RBD ChYU 4

Physics Solution – Circular Motion of Bicycle

Solution to Circular Motion Problem

1. Geometry of the System

Consider the bicycle turning about a center $O$. Let $\theta$ be the angle of the front wheel with respect to the frame. Let $\alpha$ be the angle made by the position vector of the center of mass (CO) with the rear axle line.

From the geometry of the turn:

  • $R_1$: Radius of the path of the Front Wheel.
  • $R_2$: Radius of the path of the Center of Mass (CM).
  • $R_3$: Radius of the path of the Rear Wheel.
  • $f_1$: Friction force on the Rear Wheel.
  • $f_2$: Friction force on the Front Wheel.
O R₁ R₂ R₃ f₁ f₂ C α θ

Geometric and Force Analysis showing radii $R_1, R_2, R_3$ and friction forces.

2. Force Analysis Equations

We resolve the forces along two mutually perpendicular directions: the radial direction (along the line connecting the center of rotation $O$ to the CM) and the direction perpendicular to it.

1. Equation of Motion in the Radial Direction:
The net force towards the center $O$ provides the centripetal force $\frac{mv^2}{R_2}$.

$$ f_2 \cos(\theta – \alpha) – f_1 \sin \alpha = \frac{mv^2}{R_2} \quad \dots \text{(i)} $$

2. Force Balance in the Transverse Direction:
Forces perpendicular to the radial direction must balance to maintain the turn without slipping tangentially.

$$ f_2 \sin(\theta – \alpha) = f_1 \cos \alpha \quad \dots \text{(ii)} $$

3. Solving for $f_2$:
From equation (ii), we express $f_1$ in terms of $f_2$:

$$ f_1 = f_2 \frac{\sin(\theta – \alpha)}{\cos \alpha} $$

Substitute this into equation (i):

$$ f_2 \cos(\theta – \alpha) – \left( f_2 \frac{\sin(\theta – \alpha)}{\cos \alpha} \right) \sin \alpha = \frac{mv^2}{R_2} $$

Factor out $f_2$ and take the common denominator:

$$ f_2 \left[ \frac{\cos(\theta – \alpha)\cos \alpha – \sin(\theta – \alpha)\sin \alpha}{\cos \alpha} \right] = \frac{mv^2}{R_2} $$

Using the trigonometric identity $\cos(A+B) = \cos A \cos B – \sin A \sin B$, the numerator becomes $\cos((\theta – \alpha) + \alpha) = \cos \theta$:

$$ f_2 \frac{\cos \theta}{\cos \alpha} = \frac{mv^2}{R_2} $$ $$ \implies f_2 = \frac{mv^2 \cos \alpha}{R_2 \cos \theta} $$
3. Friction Coefficient Calculation

The condition for the front wheel not to slip is determined by the limiting friction. Assuming the normal force on each wheel is $mg/2$:

$$ \mu \left( \frac{mg}{2} \right) = f_2 $$

Substitute the expression for $f_2$ derived above:

$$ \mu \frac{mg}{2} = \frac{mv^2 \cos \alpha}{R_2 \cos \theta} $$ $$ \mu = \frac{2 v^2 \cos \alpha}{g R_2 \cos \theta} $$

Geometric Substitutions:

From the geometry of the triangle formed by $O$, the rear axle, and CM:

$$ \cos \alpha = \frac{R_3}{R_2} = \frac{l \cot \theta}{R_2} $$

Substituting $\cos \alpha$ into the expression for $\mu$:

$$ \mu = \frac{2 v^2}{g R_2 \cos \theta} \cdot \left( \frac{l \cot \theta}{R_2} \right) $$ $$ \mu = \frac{2 v^2 l}{g \cos \theta R_2^2} \cdot \frac{\cos \theta}{\sin \theta} $$ $$ \mu = \frac{2 v^2 l}{g \sin \theta R_2^2} $$

We know that $R_2^2 = R_3^2 + (l/2)^2 = l^2 \cot^2 \theta + \frac{l^2}{4} = \frac{l^2}{4}(4\cot^2 \theta + 1)$. Substituting this back:

$$ \mu = \frac{2 v^2 l}{g \sin \theta \cdot \frac{l^2}{4}(4\cot^2 \theta + 1)} $$ $$ \mu = \frac{8 v^2}{g l \sin \theta (1 + 4\cot^2 \theta)} $$

Finally, using the identity $1 + 4\cot^2 \theta = 1 + \frac{4}{\tan^2 \theta} = \frac{\tan^2 \theta + 4}{\tan^2 \theta}$:

$$ \mu = \frac{8 v^2 \tan^2 \theta}{g l \sin \theta (4 + \tan^2 \theta)} $$
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