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A thin, rigid, uniform rod has a mass of \(2.00 \mathrm{kg}\) and a length of \(2.00 \mathrm{m}\). (a) Find the moment of inertia of the rod relative to an axis that is perpendicular to the rod at one end. (b) Suppose all the mass of the rod were located at a single point. Determine the perpendicular distance of this point from the axis in part (a), such that this point particle has the same moment of inertia as the rod does. This distance is called the radius of gyration of the rod.

Short Answer

Expert verified
(a) The moment of inertia is \(2.67 \: \text{kg} \cdot \text{m}^2\). (b) The radius of gyration is approximately \(1.16 \: \text{m}\).

Step by step solution

01

Understand the Physical System

We have a rod of mass \(2.00 \: \text{kg}\) and length \(2.00 \: \text{m}\). The task is to find the moment of inertia about an axis that is perpendicular to the rod and passes through one of its ends.
02

Apply the Formula for Moment of Inertia

The formula for the moment of inertia \(I\) of a rod about an axis through one end is \[I = \frac{1}{3} m L^2\] where \(m\) is the mass and \(L\) is the length of the rod.
03

Substitute the Given Values

Substitute \(m = 2.00 \: \text{kg}\) and \(L = 2.00 \: \text{m}\) into the formula to find the moment of inertia: \[I = \frac{1}{3} \times 2.00 \: \text{kg} \times (2.00 \: \text{m})^2 = \frac{1}{3} \times 2.00 \times 4.00 = \frac{8.00}{3} = 2.67 \: \text{kg} \cdot \text{m}^2\]
04

Define the Radius of Gyration

The radius of gyration \(k\) is defined such that when the entire mass is concentrated at this distance from the axis, it has the same moment of inertia. The relation is given by \(I = m k^2\).
05

Solve for the Radius of Gyration

We already have \(I = 2.67 \: \text{kg} \cdot \text{m}^2\) and \(m = 2.00 \: \text{kg}\). Substitute these into the equation \(I = m k^2\): \[2.67 = 2.00 \cdot k^2\]Solve for \(k\):\[k^2 = \frac{2.67}{2} = 1.335\] \[ k = \sqrt{1.335} \approx 1.16 \: \text{m}\].

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Key Concepts

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

Radius of Gyration
The concept of radius of gyration is essential in understanding how a body's mass distribution affects its rotational characteristics. In simple terms, it is the radius at which the entire mass of the object could be concentrated without changing the object's moment of inertia.
The radius of gyration (\( k \)) is used when we want to simplify the analysis of a complex shape into an equivalent point mass. This is particularly useful in physical system analysis where objects are often modeled as point masses to make calculations easier.
  • Formula: \( I = m k^2 \)
  • Where \( I \) is the moment of inertia and \( m \) is the total mass.
Using this concept, one can determine the effective distance from the axis of rotation where you could "compact" the mass to achieve the same rotational effect.
Uniform Rod
A uniform rod is a solid body with a consistent mass distribution along its length. This means that each segment of the rod has the same density and any segment has a proportional amount of mass compared to its length. Understanding how its mass and shape contribute to rotational inertia is important for physics, especially when analyzing rotational dynamics.
When analyzing the moment of inertia of such a rod about an axis, we often use simplified formulas due to the uniformity of the mass distribution.
  • For example, the formula for the moment of inertia about an axis through one end of the rod, and perpendicular to its length, is \( I = \frac{1}{3} m L^2 \)
  • Where \( m \) is the mass and \( L \) is the length of the rod.
This relationship helps us understand how the inertia depends not just on mass, but also significantly on how and where this mass is distributed.
Perpendicular Axis
The perpendicular axis theorem is a crucial concept when examining the rotational properties of rigid bodies. It states that the moment of inertia of a flat, two-dimensional object about an axis perpendicular to its plane can be determined by the sum of the moments of inertia about two perpendicular axes in its plane.
This principle is especially important when analyzing objects that are symmetrical and when dealing with complex systems where simplification requires understanding the effects of different rotational axes.
  • In particular, for the exercise, the axis is perpendicular to the uniform rod and passes through one end, meaning it's essential to apply the standard formula for moment of inertia at such an axis.
  • This kind of analysis simplifies complex rotational problems into manageable mathematical equations, making it easier to predict and analyze rotational behaviors.
Physical System Analysis
Physical system analysis involves evaluating the physical characteristics and behaviors of systems to understand and predict their responses under various conditions. This integrates concepts like moment of inertia, mass distribution, and axes of rotation to develop a comprehensive understanding of physical motion.
The analysis starts by understanding basic quantities such as mass, shape, and rotational properties of the system. For a uniform rod, this means looking at its mass distribution, length, and applied forces.
  • By breaking down the system into fundamental concepts, like mass concentrated at a radius of gyration, complex problems are simplified.
  • Understanding these principles allows one to assess the stability, efficiency, and expected performance of mechanical structures and predict the impact of varying loads or configurations.
Overall, effectively analyzing physical systems involves combining mathematical theory with practical insights into material behavior and engineering constraints, contributing to safer and more reliable designs.

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Most popular questions from this chapter

Three objects lie in the \(x, y\) plane. Each rotates about the \(z\) axis with an angular speed of 6.00 rad/s. The mass \(m\) of each object and its perpendicular distance \(r\) from the \(z\) axis are as follows: (1) \(m_{1}=6.00 \mathrm{kg}\) and \(r_{1}=2.00 \mathrm{m},(2) m_{2}=4.00 \mathrm{kg}\) and \(r_{2}=1.50 \mathrm{m}\) (3) \(m_{3}=3.00 \mathrm{kg}\) and \(r_{3}=3.00 \mathrm{m} .\) (a) Find the tangential speed of each object. (b) Determine the total kinetic energy of this system using the expression \(\mathrm{KE}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+\frac{1}{2} m_{3} v_{3}^{2}\) (c) Obtain the moment of inertia of the system. (d) Find the rotational kinetic energy of the system using the relation \(\mathrm{KE}_{4}=\frac{1}{2} \mathrm{I\omega}^{2}\) to verify that the answer is the same as the answer to (b).

A 15.0-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2}\), and its radius is \(0.160 \mathrm{m} .\) When the reel is turning. friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N}\), how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{kg})\) is rotating at an angular velocity of 13.1 rad/s (see the drawing). The front brake is then applied for \(3.0 \mathrm{s},\) and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{m}\). The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s} .\) Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.4 \mathrm{rad} / \mathrm{s}\). The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 7700 rev/min. The motorcycle rider forgets to throttle back, so the engine"s angular speed increases to 12.500 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 3.8 rev/min. Calculate the ratio \(I_{\mathrm{L}} / I_{\mathrm{M}}\) of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

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