Physics Solution: Bead on a Helix

Physics Solution: Q30

Dynamics of a Bead on a Helix

1. Force Analysis

The bead moves down the helix with a constant speed $v$. This implies that the net tangential force is zero (kinetic friction balances the tangential component of gravity). However, the direction of velocity changes, implying a net radial force provides the necessary centripetal acceleration.

Let $\theta$ be the angle of the helix with the horizontal. We define a local coordinate system attached to the bead:

$\hat{t}$: Tangent vector (direction of motion).
$\hat{n}$: Normal vector to the cylinder surface (radial direction, towards the axis).
$\hat{b}$: Binormal vector (perpendicular to $\hat{t}$ and $\hat{n}$, lying on the cylinder surface).

2. Equations of Motion

A. Tangential Direction ($\hat{t}$):
Since the speed is constant, tangential acceleration $a_t = 0$. The component of gravity along the tangent balances friction: $$ mg \sin \theta – f_k = 0 \implies f_k = mg \sin \theta $$ Where $f_k$ is the kinetic friction force.

B. Normal Plane Directions:
The bead undergoes circular motion in the horizontal plane with radius $R$ and horizontal speed $v_{horiz} = v \cos \theta$. The required centripetal force is directed radially inward: $$ F_{cp} = \frac{m (v \cos \theta)^2}{R} $$

We resolve the Normal force $\vec{N}$ into two orthogonal components:

$N_{rad}$: Provides the centripetal acceleration. $N_{rad} = \frac{m v^2 \cos^2 \theta}{R}$
$N_{\perp}$: Balances the normal component of gravity (similar to an inclined plane). $N_{\perp} = mg \cos \theta$

Since these components are orthogonal ($N_{rad}$ is horizontal, $N_{\perp}$ is in the vertical plane perpendicular to the wire), the total normal force magnitude is:

N = \sqrt{N_{rad}^2 + N_{\perp}^2} $$ $$ N = \sqrt{\left( \frac{m v^2 \cos^2 \theta}{R} \right)^2 + (mg \cos \theta)^2} $$ $$ N = m \cos \theta \sqrt{\left( \frac{v^2 \cos \theta}{R} \right)^2 + g^2}

3. Calculating the Coefficient of Friction

Using the law of kinetic friction $f_k = \mu N$, we equate our expressions:

mg \sin \theta = \mu \left[ m \cos \theta \sqrt{\frac{v^4 \cos^2 \theta}{R^2} + g^2} \right]

Solving for $\mu$:

\mu = \frac{g \sin \theta}{\cos \theta \sqrt{\frac{v^4 \cos^2 \theta}{R^2} + g^2}} $$ $$ \mu = \frac{g \tan \theta}{\sqrt{\frac{v^4 \cos^2 \theta + g^2 R^2}{R^2}}}

Rearranging the terms by moving $R$ from the denominator of the square root to the numerator:

Final Answer

The coefficient of kinetic friction is:

$$ \mu = \frac{g R \tan \theta}{\sqrt{g^2 R^2 + v^4 \cos^2 \theta}} $$