Further Mathematics (9665) | Vertical circular motion

Welcome to Vertical Circular Motion!

Get ready to tackle one of the most dynamic and exciting topics in Further Mechanics! Vertical circular motion looks tricky, but it’s simply a combination of two things you already know well: Newton’s Second Law for Circular Motion and the Conservation of Energy.

Understanding this chapter is essential for modelling real-world scenarios, from rollercoasters to swinging a bucket of water over your head without spilling a drop! We will focus specifically on how gravity affects motion when the path is vertical, and determine the critical speeds needed to keep the motion going.

Section 1: The Dynamics of Vertical Motion

The Crucial Difference: Gravity Matters

In horizontal circular motion, gravity acts perpendicular to the plane of movement, so it doesn't affect the speed. In a vertical circle, however, gravity is constantly pulling the object back down, meaning its speed is not constant.

This varying speed means the tension in a string (or the reaction force from a track) must also vary, creating the unique dynamics we need to analyze.

Key Concept Review: Centripetal Force
Remember that any object moving in a circle needs a net force pointing towards the center (the centripetal force, \(F_c\)). $$F_c = ma = \frac{mv^2}{r}$$ Where \(m\) is mass, \(v\) is speed, and \(r\) is the radius. In vertical motion, \(F_c\) is the vector sum of all forces acting radially (Tension/Reaction and components of Gravity).

Energy Conservation: Your Essential Tool

Since we are typically dealing with motion where the only non-conservative force (like tension) does no work, the Principle of Conservation of Mechanical Energy (CME) is vital.

We use CME to link the speed at one point in the circle (like the bottom) to the speed at any other point (like the top). $$KE_1 + PE_1 = KE_2 + PE_2$$ Where:

• Kinetic Energy (KE): \(\frac{1}{2}mv^2\)
• Gravitational Potential Energy (PE): \(mgh\) (where \(h\) is height above a chosen datum, usually the bottom of the circle).

Analogy: Think of a pendulum. It goes slowest at the highest point (max PE) and fastest at the lowest point (max KE). A vertical circle is just a continuous, high-speed pendulum swing!

Key Takeaway for Section 1: Vertical circular motion requires applying \(F_c = \frac{mv^2}{r}\) at every point, incorporating the varying effects of gravity, and using CME to relate speeds across different heights.

Section 2: Analysing Key Points on the Circle

We focus on the radial forces at three critical positions: the bottom, the top, and a general point defined by an angle \(\theta\).

1. At the Bottom of the Circle

This is the point where the speed is maximum and the tension/reaction force is maximum. Let the speed here be \(v_B\).

Forces (Radial):
The Tension or Reaction force (\(T\)) acts upwards (towards the centre).
Gravity (\(mg\)) acts downwards (away from the centre).
The net centripetal force \(F_c\) must act upwards.

Equation of Motion: $$T - mg = \frac{mv_B^2}{r}$$

Therefore, the actual tension/reaction at the bottom is: $$T = mg + \frac{mv_B^2}{r}$$ Notice that the required tension is always greater than gravity because it has to counteract gravity AND provide the necessary centripetal force.

2. At the Top of the Circle (The Critical Point)

This is the point where the speed is minimum and the tension/reaction force is minimum. Let the speed here be \(v_T\).

Forces (Radial):
Both the Tension or Reaction force (\(T\)) and Gravity (\(mg\)) act downwards (towards the centre).
The net centripetal force \(F_c\) must act downwards.

Equation of Motion: $$T + mg = \frac{mv_T^2}{r}$$

Therefore, the actual tension/reaction at the top is: $$T = \frac{mv_T^2}{r} - mg$$ The gravity helps provide the centripetal force, so the required tension/reaction is lower here.

3. At a General Point (Angle \(\theta\))

Consider the particle at an angle \(\theta\) from the lowest point. Let the speed here be \(v\).

Forces (Radial):
Gravity must be resolved into two components:

• Component acting radially inward (opposite to the tension direction): \(mg \cos \theta\)
• Component acting tangentially (changing the speed): \(mg \sin \theta\)

Equation of Motion (Radial Direction): $$T - mg \cos \theta = \frac{mv^2}{r}$$

We can rearrange this to find tension: $$T = \frac{mv^2}{r} + mg \cos \theta$$

Tip: Ensure you define \(\theta\) consistently (usually from the vertical radius pointing down) to correctly use the \(\cos \theta\) component.

Section 3: Conditions to Complete Vertical Circles

The most common examination question asks for the minimum speed required to successfully complete the circle. This relies entirely on the dynamics at the Top Point.

What Does "Completing the Circle" Mean?

For a particle attached to a string, the string must remain taut (Tension \(T \ge 0\)). If \(T=0\), the string is 'slack'.

For a particle moving on a track (like a rollercoaster), the particle must remain in contact with the track (Reaction force \(R \ge 0\)). If \(R=0\), the particle leaves the track.

The condition to *just* complete the circle is that the tension \(T\) or reaction \(R\) becomes zero exactly at the highest point. This minimum speed is called the critical speed.

Step 1: Finding the Critical Speed at the Top (\(v_{min, T}\))

We use the equation of motion at the top point and set \(T = 0\):

$$T + mg = \frac{mv_{min, T}^2}{r}$$ Setting \(T=0\): $$0 + mg = \frac{mv_{min, T}^2}{r}$$

The mass \(m\) cancels out! $$g = \frac{v_{min, T}^2}{r}$$ $$v_{min, T}^2 = gr$$

The minimum speed at the top required to complete the circle is: $$v_{min, T} = \sqrt{gr}$$

Note: If the speed at the top is less than \(\sqrt{gr}\), the required centripetal force is less than gravity, meaning the object will fall before reaching the full height (or the string will go slack).

Step 2: Finding the Minimum Speed at the Bottom (\(v_{min, B}\))

Usually, you are asked for the speed needed at the starting point (the bottom, \(B\)) to achieve this critical speed at the top (\(T\)). We use the Conservation of Energy between B and T.

Let the datum (\(h=0\)) be the bottom of the circle. The height at the top is \(h_T = 2r\).

$$KE_B + PE_B = KE_T + PE_T$$ $$\frac{1}{2}m v_{min, B}^2 + 0 = \frac{1}{2}m v_{min, T}^2 + mg(2r)$$

Divide through by mass \(m\): $$\frac{1}{2} v_{min, B}^2 = \frac{1}{2} v_{min, T}^2 + 2gr$$

Now substitute the critical speed found in Step 1, \(v_{min, T}^2 = gr\): $$\frac{1}{2} v_{min, B}^2 = \frac{1}{2} (gr) + 2gr$$ $$\frac{1}{2} v_{min, B}^2 = \frac{1}{2} gr + \frac{4}{2} gr$$ $$\frac{1}{2} v_{min, B}^2 = \frac{5}{2} gr$$

Multiply by 2: $$v_{min, B}^2 = 5gr$$

The minimum speed at the bottom required to complete the circle is: $$v_{min, B} = \sqrt{5gr}$$

Section 4: Leaving the Circular Path

Sometimes, the particle does not complete the circle. It might be launched with a speed \(v_B\) such that it slows down and the string goes slack (or it leaves the track) somewhere between the bottom and the top.

Finding Where the Particle Leaves the Path

If the speed \(v_B\) is too low (\(v_B < \sqrt{5gr}\)), the particle will leave the path at some angle \(\theta_{crit}\) (where \(\theta\) is measured from the bottom).

This happens when the tension \(T\) (or reaction \(R\)) drops to zero before reaching the top.

Step 1: Set \(T=0\) at the general point.
The radial equation is: $$T = \frac{mv^2}{r} + mg \cos \theta$$ Setting \(T=0\) gives the critical condition for leaving the path: $$0 = \frac{mv_{crit}^2}{r} + mg \cos \theta_{crit}$$ $$v_{crit}^2 = -gr \cos \theta_{crit}$$ Note: Since \(v^2\) must be positive, this equation only makes sense for angles \(\theta\) greater than 90° (i.e., \(\cos \theta\) is negative), confirming the particle leaves the path in the upper half of the circle.

Step 2: Use CME to relate \(v_{crit}\) back to the initial speed \(v_B\).
The height \(h\) of the particle at angle \(\theta\) (measured from the bottom) is: $$h = r - r \cos \theta = r(1 - \cos \theta)$$ Using CME between B and \(\theta\): $$\frac{1}{2}m v_B^2 = \frac{1}{2}m v_{crit}^2 + mgh$$ $$\frac{1}{2} v_B^2 = \frac{1}{2} v_{crit}^2 + gr(1 - \cos \theta_{crit})$$

Step 3: Combine the two equations to solve for \(\cos \theta_{crit}\).
Substitute \(v_{crit}^2 = -gr \cos \theta_{crit}\) into the CME equation: $$\frac{1}{2} v_B^2 = \frac{1}{2} (-gr \cos \theta_{crit}) + gr - gr \cos \theta_{crit}$$ Multiply by 2 and group the \(\cos \theta\) terms: $$v_B^2 = -gr \cos \theta_{crit} + 2gr - 2gr \cos \theta_{crit}$$ $$v_B^2 = 2gr - 3gr \cos \theta_{crit}$$

The final required expression to find the critical angle is: $$\cos \theta_{crit} = \frac{2gr - v_B^2}{3gr}$$

If you are given the initial speed \(v_B\), you can use this formula to find the precise angle at which the object loses contact or the string goes slack.

Key Takeaway for Section 4: If the minimum speed for completion (\(\sqrt{5gr}\)) isn't met, the object leaves the path when the radial inward force (Tension/Reaction) hits zero. We use both the radial force equation (\(T=0\)) and CME to solve for the exact angle.

Summary and Final Checklist

Mastering vertical circular motion relies on your ability to switch fluently between two analytical methods:

1. Analysis of Forces (Newton's Second Law):

Used to find Tension (T) or Reaction (R) at a specific point based on the speed \(v\) at that point.
Rule: Net force towards the centre = \(\frac{mv^2}{r}\).

2. Analysis of Energy (CME):

Used to find the speed \(v\) at one point based on the speed at another, incorporating the change in height \(h\).
Rule: \(\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2\).

Critical Conditions Checklist:

• To just complete the circle: \(T\) or \(R\) must be \(\ge 0\) at all times. The critical condition is \(T=0\) (or \(R=0\)) at the top.
• Minimum speed at the top: \(v_{min, T} = \sqrt{gr}\)
• Minimum speed at the bottom: \(v_{min, B} = \sqrt{5gr}\)

Keep practicing those step-by-step derivations, and don't worry if the algebra gets heavy—always keep track of your kinetic and potential energy terms! You've got this!