Welcome to Further Circular Motion!
Have you ever been on a rollercoaster doing a loop-the-loop and wondered why you don't fall out at the very top? Or have you ever swung a bucket of water around in a vertical circle without spilling a single drop? In this chapter, we take the ideas of circular motion into the vertical plane.
Unlike horizontal circular motion (where speed stays constant), an object moving vertically speeds up as it falls and slows down as it climbs due to gravity. Don't worry if this sounds intimidating at first! We can solve almost every single vertical circular motion problem using just two master steps:
1. Conservation of Energy (to find the speed).
2. Newton's Second Law along the radius (to find the tension or normal reaction).
Did you know? The first loop-the-loop rollercoasters were built with circular loops, but the extreme forces caused whiplash for riders. Modern rollercoasters use teardrop-shaped loops (called clothoids) to reduce the sudden centripetal acceleration on entry!
1. The Two Golden Rules of Vertical Circular Motion
Whenever a particle moves in a vertical circle of radius \(r\), gravity continuously changes its speed. To model this motion, we apply two fundamental physical principles at any general position defined by an angle \(\theta\).
Rule 1: Conservation of Mechanical Energy
Assuming no air resistance or friction, total mechanical energy is conserved:
\(\text{Total Energy at Start} = \text{Total Energy at New Position}\)
\(\frac{1}{2}mu^2 + mg h_1 = \frac{1}{2}mv^2 + mg h_2\)
where \(u\) is the initial speed, \(v\) is the speed at the new height, and \(h\) is the vertical height above an agreed reference level.
Rule 2: Equation of Motion Along the Radius (\(F = ma\))
At any point on the circle, the net force directed towards the center of the circle provides the necessary centripetal acceleration:
\(\Sigma F_{\text{radial}} = \frac{m v^2}{r}\)
Key Takeaway: Always start your vertical circle problems by writing down the energy equation first to get an expression for \(v^2\), and then substitute \(v^2\) into the radial force equation.
2. Motion in a Vertical Circle: Particle on a String
Imagine a particle of mass \(m\) attached to a light, inextensible string of length \(r\), suspended from a fixed point \(O\). The particle is projected horizontally from the lowest point with an initial speed \(u\).
Step-by-Step Derivation of Velocity and Tension
Let \(\theta\) be the angle the string makes with the downward vertical.
Step 1: Find the height gained
Taking the lowest point as the zero-height level (\(h = 0\)):
Height at angle \(\theta\): \(h = r - r\cos\theta = r(1 - \cos\theta)\)
Step 2: Apply Conservation of Energy
\(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgr(1 - \cos\theta)\)
Dividing by \(\frac{1}{2}m\):
\(v^2 = u^2 - 2gr(1 - \cos\theta)\)
Step 3: Resolve forces towards the center
The forces acting on the particle are tension \(T\) acting towards \(O\), and weight \(mg\) acting straight down. The component of weight acting away from the center along the string is \(mg\cos\theta\).
\(T - mg\cos\theta = \frac{mv^2}{r}\)
Rearranging for Tension \(T\):
\(T = \frac{mv^2}{r} + mg\cos\theta\)
Step 4: Combine the two equations
Substitute \(v^2 = u^2 - 2gr + 2gr\cos\theta\) into the tension equation:
\(T = \frac{m}{r}\left(u^2 - 2gr + 2gr\cos\theta\right) + mg\cos\theta\)
\(T = \frac{m}{r}\left(u^2 - 2gr + 3gr\cos\theta\right)\)
Conditions for Completing a Full Circle on a String
For a string, the particle will complete full circles only if the string remains taut throughout the motion. The most critical point is the highest point (where \(\theta = 180^\circ\), so \(\cos 180^\circ = -1\)):
• At the top: \(T_{\text{top}} = \frac{m}{r}(u^2 - 5gr)\)
• For the string not to go slack: \(T_{\text{top}} \ge 0\)
• Therefore: \(u^2 - 5gr \ge 0 \implies u \ge \sqrt{5gr}\)
What Happens for Other Initial Speeds?
• If \(u \le \sqrt{2gr}\): The particle comes to rest (\(v = 0\)) before reaching horizontal level (\(\theta \le 90^\circ\)). The string never goes slack; the particle simply oscillates back and forth like a simple pendulum.
• If \(\sqrt{2gr} < u < \sqrt{5gr}\): The particle rises above the horizontal (\(\theta > 90^\circ\)), but tension drops to zero (\(T = 0\)) while \(v > 0\). The string becomes slack, and the particle leaves the circular path to travel as a projectile under gravity until the string becomes taut again.
Key Takeaway: For a flexible string, complete circular motion requires \(T \ge 0\) at the very top, giving a minimum launch speed of \(u = \sqrt{5gr}\).
3. Motion on a Light Rigid Rod or Smooth Wire Ring
Now, what if the string is replaced by a light rigid rod of length \(r\), or the particle is a bead threaded onto a smooth vertical circular wire?
A rod is fundamentally different from a string: a rod can push as well as pull. It can provide a thrust (compression) as well as a tension.
Condition to Complete a Full Circle on a Rod
Because the rod cannot go "slack", the particle doesn't need a positive tension to stay on the circle. It just needs enough kinetic energy to reach the highest point without stopping early!
• Condition at the top: \(v_{\text{top}} \ge 0\)
• Using energy from bottom to top (height gained is \(2r\)):
\(\frac{1}{2}mu^2 \ge mg(2r)\)
\(u^2 \ge 4gr \implies u \ge \sqrt{4gr} = 2\sqrt{gr}\)
Quick Review: String vs. Rod
• Minimum speed at bottom to complete circle with a string: \(u = \sqrt{5gr}\)
• Minimum speed at bottom to complete circle with a rod: \(u = 2\sqrt{gr}\)
4. Particle Moving on the Outside of a Smooth Sphere
Consider a particle of mass \(m\) resting at the top of a smooth fixed sphere of radius \(r\). If it is slightly displaced (or projected with speed \(u\)), it will slide down the surface until it loses contact.
Finding Where the Particle Leaves the Surface
Let \(\theta\) be the angle made with the upward vertical as the particle slides down.
Step 1: Energy equation
Vertical distance fallen: \(h = r - r\cos\theta = r(1 - \cos\theta)\)
By conservation of energy (assuming starting from rest at the top, \(u = 0\)):
\(\frac{1}{2}mv^2 = mgr(1 - \cos\theta) \implies v^2 = 2gr(1 - \cos\theta)\)
Step 2: Radial equation of motion
The forces acting radially are the normal reaction \(R\) outwards and the component of weight \(mg\cos\theta\) inwards (towards the center):
\(mg\cos\theta - R = \frac{mv^2}{r}\)
Rearranging for \(R\):
\(R = mg\cos\theta - \frac{mv^2}{r}\)
Step 3: Condition for leaving the surface
The particle leaves the surface at the precise moment the normal reaction becomes zero (\(R = 0\)):
\(mg\cos\theta = \frac{mv^2}{r} \implies v^2 = gr\cos\theta\)
Step 4: Solve for \(\theta\)
Substitute \(v^2 = 2gr(1 - \cos\theta)\) into the condition:
\(2gr(1 - \cos\theta) = gr\cos\theta\)
\(2 - 2\cos\theta = \cos\theta\)
\(3\cos\theta = 2 \implies \cos\theta = \frac{2}{3}\)
Thus, the particle loses contact when \(\theta = \arccos\left(\frac{2}{3}\right) \approx 48.2^\circ\). After this point, it moves freely as a projectile under gravity!
Key Takeaway: For objects on the outside of curved surfaces, "leaving the surface" occurs when the normal reaction drops to zero (\(R = 0\)).
5. Common Mistakes to Avoid
• Mixing up angle definitions: Always check whether \(\theta\) is measured from the downward vertical (common for swinging pendulums) or the upward vertical (common for spheres). This changes whether the height is \(r(1 - \cos\theta)\) or \(r(1 + \cos\theta)\).
• Using constant acceleration equations (SUVAT): Never use SUVAT in vertical circular motion! The acceleration changes at every instant because speed and direction both vary.
• Confusing the conditions for full circles: Remember that strings need \(T \ge 0\) at the top (\(u \ge \sqrt{5gr}\)), whereas rods only need \(v \ge 0\) at the top (\(u \ge 2\sqrt{gr}\)).
• Forgetting projectile motion after string goes slack: If a string goes slack at an angle \(\theta > 90^\circ\), its subsequent path is parabolic, with initial velocity \(v\) perpendicular to the string position at that instant.
Summary Checklist
To master vertical circular motion questions, make sure you can:
• Apply conservation of energy to find speed \(v\) at any height.
• Resolve forces towards the center to set up \(\Sigma F_{\text{radial}} = \frac{mv^2}{r}\).
• State and apply the condition for a string to stay taut (\(T \ge 0\)).
• State and apply the condition for a rod to reach the top (\(v \ge 0\)).
• Identify the point where a particle leaves a surface by setting \(R = 0\).