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Measuring \(I .\) As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.740 \(\mathrm{m}\) and find that it weighs 280 \(\mathrm{N.}\) You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang an 8.00 -kg mass from the free end of the rope, as shown in Fig. 9.17 . You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 5.00 \(\mathrm{m} / \mathrm{s}\) after it has descended 2.00 \(\mathrm{m} .\) (a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center? (b) Your boss tells you that a larger \(I\) is needed. He asks you to design a wheel of the same mass and radius that has \(I=19.0 \mathrm{kg} \cdot \mathrm{m}^{2} .\) How do you reply?

Step by step solution

01

Understanding the problem

We have a wheel with a known diameter of 0.740 m, a weight of 280 N, and an 8.00 kg mass attached. The mass descends 2.00 meters and reaches a speed of 5.00 m/s. We need to compute the moment of inertia, \(I\), for part (a). In part (b), we'll consider changes to the design based on the required moment of inertia.

02

Finding the wheel's radius

To find the radius \(r\) of the wheel, divide the diameter by 2: \(r = \frac{0.740}{2} = 0.370\) m.

03

Calculating gravitational force

The gravitational force \(F_g\) on the mass is given by \(F_g = m \cdot g = 8.00 \cdot 9.81 = 78.48\) N.

04

Applying the work-energy principle

The work-energy principle states \( \Delta K = W_{net} \), where \( K \) is kinetic energy and \( W_{net} = F_g \cdot h \). The change in kinetic energy is \( \Delta K = \frac{1}{2} mv^2 = \frac{1}{2} \times 8.00 \times (5.00)^2 = 100 \text{ J} \). Therefore, \( F_g \cdot h = 100 \text{ J} \).

05

Calculating moment of inertia

Using \( T = I \alpha \) and \( T = Fr - mgh = I \frac{a}{r} \), solve for \( a = v^2/2h \). Apply this acceleration to solve for \( I \). Substituting, \( I = \frac{(F_g \cdot h - mgh)}{(r^2)} = 0.678 \text{ kg} \cdot \text{m}^2 \).

06

Analyzing Part (b)

The desired moment of inertia is 19.0 kg \cdot m^2 for a wheel with the same mass and radius. Since the mass distribution of a solid cylinder cannot be changed, adding more mass or altering shape would be required to achieve this \(I\).

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Key Concepts

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

Work-Energy Principle

The work-energy principle is essential in understanding the physics of motion. It tells us how energy transforms into motion, by stating that the work done on an object results in a change in its kinetic energy. To grasp this, imagine the wheel as a system where different forces act to do work. In the scenario, gravitational force acts on the hanging mass causing it to descend, which in turn rotates the wheel. The work done by this force is what eventually changes the system's energy.
In this exercise, we calculate the work done as the product of gravitational force and the height the mass descends:

• Work = Gravitational Force imes Height = 78.48 ext{ N} imes 2 ext{ m} = 156.96 ext{ J}

This value tells us how much energy was converted into kinetic energy as the mass fell and the wheel rotated. The principle shows how forces lead to motion, which is crucial in engineering designs like our wheel setup.

Kinetic Energy

Kinetic energy is the energy of motion, and it's vital to understand how it applies to rotating systems like the wheel in our problem. For any object in motion, kinetic energy is given by

• Kinetic Energy (K) = \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity.

The hanging mass gains speed as it descends, converting gravitational potential energy into kinetic energy. Upon descent, the system gained kinetic energy measured by the increase in velocity:

• \(K = \frac{1}{2} \times 8.00 \times (5.00)^2 = 100 \text{ J}\).

This energy relates directly to the angular velocity of the wheel and demonstrates how energy conservation applies to both linear and rotational dynamics. In designs, knowing how kinetic energy translates to system movements aids in predicting system behavior and optimizing performance.

Gravitational Force

Gravitational force is the pulling force that acts between any two masses. In simpler terms, it is the reason why the 8.00-kg mass hanging from the wheel descends. This force can be calculated with the formula:

• Gravitational Force (\(F_g\)) = mass \(m \times\) acceleration due to gravity \(g\).

For the problem, that gravitational force lifting the wheel is

• \(F_g = 8.00 \times 9.81 = 78.48 \text{ N}\).

This constant force is crucial in rotational systems as it provides the torque necessary to start turning the wheel. In any wheel-based engineering design, analyzing gravitational effects ensures that mechanical systems can handle the expected forces without failing, ensuring safety and efficiency.

Wheel Design

Designing a wheel with a specific moment of inertia involves understanding how mass distribution affects rotational dynamics. The moment of inertia, \(I\), depends not just on mass, but on how that mass is positioned relative to the rotation axis.

• The formula for a solid cylinder, like a wheel, is \(I = \frac{1}{2} m r^2\).

Designing a wheel with a larger \(I\) means altering its mass distribution. Options include:

• Adding more mass to the rim, which shifts mass distribution outward.
• Increasing the wheel's radius.

For engineers, this involves material selection and geometry to achieve desired specifications without exceeding weight or size limitations. Understanding wheel design principles allows for optimizing systems for stability and performance, critical in applications ranging from vehicles to turbines.