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Chapter 6: Problem 43

ILW What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is \(29 \mathrm{~km} / \mathrm{h}\) and the \(\mu_{s}\) between tires and track is \(0.32 ?\)

Short Answer

Expert verified
The smallest radius is approximately 20.57 meters.

Step by step solution

01

Understanding the Problem

We need to find the smallest radius of a flat track where a bicyclist can travel safely at a given speed, considering the static friction coefficient between the tires and the track.
02

Convert Speed Units

We convert the speed from km/h to m/s for easier calculation with SI units: \[ v = 29 \text{ km/h} = \frac{29 \times 1000}{3600} \text{ m/s} \approx 8.06 \text{ m/s} \]
03

Identify the Forces Involved

When a cyclist travels in a circular path, centripetal force is needed, which is provided by static friction between the tires and the track. The centripetal force is given by:\[ f_c = \frac{m v^2}{r} \]where \(m\) is the mass, \(v\) is the velocity, and \(r\) is the radius of the circle.
04

Relate Friction to Centripetal Force

The maximum static friction force that can be generated is given by:\[ f_s = \mu_s \times m \times g \]where \(\mu_s\) is the coefficient of static friction and \(g\) is the acceleration due to gravity \(9.8 \text{ m/s}^2\). Equating this with the centripetal force gives:\[ \mu_s \times m \times g = \frac{m \times v^2}{r} \]
05

Solve for Radius

We cancel \(m\) from both sides and solve for \(r\):\[ r = \frac{v^2}{\mu_s \times g} \]Substitute the known values:\[ r = \frac{(8.06)^2}{0.32 \times 9.8} \approx 20.57 \text{ meters} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Static Friction in Circular Motion
Static friction acts as the glue keeping our bicyclist from slipping off the track. In circular motion, static friction is the force that holds the tire to the surface of the track, preventing the bicycle from sliding outwards.
  • Static friction is the force that resists movement between two surfaces that are in contact but not moving relative to each other.
  • In the context of a bicycle on an unbanked track, it is the maximum force that keeps the bike in motion around the curve without sliding.
  • The frictional force depends on the coefficient of static friction (\[\mu_s\]) between the track and the tires.
Static friction is crucial, as it provides the necessary centripetal force needed for circular motion.
The Role of Centripetal Force
Centripetal force is an essential concept for understanding how objects move in a circular path.It is the force required to keep an object moving in a circle and acts towards the center of that circle.
  • This force is not a separate force but arises from other forces, like static friction, tension, or gravity.
  • In our exercise, static friction provided the centripetal force necessary for the bicycle's circular path.
  • The equation for centripetal force is given by \( f_c = \frac{m v^2}{r}\), where \(m\) represents mass, \(v\) is velocity, and \(r\) is the radius of the circular path.
Balancing this force with the frictional force allows us to calculate the smallest radius a bicyclist can navigate safely.
Understanding an Unbanked Track
An unbanked track is a flat path with no incline to assist in turning around curves. This type of track relies entirely on static friction to keep a bicyclist moving in a circle.
  • Unlike banked tracks, where the incline of the track helps provide centripetal force, unbanked tracks require the frictional force between the tires and the track for circular motion.
  • The biggest challenge on an unbanked track is maintaining enough static friction to counteract the centrifugal force pushing the bicyclist outward.
  • By analyzing factors like speed and the coefficient of static friction, we can determine the smallest possible radius for safe travel, as shown by \(r = \frac{v^2}{\mu_s \times g}\).
Relying on static friction solely highlights the importance of maintaining proper tire and track conditions for safety.

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Most popular questions from this chapter

(co A \(3.5 \mathrm{~kg}\) block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

A police officer in hot pursuit drives her car through a circular turn of radius \(300 \mathrm{~m}\) with a constant speed of \(80.0 \mathrm{~km} / \mathrm{h}\). Her mass is \(55.0 \mathrm{~kg}\). What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

During an Olympic bobsled run, the Jamaican team makes a turn of radius \(7.6 \mathrm{~m}\) at a speed of \(96.6 \mathrm{~km} / \mathrm{h}\). What is their acceleration in terms of \(g\) ?

An airplane is flying in a horizontal circle at a speed of \(480 \mathrm{~km} / \mathrm{h}\) (Fig. 6-41). If its wings are tilted at angle \(\theta=40^{\circ}\) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

In the early afternoon, a car is parked on a street that runs down a steep hill, at an angle of \(35.0^{\circ}\) relative to the horizontal. Just then the coefficient of static friction between the tires and the street surface is \(0.725 .\) Later, after nightfall, a sleet storm hits the area, and the coefficient decreases due to both the ice and a chemical change in the road surface because of the temperature decrease. By what percentage must the coefficient decrease if the car is to be in danger of sliding down the street?
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