Problem 4: Radii of Bicycle Tracks
Geometric Constraints
When a bicycle moves in a circular path of constant speed, the instantaneous center of rotation $O$ lies along the extension of the rear axle (since the rear wheel is fixed in orientation relative to the frame). The front wheel is steered such that its axle is perpendicular to the radius from $O$.
This creates a right-angled triangle formed by:
- The radius of the rear wheel’s track, $R_{rear}$.
- The wheelbase of the bicycle, $l$.
- The radius of the front wheel’s track, $R_{front}$.
From the Pythagorean theorem:
$$ R_{front}^2 = R_{rear}^2 + l^2 $$
Kinematic Relationship
The speed of a wheel in pure rolling motion is $v = \omega a$, where $a$ is the wheel radius and $\omega$ is the angular velocity of the wheel.
The linear speed of the axle along the circular track is given by $v = \Omega R_{track}$, where $\Omega$ is the angular velocity of the bike around the turn center $O$.
We are given that the front wheel rotates $\eta$ times faster than the rear wheel:
$$ \omega_{front} = \eta \omega_{rear} $$
Since the wheels are identical (same radius $a$), their linear speeds are proportional to their angular speeds:
$$ v_{front} = \eta v_{rear} $$
Substituting $v = \Omega R_{track}$:
$$ \Omega R_{front} = \eta (\Omega R_{rear}) \implies R_{front} = \eta R_{rear} $$
Solving for Radii
Substitute $R_{front} = \eta R_{rear}$ into the geometric equation:
$$ (\eta R_{rear})^2 = R_{rear}^2 + l^2 $$
$$ R_{rear}^2 (\eta^2 – 1) = l^2 $$
$$ R_{rear} = \frac{l}{\sqrt{\eta^2 – 1}} $$
Similarly for the front wheel:
$$ R_{front} = \eta R_{rear} = \frac{\eta l}{\sqrt{\eta^2 – 1}} $$
$$ \text{Rear wheel radius: } \frac{l}{\sqrt{\eta^2 – 1}} $$
$$ \text{Front wheel radius: } \frac{\eta l}{\sqrt{\eta^2 – 1}} $$
Geometric Constraints
When a bicycle moves in a circular path of constant speed, the instantaneous center of rotation $O$ lies along the extension of the rear axle (since the rear wheel is fixed in orientation relative to the frame). The front wheel is steered such that its axle is perpendicular to the radius from $O$.
This creates a right-angled triangle formed by:
- The radius of the rear wheel’s track, $R_{rear}$.
- The wheelbase of the bicycle, $l$.
- The radius of the front wheel’s track, $R_{front}$.
From the Pythagorean theorem:
$$ R_{front}^2 = R_{rear}^2 + l^2 $$Kinematic Relationship
The speed of a wheel in pure rolling motion is $v = \omega a$, where $a$ is the wheel radius and $\omega$ is the angular velocity of the wheel.
The linear speed of the axle along the circular track is given by $v = \Omega R_{track}$, where $\Omega$ is the angular velocity of the bike around the turn center $O$.
We are given that the front wheel rotates $\eta$ times faster than the rear wheel:
$$ \omega_{front} = \eta \omega_{rear} $$Since the wheels are identical (same radius $a$), their linear speeds are proportional to their angular speeds:
$$ v_{front} = \eta v_{rear} $$Substituting $v = \Omega R_{track}$:
$$ \Omega R_{front} = \eta (\Omega R_{rear}) \implies R_{front} = \eta R_{rear} $$Solving for Radii
Substitute $R_{front} = \eta R_{rear}$ into the geometric equation:
$$ (\eta R_{rear})^2 = R_{rear}^2 + l^2 $$ $$ R_{rear}^2 (\eta^2 – 1) = l^2 $$ $$ R_{rear} = \frac{l}{\sqrt{\eta^2 – 1}} $$Similarly for the front wheel:
$$ R_{front} = \eta R_{rear} = \frac{\eta l}{\sqrt{\eta^2 – 1}} $$