2}{r} + mg\cos\theta\).

Q: What is the minimum initial speed \(u\) at the bottom for a particle on a light string to complete full vertical circles?

\(u \ge \sqrt{5gr}\), corresponding to a minimum speed at the top of \(v_{\text{top}} \ge \sqrt{gr}\).

Q: What are the conditions on the projection speed \(u\) for pendulum oscillations versus leaving the circular path on a light string?

Oscillations occur when \(u \le \sqrt{2gr}\) (the particle stops before or at the horizontal); the string goes slack and the particle leaves the circle when \(\sqrt{2gr}

Q: How does vertical circular motion inside a smooth hollow sphere compare to motion on a light string?

It is mathematically identical: replace tension \(T\) with normal reaction \(R\), requiring \(R \ge 0\) and \(u \ge \sqrt{5gr}\) to complete full loops.

Q: What is the condition for a particle attached to a light rigid rod to complete full vertical circles?

The particle only needs to reach the top with non-negative speed (\(v_{\text{top}} \ge 0\)), requiring \(u \ge 2\sqrt{gr}\).

Q: At what angle does a particle released from rest at the top of a smooth sphere lose contact with the outer surface?

At \(\cos\theta = \frac{2}{3}\) (or \(\theta = \arccos\left(\frac{2}{3}\right) \approx 48.2^{\circ}\)), measured from the upward vertical.

Further Mathematics 2330 · CCEA A Level · 8 min read · Updated: 31 Aug 2026

Welcome to Further Circular Motion!

Welcome to one of the most exciting topics in Mechanics! In earlier studies, you explored circular motion in a horizontal plane where the speed stays constant. But what happens when you ride a rollercoaster loop-the-loop, swing a bucket of water over your head, or watch a skateboarder ride over a curved hill? The speed changes continuously because gravity is either helping pull the object down or fighting against it as it rises. That is the essence of vertical circular motion.

Don't worry if this seems a bit daunting at first. Even though the motion changes constantly, every single problem in this chapter relies on combining just two master principles:
1. Conservation of Mechanical Energy (relating speed and height).
2. Newton's Second Law along the radius (resolving forces towards the centre to find tension or normal reaction).
Master these two steps, and you will be able to solve any question with confidence!

1. Quick Prerequisite Review: The Foundations

Before diving into vertical circles, let's refresh the essentials of circular motion:

• Angular speed (\(\omega\)): The rate of change of angle, measured in radians per second (\(\text{rad s}^{-1}\)), related to linear speed \(v\) by \(v = r\omega\), where \(r\) is the radius of the circular path.

• Centripetal acceleration (\(a\)): Any object moving in a circle of radius \(r\) with instantaneous speed \(v\) experiences an acceleration directed strictly towards the centre of the circle:
\(a = \frac{v^2}{r} = r\omega^2\)

• Centripetal force: This is not a new physical force! It is simply the resultant force acting towards the centre:
\(F_{\text{net}} = ma = \frac{mv^2}{r}\)

Did you know? Centripetal means "centre-seeking" in Latin. It is the real inward force (like tension or gravity) that keeps an object turning rather than flying off in a straight line.

2. The Universal Two-Step Method for Vertical Circular Motion

Whenever a particle moves in a vertical circle, its speed \(v\) changes as its height changes. To analyse the motion at any general point where the radius makes an angle \(\theta\) with the vertical:

Step 1: Apply Conservation of Mechanical Energy

Take a convenient reference level (usually the lowest point of the circle or the centre of the circle) where gravitational potential energy is defined.

If a particle of mass \(m\) is projected from the lowest point with initial speed \(u\), and has speed \(v\) at a height \(h\) above the initial position:
\(\text{Initial Energy} = \text{Final Energy}\)
\(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh\)
Dividing by \(m\) and rearranging gives:
\(v^2 = u^2 - 2gh\)

If \(\theta\) is the angle that the string or radius makes with the downward vertical, the height above the lowest point is:
\(h = r - r\cos\theta = r(1 - \cos\theta)\)
Substituting this into our energy equation gives:
\(v^2 = u^2 - 2gr(1 - \cos\theta)\)

Step 2: Apply \(F = ma\) towards the Centre

Draw a clear free-body diagram showing all forces acting on the particle: the contact force (tension \(T\) or normal reaction \(R\)) and the weight \(mg\).

Resolve all forces along the radius towards the centre of the circle:
\(T - mg\cos\theta = \frac{mv^2}{r}\)
\(\implies T = \frac{mv^2}{r} + mg\cos\theta\)

Key Takeaway for the Universal Method: Step 1 finds the speed \(v\) at any position. Step 2 finds the tension \(T\) or normal contact force \(R\) at that same position by substituting \(v^2\).

3. Motion on a Light Inextensible String

Imagine swinging a ball on a light string in a vertical circle of radius \(r\), projected from the bottom with speed \(u\).

What happens as the particle moves?

• At the bottom (\(\theta = 0^{\circ}\)): The tension is at its maximum:
\(T_{\text{bottom}} = \frac{mu^2}{r} + mg\)

• At the top (\(\theta = 180^{\circ}\)): Here \(\cos(180^{\circ}) = -1\), and \(h = 2r\). The speed at the top \(v_{\text{top}}\) satisfies:
\(v_{\text{top}}^2 = u^2 - 4gr\)
The equation towards the centre at the top is:
\(T_{\text{top}} + mg = \frac{mv_{\text{top}}^2}{r} \implies T_{\text{top}} = \frac{mv_{\text{top}}^2}{r} - mg\)

Conditions for Different Types of Motion:

A string can only pull; it cannot push! Therefore, a string stays taut as long as \(T \ge 0\). If \(T < 0\), the string goes slack and the particle leaves the circular path, continuing under gravity as a projectile.

1. Condition to complete full vertical circles:
The string must not go slack at the very top (the most critical point where tension is minimum).
Set \(T_{\text{top}} \ge 0\):
\(\frac{mv_{\text{top}}^2}{r} - mg \ge 0 \implies v_{\text{top}}^2 \ge gr \implies v_{\text{top}} \ge \sqrt{gr}\)
Using \(v_{\text{top}}^2 = u^2 - 4gr \ge gr\):
\(u^2 \ge 5gr \implies u \ge \sqrt{5gr}\)
Memory Aid: Remember the number 5! The minimum projection speed at the bottom to complete a full loop on a string is always \(\sqrt{5gr}\).

2. Condition for oscillations (pendulum motion):
The particle comes to instantaneous rest (\(v = 0\)) at or before reaching the horizontal level (\(\theta \le 90^{\circ}\)). Since \(\theta \le 90^{\circ}\), \(\cos\theta \ge 0\), so the tension \(T = \frac{mv^2}{r} + mg\cos\theta\) remains strictly positive throughout.
At the horizontal level (\(h = r\)):
\(0 = u^2 - 2gr \implies u \le \sqrt{2gr}\)

3. Condition for leaving the circle (string goes slack):
If \(\sqrt{2gr} < u < \sqrt{5gr}\), the particle rises above the horizontal level (\(\theta > 90^{\circ}\)), but the tension drops to zero (\(T = 0\)) while \(v > 0\). The string goes slack, and the particle moves as a projectile under gravity until the string pulls taut again.

4. Motion Inside a Smooth Hollow Sphere or Cylinder

A particle moving on the smooth inside surface of a sphere of radius \(r\) behaves mathematically identically to a particle on a string!

• Replace tension \(T\) with the normal reaction force \(R\).
• The surface can only push inward on the particle, so \(R \ge 0\).
• If \(R = 0\) above the horizontal, the particle leaves the surface and flies freely as a projectile.
• Minimum speed at lowest point to complete a full loop: \(u \ge \sqrt{5gr}\).

5. Motion Attached to a Light Rigid Rod

What if the particle is attached to a light rigid rod instead of a string?

This is a crucial distinction that frequently appears in CCEA exams!

• A rod is rigid, meaning it cannot go slack. It can support compression (push) as well as tension (pull).
• Therefore, the particle does not need a minimum tension at the top to stay on the circular path. It only needs enough energy to reach the top without stopping beforehand!
• Condition to complete full circles: speed at the top \(v_{\text{top}} \ge 0\).
Using conservation of energy from the bottom to the top:
\(v_{\text{top}}^2 = u^2 - 4gr \ge 0 \implies u^2 \ge 4gr \implies u \ge 2\sqrt{gr}\)

Comparison Summary Box:
• String / Inside smooth sphere: Needs \(T_{\text{top}} \ge 0 \implies u \ge \sqrt{5gr}\) (Speed at top is at least \(\sqrt{gr}\)).
• Light rigid rod / Smooth circular tube: Needs \(v_{\text{top}} \ge 0 \implies u \ge \sqrt{4gr} = 2\sqrt{gr}\) (Speed at top can be \(0\)).

6. Motion on the Outside of a Smooth Sphere

Now consider a particle placed at the top of a fixed, smooth sphere of radius \(r\) and slightly disturbed so that it slides down the outside surface.

Setting up the Equations:

Let \(\theta\) be the angle the radius makes with the upward vertical as the particle slides down.

• Conservation of Energy:
The particle starts from rest at the top (\(\theta = 0^{\circ}\), height \(h_0 = r\)).
At angle \(\theta\), the height is \(h = r\cos\theta\).
\(\frac{1}{2}mv^2 = mg(r - r\cos\theta) \implies v^2 = 2gr(1 - \cos\theta)\)

• Equation of Motion towards the centre:
The forces acting along the radius are the component of weight \(mg\cos\theta\) (inwards) and the normal reaction \(R\) (outwards).
\(mg\cos\theta - R = \frac{mv^2}{r} \implies R = mg\cos\theta - \frac{mv^2}{r}\)

Finding Where the Particle Leaves the Surface:

The particle remains in contact with the sphere as long as the surface pushes against it (\(R > 0\)). It loses contact (leaves the sphere) precisely when \(R = 0\).

Set \(R = 0\):
\(mg\cos\theta = \frac{mv^2}{r} \implies v^2 = gr\cos\theta\)

Now substitute our expression for \(v^2\) from the energy equation:
\(2gr(1 - \cos\theta) = gr\cos\theta\)
Dividing both sides by \(gr\):
\(2(1 - \cos\theta) = \cos\theta\)
\(2 - 2\cos\theta = \cos\theta \implies 3\cos\theta = 2 \implies \cos\theta = \frac{2}{3}\)

The angle at which the particle flies off is:
\(\theta = \arccos\left(\frac{2}{3}\right) \approx 48.2^{\circ}\)

The speed at the moment of leaving the surface is:
\(v = \sqrt{gr\cos\theta} = \sqrt{\frac{2}{3}gr}\)

After leaving the surface, the particle moves under gravity alone along a parabolic path (standard projectile motion).

7. Common Mistakes to Avoid

• Mixing up angles: Always check whether \(\theta\) is measured from the upward vertical or the downward vertical! Drawing a large, clear diagram prevents sign errors with \(\cos\theta\).

• Assuming \(v = 0\) when a string goes slack: A string goes slack when \(T = 0\), not when \(v = 0\). In fact, the particle still has speed and becomes a projectile.

• Confusing rods and strings: Remember, rods can push (\(T < 0\) acts outwards as a thrust), so a rod only requires \(v \ge 0\) at the top to complete a loop, while a string requires \(T \ge 0\).

• Forgetting \(g\) in centripetal equations: Always double-check your units and dimensions. \(\frac{mv^2}{r}\) has units of force (Newtons), exactly like \(mg\).

8. Summary of Key Formulas and Results

• Centripetal Acceleration: \(a = \frac{v^2}{r} = r\omega^2\)

• Energy Relation (from lowest point): \(v^2 = u^2 - 2gr(1 - \cos\theta)\) where \(\theta\) is from the downward vertical.

• Full loop on string / inside sphere: \(u \ge \sqrt{5gr}\) (Condition: \(T_{\text{top}} \ge 0\))

• Full loop on rigid rod: \(u \ge 2\sqrt{gr}\) (Condition: \(v_{\text{top}} \ge 0\))

• Oscillations on string: \(u \le \sqrt{2gr}\) (Particle stops before/at horizontal)

• Leaving outside of sphere from rest at top: Leaves surface when \(\cos\theta = \frac{2}{3}\) where \(\theta\) is from the upward vertical.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are the definitions and formulas for angular speed, centripetal acceleration, and centripetal force?

Angular speed is \(\omega = \frac{v}{r}\), centripetal acceleration is directed towards the centre with magnitude \(a = \frac{v^2}{r} = r\omega^2\), and centripetal force is the resultant inward force \(F_{\text{net}} = \frac{mv^2}{r}\).

How is Conservation of Mechanical Energy applied to find the speed of a particle at any point in vertical circular motion?

\(v^2 = u^2 - 2gr(1 - \cos\theta)\), where \(u\) is the speed at the lowest point and \(\theta\) is the angle measured from the downward vertical.

How do you formulate the radial equation of motion for a particle attached to a string in vertical circular motion?

Resolve forces towards the centre: \(T - mg\cos\theta = \frac{mv^2}{r}\), which gives \(T = \frac{mv^2}{r} + mg\cos\theta\).

What is the minimum initial speed \(u\) at the bottom for a particle on a light string to complete full vertical circles?

\(u \ge \sqrt{5gr}\), corresponding to a minimum speed at the top of \(v_{\text{top}} \ge \sqrt{gr}\).

What are the conditions on the projection speed \(u\) for pendulum oscillations versus leaving the circular path on a light string?

Oscillations occur when \(u \le \sqrt{2gr}\) (the particle stops before or at the horizontal); the string goes slack and the particle leaves the circle when \(\sqrt{2gr} < u < \sqrt{5gr}\).

How does vertical circular motion inside a smooth hollow sphere compare to motion on a light string?

It is mathematically identical: replace tension \(T\) with normal reaction \(R\), requiring \(R \ge 0\) and \(u \ge \sqrt{5gr}\) to complete full loops.

What is the condition for a particle attached to a light rigid rod to complete full vertical circles?

The particle only needs to reach the top with non-negative speed (\(v_{\text{top}} \ge 0\)), requiring \(u \ge 2\sqrt{gr}\).

At what angle does a particle released from rest at the top of a smooth sphere lose contact with the outer surface?

At \(\cos\theta = \frac{2}{3}\) (or \(\theta = \arccos\left(\frac{2}{3}\right) \approx 48.2^{\circ}\)), measured from the upward vertical.

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