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Chapter 10: Problem 89

A bicyclist of mass \(70 \mathrm{~kg}\) puts all his mass on each downwardmoving pedal as he pedals up a steep road. Take the diameter of the circle in which the pedals rotate to be \(0.40 \mathrm{~m},\) and determine the magnitude of the maximum torque he exerts about the rotation axis of the pedals.

Short Answer

Expert verified
The maximum torque exerted is 137.34 N路m.

Step by step solution

01

Analyze the Problem

We need to find the maximum torque exerted by the bicyclist. Torque is given by the formula: \( \tau = F \times r \), where \( F \) is the force and \( r \) is the lever arm (the radius of the pedal rotation). The force here is the weight exerted by the bicyclist when he applies all his weight on the pedal. The lever arm, in this case, is half of the pedal rotation diameter.
02

Calculate the Force

The force exerted by the bicyclist is his weight, which can be calculated by multiplying his mass by the acceleration due to gravity. Thus, \( F = m \times g = 70 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 686.7 \, \text{N} \).
03

Determine the Lever Arm

The diameter of the pedal rotation is given as 0.40 m. Therefore, the radius (lever arm \( r \)) is half of the diameter: \( r = \frac{0.40}{2} = 0.20 \, \text{m} \).
04

Calculate Torque

Substitute the values obtained for force \( F \) and lever arm \( r \) into the torque formula: \( \tau = F \times r = 686.7 \, \text{N} \times 0.20 \, \text{m} = 137.34 \, \text{N} \cdot \text{m} \).

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Key Concepts

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

Understanding Bicycle Mechanics
Bicycle mechanics involve a fascinating interplay of physics principles, one of which is torque. When riding, a cyclist uses their body weight to apply force to the pedals, creating a rotational effect around the pedal axle. This force is crucial, especially when riding uphill, as it determines how effectively the bicycle moves.

The pedals of a bicycle rotate in a circular path, which makes the circle's diameter essential in calculating movements like torque. By applying force on the pedals, the cyclist is interacting with various elements of physics, seemingly participating in a dance of leverage and gravitational forces. Understanding this interaction helps in designing better bicycles that enhance performance and efficiency.

In essence, cycling is not just about pedaling hard but understanding how each mechanical aspect contributes to your ride, from the chain tension to the gears' adjustment, every component playing its role.
Exploring the Lever Arm Concept
The concept of the lever arm is pivotal in understanding how torque functions. The lever arm is the perpendicular distance from the pivot point (rotation axis) to the line along which the force acts. In the case of a bicycle, the pedal's rotation serves as the axis, and the pedal itself acts as the lever arm.

For this exercise, the pedal rotation forms part of a circle, and the lever arm equates to the circle's radius. Knowing the diameter of pedal rotation gives insight into calculating effective leverage. A shorter lever arm requires more force to create the same torque, whereas a longer lever arm requires less force:
  • The diameter of the pedal rotation is given as 0.40m.
  • The radius, or lever arm, is half the diameter: 0.20m.
These measurements are essential, allowing you to determine the torque exerted by the cyclist effectively.
Force Calculation in Torque Problems
Calculating force is the first step in solving problems related to torque. Force, in physics, is usually the product of mass and acceleration (in this case, due to gravity). For the cyclist, their weight acts as the force applied to the pedals.

The formula for force due to gravity is:\[ F = m \times g \]where:
  • m represents mass (70 kg in this case).
  • g is the acceleration due to gravity, approximately 9.81 m/s虏.
Calculating this gives:\[ F = 70 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 686.7 \, \text{N} \]

In torque applications, understanding the magnitude of this force is necessary for calculating how much rotational effect can be achieved. Boldly put, the more effective the force exerted, the greater the torque generated.

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Most popular questions from this chapter

The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3},\) where \(\theta\) is in radians and \(t\) is in seconds. At \(t=0,\) what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s} ?\) (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\). (e) Is its angular acceleration constant?

Two uniform solid cylinders, each rotating about its central (longitudinal) axis at \(235 \mathrm{rad} / \mathrm{s},\) have the same mass of \(1.25 \mathrm{~kg}\) but differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius \(0.25 \mathrm{~m},\) and \((\mathrm{b})\) the larger cylinder, of radius \(0.75 \mathrm{~m} ?\)

A record turntable rotating at \(33 \frac{1}{3}\) rev \(/\) min slows down and stops in \(30 \mathrm{~s}\) after the motor is turned off. (a) Find its (constant) angular acceleration in revolutions per minute-squared. (b) How many revolutions does it make in this time?

A disk rotates at constant angular acceleration, from angular position \(\theta_{1}=10.0\) rad to angular position \(\theta_{2}=70.0\) rad in \(6.00 \mathrm{~s}\). Its angular velocity at \(\theta_{2}\) is \(15.0 \mathrm{rad} / \mathrm{s}\). (a) What was its angular velocity at \(\theta_{1} ?\) (b) What is the angular acceleration? (c) At what angular position was the disk initially at rest? (d) Graph \(\theta\) versus time \(t\) and angular speed \(\omega\) versus \(t\) for the disk, from the beginning of the motion \((\) let \(t=0\) then \()\)

A \(32.0 \mathrm{~kg}\) wheel, essentially a thin hoop with radius \(1.20 \mathrm{~m}\) is rotating at 280 rev/min. It must be brought to a stop in \(15.0 \mathrm{~s}\). (a) How much work must be done to stop it? (b) What is the required average power?
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