91Ó°ÊÓ

Friction is a force that resists the relative motion of solid surfaces. It plays a crucial role in circular motion, especially in problems involving circular tracks or curves. The coefficient of static friction (\( \text{μ}_s \)) is a dimensionless number that characterizes the frictional force between two surfaces at rest relative to each other.
Static friction prevents slipping and is usually higher than kinetic friction, which comes into play when objects are in motion.

In our exercise, the coefficient of static friction between the bicycle tires and the track is given as 0.32. This value determines how much frictional force is available to keep the bicyclist from sliding off while negotiating a turn.

Remember, the frictional force can be calculated using the formula:
\( F_f = \text{μ}_s \times N \)
where \( N \) is the normal force, which in this case equals the weight of the bicyclist (\( mg \)).
Consequently, the frictional force (\( F_f \)) providing the centripetal force (\( F_c \)) is: \( F_f = \text{μ}_s mg \).

This frictional force allows the bicycle to travel in a circular path without slipping outwards.

Centripetal force is the force that keeps an object moving in a circular path. It acts as a 'pull' towards the center of the circle. For a bicyclist on a curved track, this force is crucial in maintaining their circular motion.

The equation for centripetal force (\( F_c \)) is: \[ F_c = \frac{mv^2}{r} \] where:

• \( m \) = mass of the bicyclist
• \( v \) = velocity of the bicyclist
• \( r \) = radius of the track

This relation tells us that centripetal force depends directly on the mass and the square of the velocity, and inversely on the radius of the circular path.
In our problem, the frictional force (\( F_f \)) supplies the necessary centripetal force (\( F_c \)).
We equate \( F_c \) to \( F_f \) to ensure that the bicyclist stays in motion without slipping: \[ F_f = F_c \] By substituting \( F_f = \text{μ}_s mg \) and \( F_c = \frac{mv^2}{r} \), and solving for the smallest radius (\( r \)), we ensure the correct balance of forces.

Unit conversion is essential in physics to make sure all calculations are consistent and correct. Units need to be in agreement to use standard formulas properly. In our exercise, the speed is given in kilometers per hour (km/h), but we need it in meters per second (m/s) to fit into our equations.

The conversion factor from km/h to m/s is \( 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} \).
So, \( 29 \text{ km/h} \) is converted as:
\[ 29 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 8.06 \text{ m/s} \]
This is a common conversion used in physics to ensure that speeds are compatible with other units like meters and seconds, which are standard for most equations.
Proper unit conversion prevents errors and makes equations easier to handle and understand. Always double-check your units, especially when substituting values into formulas.
Conversions make calculations smoother and more precise, leading to accurate solutions like the one we achieved for the radius of the circular path in this problem.