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Chapter 9: Problem 2

The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is applied in the same direction to each steering wheel. What is the ratio of the torque produced by this force in the truck to the torque produced in the car?

Short Answer

Expert verified
The ratio of the torques is approximately 1.32.

Step by step solution

01

Define Torque Formula

Torque (\( \tau \)) is defined as the product of the force (\( F \)) and the radius (\( r \)) of the wheel, and it can be expressed using the formula: \( \tau = r \times F \).
02

Calculate Torque for the Car

For the car, with a radius \( r_{ ext{car}} = 0.19 \) m, the torque produced is \( \tau_{ ext{car}} = 0.19 \times F \).
03

Calculate Torque for the Truck

For the truck, with a radius \( r_{ ext{truck}} = 0.25 \) m, the torque produced is \( \tau_{ ext{truck}} = 0.25 \times F \).
04

Calculate the Ratio of Torques

The ratio of the torque produced by the truck to that produced by the car is \( \frac{\tau_{ ext{truck}}}{\tau_{ ext{car}}} = \frac{0.25 \times F}{0.19 \times F} = \frac{0.25}{0.19} \).
05

Simplify the Ratio

Simplifying \( \frac{0.25}{0.19} \) gives approximately 1.32.

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Key Concepts

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

Torque Formula
Torque is a concept in physics that helps describe how forces cause objects to rotate around an axis. To calculate torque, you use the formula:
  • \[ \tau = r \times F \]
  • Where:
    • \( \tau \) is the torque
    • \( r \) is the radius (distance from the pivot point to the point where force is applied)
    • \( F \) is the force applied
Imagine holding a wrench and turning a bolt. The distance from your hand to the bolt (akin to the radius \( r \)) and the force you apply \( F \) both affect how effectively you can turn the bolt.
If you apply a large force or use a longer wrench, you get more torque, making it easier to turn the bolt.
Radius Effect on Torque
The radius plays a critical role in determining the torque. Imagine a seesaw. The further you sit away from the pivot, the more effectively you can make the seesaw move. In the context of the steering wheel:
  • A larger radius means a greater ability to apply torque for the same force.
  • Even with a small force, a large radius can create a significant torque.
In the exercise, the truck's steering wheel has a larger radius than the car's steering wheel:
  • Truck's radius: 0.25 m
  • Car's radius: 0.19 m
Because of this, even though the same force is applied, the truck's wheel produces more torque.
That's because torque depends directly on both the radius and the force: increasing one will increase the torque. It's akin to giving power to your push!
Force and Torque Relationship
The relationship between force and torque is straightforward. Torque is the rotational counterpart of linear force. Just as force can move an object in a straight line, torque can cause an object to rotate.
In our exercise example, the same force is applied to both steering wheels. Here's how it breaks down:
  • Applying a force to an object at a greater distance from the pivot point results in greater torque.
  • The same amount of force, when applied at different radii, results in different torques.
So, when you apply the same force to a larger radius (like a truck's steering wheel), you generate more torque compared to a smaller radius (the car's steering wheel). This brings efficiency in controlling larger vehicles with bigger wheels, reflecting the importance of understanding this relationship in practical applications.

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Most popular questions from this chapter

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh \(438 \mathrm{N}\) and have a center of gravity that is \(1.28 \mathrm{m}\) above the floor. His upper legs weigh \(144 \mathrm{N}\) and have a center of gravity that is \(0.760 \mathrm{m}\) above the floor. Finally, his lower legs and feet together weigh \(87 \mathrm{N}\) and have a center of gravity that is \(0.250 \mathrm{m}\) above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m},\) an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{kg} \cdot \mathrm{m}^{2} .\) A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{s} .\) (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

A particle is located at each corner of an imaginary cube. Each edge of the cube is \(0.25 \mathrm{m}\) long. and each particle has a mass of \(0.12 \mathrm{kg}\) What is the moment of inertia of these particles with respect to an axis that lies along one edge of the cube?

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius \(R\) that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is \(4.0 R\). Assume that all of the star's original mass is contained in the shell.

In outer space two identical space modules are joined together by a massless cable. These modules are rotating about their center of mass, which is at the center of the cable because the modules are identical (see the drawing). In each module, the cable is connected to a motor, so that the modules can pull each other together. The initial tangential speed of each module is \(v_{0}=17 \mathrm{m} / \mathrm{s} .\) Then they pull together until the distance between them is reduced by a factor of two. Each module has a final tangential speed of \(v_{\mathrm{r}}\). Find the value of \(v_{\mathrm{f}}\)
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