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Chapter 6: Problem 81

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 net force on the bicycle from the road.

Short Answer

Expert verified
(a) The force of friction is 275.4 N. (b) The net force is 275.4 N.

Step by step solution

01

Understand the Problem

We need to calculate two things: (a) the force of friction on the bicycle from the road, and (b) the net force on the bicycle from the road. The bicyclist is moving in a circle, so we will use concepts of circular motion.
02

Find the Centripetal Force Required

To move in a circle, the bicycle must have a centripetal force acting towards the center of the circle. The formula for centripetal force is \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is speed, and \( r \) is the radius of the circle.Substituting in the given values:\( m = 85.0 \, \text{kg}, \, v = 9.00 \, \text{m/s}, \, r = 25.0 \, \text{m} \).\[ F_c = \frac{85.0 \times 9.00^2}{25.0} \]
03

Calculate the Force of Friction

The force of friction is what provides the centripetal force for the bicycle to move in a circle. Therefore, the magnitude of the force of friction is equal to the centripetal force.Calculate \( F_c \):\[ F_c = \frac{85.0 \times 81.0}{25.0} = \frac{6885.0}{25.0} = 275.4 \, \text{N} \]So, the force of friction is \( 275.4 \, \text{N} \).
04

Determine the Net Force

In this scenario, the net force on the bicycle from the road is due to the centripetal force needed to maintain the motion in a circle. Therefore, the net force is also \( 275.4 \, \text{N} \) acting towards the center of the circle.

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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 circle at a constant speed, it is said to be in circular motion. This type of motion is fascinating because, even though the speed is constant, the direction of the velocity keeps changing, which affects the overall dynamics. In circular motion, a force is required to keep the object moving along its curved path, and this force is directed towards the center of the circle.
  • Centripetal Force: Centripetal force is necessary to maintain an object's circular path. It pulls the object inward, towards the center of the circle.
  • Velocity Change: Although the object maintains a constant speed, the continuous change in direction means the object is constantly accelerating. This is known as centripetal acceleration.

In our bicycle example, the bike travels in a circle and requires a centripetal force to sustain this motion without drifting away from the path.
Force of Friction
The force of friction is often the unsung hero in problems involving circular motion. It is the key factor in preventing slippage and allowing objects to move in a circular track, like our bicyclist moving around the circle.
  • Role of Friction: In circular motion, friction acts as the force that supplies the necessary centripetal force to keep the object moving in a circle.
  • Friction and Surfaces: The roughness between the contact surfaces influences the amount of friction. Greater roughness usually means greater friction.

In the exercise, the force of friction is what keeps the bicycle moving in a circle. This frictional force must match the centripetal force calculated using the formula: \[ F_c = \frac{mv^2}{r} \], which in this scenario equals 275.4 N.
Net Force
The net force is essentially the sum of all forces acting on an object. In the case of circular motion, when an object moves at a constant speed along a circular path, the net force is the centripetal force itself, moving the object inward.
  • Net Force and Centripetal Force: In our scenario, the net force essentially is the centripetal force, which means our calculated net force is the same 275.4 N directed towards the circle's center.
  • Analysis of Forces: Considering other forces acting on the bicycle, like gravity and the normal force, these counterbalance each other in the vertical direction, affecting only the radial motion.

In essence, to maintain a cyclist's circular motion, the net force, which matches the centripetal force, is crucial. This net force ensures the bicycle's continuous path around the circle.

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Most popular questions from this chapter

An old streetcar rounds a flat corner of radius \(9.1 \mathrm{~m},\) at \(16 \mathrm{~km} / \mathrm{h} .\) What angle with the vertical will be made by the loosely hanging hand straps?

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 \((\mathrm{c}) F_{h}\) 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 \((\mathrm{g}) 0^{\circ}\) (h) \(30.0^{\circ}\) and (i) \(60.0^{\circ}\) ?

In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. (a) Suppose the slope angle is \(\theta=40.0^{\circ},\) the snow is dry snow with a coefficient of kinetic friction \(\mu_{k}=0.0400,\) the mass of the skier and equipment is \(m=85.0 \mathrm{~kg}\), the cross-sectional area of the (tucked) skier is \(A=1.30 \mathrm{~m}^{2}\), the drag coefficient is \(C=0.150\), and the air density is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the terminal speed? (b) If a skier can vary \(C\) by a slight amount \(d C\) by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed?

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?

What is the terminal speed of a \(6.00 \mathrm{~kg}\) spherical ball that has a radius of \(3.00 \mathrm{~cm}\) and a drag coefficient of \(1.60 ?\) The density of the air through which the ball falls is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\).
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