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Chapter 6: Problem 43

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 20.72 meters.

Step by step solution

01

Convert Speed to Meters per Second

The bicyclist's speed is given as \(29 \text{ km/h}\). First, convert this speed to meters per second (m/s) since this is the standard unit for velocity in physics problems. \[29 \text{ km/h} = \frac{29 \times 1000}{3600} \text{ m/s} = 8.06 \text{ m/s}\]
02

Understand the Formula for Centripetal Force

For an object traveling in a circle, the centripetal force required to keep it in a circular path is provided by the frictional force. The centripetal force equation is given by \[F_c = \frac{mv^2}{r}\]where \(m\) is the mass of the object, \(v\) is the speed, and \(r\) is the radius of the circle.
03

Relate Frictional Force to Centripetal Force

The frictional force \(F_f\) that prevents slipping is given by:\[F_f = \mu_s \cdot m \cdot g\]where \(\mu_s = 0.32\) is the coefficient of static friction, \(m\) is the mass of the bicyclist and bicycle, and \(g = 9.8 \text{ m/s}^2\) is the acceleration due to gravity.
04

Set Frictional Force Equal to Centripetal Force

Since the maximum frictional force provides the necessary centripetal force, set \(F_f\) equal to \(F_c\):\[\mu_s \cdot m \cdot g = \frac{mv^2}{r}\]The mass \(m\) cancels out from both sides because it is not given, simplifying the equation to:\[\mu_s \cdot g = \frac{v^2}{r}\]
05

Solve for the Radius \(r\)

Rearrange the equation to solve for the radius \(r\):\[r = \frac{v^2}{\mu_s \cdot g}\]Substitute the known values into the equation:\[r = \frac{(8.06)^2}{0.32 \times 9.8}\]Calculate:\[r = \frac{64.96}{3.136} \approx 20.72 \text{ meters}\]
06

Conclusion

The smallest radius of an unbanked track around which the bicyclist can travel without slipping is approximately 20.72 meters.

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Key Concepts

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

Friction
Friction is a force that resists the relative motion of two surfaces in contact. It is crucial in daily life, providing grip and control while walking or driving. There are two main types: kinetic friction, which acts on moving objects, and static friction, which prevents objects from starting to move.
  • In our exercise, static friction is crucial because it prevents the bicycle from slipping off the track.
  • The frictional force is proportional to the normal force and the coefficient of static friction, often represented by \( F_f = \mu_s \cdot F_n \),where \( F_f \)is the frictional force, \( \mu_s \)is the coefficient of static friction, and \( F_n \)is the normal force.
  • The normal force,usually, equals the object's weight on a flat surface and is calculated as \( m \cdot g \),where \( m \)is mass, and \( g \)is the acceleration due to gravity \( (9.8 \, \text{m/s}^2) \).
As explained in the solution, the frictional force is what allows the cyclist to travel in a circle without slipping off.
Circular Motion
Circular motion occurs when an object travels along a circular path. This movement requires a force directed towards the center of the circle, known as centripetal force. Without this force, an object would continue moving in a straight line due to inertia.
  • In the bicycle exercise, centripetal force is necessary to keep the bicycle on its circular path.
  • The formula for calculating centripetal force is \( F_c = \frac{mv^2}{r} \),where \( m \)is mass, \( v \)is speed, and \( r \)is the radius of the circle.
  • The role of centripetal force can be fulfilled by different forces, like tension, gravity, or friction. In our exercise, it is static friction that provides this force.

By understanding circular motion, one can grasp why forces must act towards the center to keep a bicyclist moving in a circle rather than shooting out tangentially.
Coefficient of Static Friction
The coefficient of static friction, denoted by \( \mu_s \),is a dimensionless quantity that represents how much frictional force can be exerted without movement. It varies depending on the materials in contact.
  • A higher \( \mu_s \)indicates that the surfaces have better grip and require a larger force to initiate sliding.
  • In the problem, \( \mu_s = 0.32 \),meaning the tires need to overcome 32% of their own weight to start slipping.
  • Using the formula \( F_f = \mu_s \cdot m \cdot g \),we see how this coefficient limits the maximum possible static friction.
The coefficient fundamentally affects how tightly a bicyclist can turn at a given speed without slipping. In the exercise, it dictates the minimum radius possible for safe circular motion. Understanding this can help in assessing risks when making sharp turns on different surfaces.

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Most popular questions from this chapter

A loaded penguin sled weighing \(80 \mathrm{~N}\) rests on a plane inclined at angle \(\theta=20^{\circ}\) to the horizontal (Fig. 6 -23). Between the sled and the plane, the coefficient of static friction is \(0.25\), and the coefficient of kinetic friction is \(0.15\). (a) What is the least magnitude of the force \(\vec{F}\), parallel to the plane, that will prevent the sled from slipping down the mum magnitude \(F\) that will start the What value of \(F\) is required to move the sled up the plane at constant velocity?

Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(C=0.75\).

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.

ssm A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68\), and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) \(222 \mathrm{~N}\), (b) \(334 \mathrm{~N}\), (c) \(445 \mathrm{~N}\), and (d) \(556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

A \(100 \mathrm{~N}\) force, directed at an angle \(\theta\) above a horizontal floor, is applied to a \(25.0 \mathrm{~kg}\) chair sitting on the floor. If \(\theta=0^{\circ}\), what are (a) the horizontal component \(F_{h}\) of the applied force and (b) the magnitude \(F_{N}\) of the normal force of the floor on the chair? If \(\theta=30.0^{\circ}\), what are (c) \(F_{h}\) and \((\) d \() F_{N}\) ? If \(\theta=60.0^{\circ}\), what are (e) \(F_{h}\) and (f) \(F_{N}\) ? Now assume that the coefficient of static friction between chair and floor is \(0.420\). Does the chair slide or remain at rest if \(\theta\) is \((\mathrm{g}) 0^{\circ},(\mathrm{h}) 30.0^{\circ}\), and (i) \(60.0^{\circ} ?\)
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