The Conical Pendulum with Air Resistance

Problem Statement

A conical pendulum of mass m and length L rotates with initial angular velocity ω₀. Air resistance provides a force F = -kv² opposite to the velocity direction. Find the rate at which the string angle θ decreases with time.

Given Data

• Mass of pendulum bob: m
• Length of string: L
• Initial angular velocity: ω₀
• Air resistance force: Fₐᵢᵣ = -kv²
• Gravitational acceleration: g

Concepts & Approach

We'll use:

1. Newton's Laws in radial and vertical directions
2. Energy considerations for the damping effect
3. Differential equations to describe the motion

Step-by-Step Solution

Step 1: Forces in the Conical Pendulum

For a conical pendulum without air resistance:

• Tension T along the string
• Gravity mg downward
• Centripetal force required for circular motion

The geometry gives:

Radius of circle: $$ r = L\sin\theta $$

Height: $$ h = L\cos\theta $$

Step 2: Force Balance Equations

Vertical direction:

$$ T\cos\theta = mg $$

Radial direction (without drag):

$$ T\sin\theta = m\omega^2 r = m\omega^2 L\sin\theta $$

From these, we get the standard conical pendulum relation:

$$ \omega^2 = \frac{g}{L\cos\theta} $$

Step 3: Including Air Resistance

With air resistance F = -kv² opposing velocity, the tangential velocity decreases over time.

The velocity magnitude is:

$$ v = \omega r = \omega L\sin\theta $$

The tangential force equation becomes:

$$ m\frac{dv}{dt} = -kv^2 $$

Substituting v = ωLsinθ:

$$ mL\sin\theta \frac{d\omega}{dt} = -k(\omega L\sin\theta)^2 $$

Step 4: Relating ω and θ

From the force balance, we have:

$$ \omega^2 = \frac{g}{L\cos\theta} $$

Differentiating with respect to time:

$$ 2\omega \frac{d\omega}{dt} = \frac{g}{L} \cdot \frac{\sin\theta}{\cos^2\theta} \frac{d\theta}{dt} $$

Step 5: Solving the System

From Step 3:

$$ \frac{d\omega}{dt} = -\frac{kL\sin\theta}{m} \omega^2 $$

Substitute into the differentiated relation:

$$ 2\omega \left(-\frac{kL\sin\theta}{m} \omega^2\right) = \frac{g}{L} \cdot \frac{\sin\theta}{\cos^2\theta} \frac{d\theta}{dt} $$

Simplify:

$$ -\frac{2kL\sin\theta}{m} \omega^3 = \frac{g}{L} \cdot \frac{\sin\theta}{\cos^2\theta} \frac{d\theta}{dt} $$

Cancel sinθ (for θ ≠ 0):

$$ -\frac{2kL}{m} \omega^3 = \frac{g}{L\cos^2\theta} \frac{d\theta}{dt} $$

Step 6: Expressing in Terms of θ Only

Substitute $$ \omega^2 = \frac{g}{L\cos\theta} $$, so $$ \omega^3 = \left(\frac{g}{L\cos\theta}\right)^{3/2} $$

Then:

$$ -\frac{2kL}{m} \left(\frac{g}{L\cos\theta}\right)^{3/2} = \frac{g}{L\cos^2\theta} \frac{d\theta}{dt} $$

Step 7: Final Differential Equation

Solving for dθ/dt:

$$ \frac{d\theta}{dt} = -\frac{2kL}{m} \left(\frac{g}{L\cos\theta}\right)^{3/2} \cdot \frac{L\cos^2\theta}{g} $$

Simplify:

$$ \frac{d\theta}{dt} = -\frac{2k}{m} \sqrt{\frac{g}{L}} \cdot \frac{L\cos^2\theta}{\sqrt{\cos\theta}} $$

Final form:

$$ \frac{d\theta}{dt} = -\frac{2kL}{m} \sqrt{\frac{g}{L}} \cos^{3/2}\theta $$

Final Answer

The rate at which the string angle decreases is:

$$ \boxed{\frac{d\theta}{dt} = -\frac{2kL}{m} \sqrt{\frac{g}{L}} \cos^{3/2}\theta} $$

Physical Interpretation

1. Negative sign confirms that θ decreases over time (pendulum slows down and settles)
2. Strong dependence on θ: The rate of change is proportional to $\cos^{3/2}\theta$
3. Parameters effect:
   • Larger drag coefficient k → faster decay
   • Longer string L → faster angular change
   • Larger mass m → slower response to drag

This problem beautifully shows how dissipative forces transform a steady circular motion into a decaying oscillation, ultimately reaching the stable hanging position.

Tags: #Physics #NewtonLaws #ConicalPendulum #DifferentialEquations #AirResistance #ClassicalMechanics