/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 A bicyclist travels in a circle ... [FREE SOLUTION] | 91Ó°ÊÓ

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A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg} .\) Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the total force on the bicycle from the road.

Short Answer

Expert verified
(a) 275.4 N; (b) 877.89 N

Step by step solution

01

Understand the given values

Identify the provided values needed for the calculations: - Radius of the circle: \[ r = 25.0 \, \text{m} \] - Constant speed: \[ v = 9.00 \, \text{m/s} \] - Mass of the bicycle and rider: \[ m = 85.0 \, \text{kg} \]
02

Calculate the centripetal force

Use the formula for centripetal force:\[ F_c = \frac{m \times v^2}{r} \]Substitute the known values:\[ F_c = \frac{85.0 \, \text{kg} \times (9.00 \, \text{m/s})^2}{25.0 \, \text{m}} \]Calculate the result:\[ F_c = \frac{85.0 \times 81.0}{25.0} \, \text{N} \]\[ F_c = 275.4 \, \text{N} \]
03

Determine the force of friction (a)

Since the force of friction is providing the centripetal force, it is equal in magnitude:\[ F_{\text{friction}} = 275.4 \, \text{N} \]
04

Calculate the normal force

The normal force equals the gravitational force acting on the bicycle and rider. Use the formula:\[ F_N = m \times g \]where \[ g = 9.81 \, \text{m/s}^2 \]Substitute the values:\[ F_N = 85.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \]Calculate the result:\[ F_N = 833.85 \, \text{N} \]
05

Calculate the total force on the bicycle from the road (b)

The total force is the vector sum of the normal force and the friction force. Use Pythagoras’ theorem:\[ F_{\text{total}} = \sqrt{F_N^2 + F_{\text{friction}}^2} \]Substitute the known values:\[ F_{\text{total}} = \sqrt{(833.85 \, \text{N})^2 + (275.4 \, \text{N})^2} \]Calculate the result:\[ F_{\text{total}} = \sqrt{695309.32 + 75839.16} \, \text{N} \]\[ F_{\text{total}} = \sqrt{771148.48} \, \text{N} \]\[ F_{\text{total}} = 877.89 \, \text{N} \]

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Key Concepts

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

circular motion
When an object moves in a circular path at constant speed, it is said to be in circular motion. Despite constant speed, the object is accelerating because its direction is continuously changing.
The acceleration that acts towards the center of the circle is called centripetal acceleration, and the force that causes this acceleration is the centripetal force.
For example, in the bicycle problem, the cyclist is moving in a circle of radius 25 meters at a speed of 9 meters per second. The centripetal force required to keep the cyclist moving in a circular path can be calculated using the formula \( F_c = \frac{m \times v^2}{r} \). This force is primarily provided by the friction between the tires and the road surface. In this particular case, the centripetal force turned out to be 275.4 Newtons.
friction force
Friction is the force that resists the motion of objects sliding against each other. There are different types of friction, but when it comes to circular motion, especially when dealing with vehicles, the most relevant is kinetic friction.
Kinetic friction comes into play when an object is moving. It provides the necessary force to keep a vehicle or object in motion along a path, in this case: a circular path.
In the bicycle exercise, the friction force exactly equals the centripetal force needed to keep the bicycle on its circular path. Therefore, the force of friction on the bicycle from the road is 275.4 Newtons. This is calculated using the centripetal force formula we discussed previously.
normal force
The normal force is the force exerted perpendicularly by a surface against an object resting upon it.
It balances the gravitational force acting on the object to keep it stationary in the vertical direction. The magnitude of the normal force is equal to the weight of the object as long as the surface itself is horizontal.
Using the exercise again, the gravitational force acting on the bicycle and rider is given by \( F_N = m \times g \), where \( g \) is the acceleration due to gravity (9.81 m/s²). For a mass of 85 kg, the normal force works out to be 833.85 Newtons.
Newton's laws of motion
Newton's laws of motion are fundamental principles that describe the relationship between the movement of an object and the forces acting on it:
  • **First Law (Inertia)**: An object at rest remains at rest, and an object in motion remains in motion with the same speed and direction, unless acted upon by an unbalanced force.
  • **Second Law (Acceleration)**: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is often written as \( F = m \times a \).
  • **Third Law (Action and Reaction)**: For every action, there is an equal and opposite reaction.
In the context of the bicycle exercise, the second law is particularly relevant. It explains how the centripetal force (friction between the tires and the road) causes the cyclist to follow a circular path. The normal force is an example of the third law, counteracting the gravitational force to keep the bicycle in equilibrium vertically.

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