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A bicycle wheel has a radius of \(0.330 \mathrm{~m}\) and a rim whose mass is \(1.20 \mathrm{~kg} .\) The wheel has 50 spokes, each with a mass of \(0.010 \mathrm{~kg}\). (a) Calculate the moment of inertia of the rim about the axle. (b) Determine the moment of inertia of any one spoke, assuming it to be a long, thin rod that can rotate about one end. (c) Find the total moment of inertia of the wheel, including the rim and all 50 spokes.

Short Answer

Expert verified
The total moment of inertia of the wheel is 0.14883 kg⋅m².

Step by step solution

01

Calculate the Moment of Inertia of the Rim

The moment of inertia of the rim is calculated using the formula for a thin ring: \( I = m r^2 \),where \( m \) is the mass of the rim and \( r \) is the radius. In this case, \( m = 1.20 \text{ kg} \) and \( r = 0.330 \text{ m} \).Thus, the moment of inertia of the rim is:\[ I_{ ext{rim}} = 1.20 imes (0.330)^2 = 0.13068 \text{ kg} \cdot \text{m}^2. \]
02

Calculate the Moment of Inertia of One Spoke

To find the moment of inertia of a spoke, treated as a long thin rod rotating about one end, we use the formula: \( I = \frac{1}{3} m L^2 \),where \( m \) is the mass of the spoke and \( L \) is the length (equal to the radius of the wheel).Given \( m = 0.010 \text{ kg} \) and \( L = 0.330 \text{ m} \), the moment of inertia is:\[ I_{ ext{spoke}} = \frac{1}{3} \times 0.010 \times (0.330)^2 = 0.000363 \text{ kg} \cdot \text{m}^2. \]
03

Calculate the Total Moment of Inertia of All Spokes

The total moment of inertia for all 50 spokes is the moment of inertia for one spoke times 50. Thus:\[ I_{ ext{total spokes}} = 50 \times 0.000363 = 0.01815 \text{ kg} \cdot \text{m}^2. \]
04

Determine the Total Moment of Inertia of the Wheel

The total moment of inertia of the wheel is the sum of the moment of inertia of the rim and the total moment of inertia for all spokes:\[ I_{ ext{total}} = I_{ ext{rim}} + I_{ ext{total spokes}} = 0.13068 + 0.01815 = 0.14883 \text{ kg} \cdot \text{m}^2. \]

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Key Concepts

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

Bicycle Wheel
The bicycle wheel is a critical component in understanding rotational dynamics. It's more than just a device for transport; it's an ideal example of rotational motion in physics. The wheel consists mainly of three parts: the rim, spokes, and hub.

The rim is the outer circular band of the wheel and is typically responsible for most of the mass. Understanding its moment of inertia is important because this defines how hard it is to spin the wheel about its axle. The moment of inertia for the rim is computed using the formula for a thin ring, which considers its mass and radius.

Spokes are thin cylindrical rods that connect the rim to the hub. While their mass may seem negligible individually, collectively they contribute significantly to the wheel's inertia. Each spoke can be modeled as a long, thin rod, and its rotational inertia can be calculated based on its mass and length. The combination of these elements creates a system that efficiently transfers rotational force into motion. Understanding these components is essential when analyzing problems involving physical systems in rotation.
Physics Problem Solving
Physics problem solving requires a strategic approach that involves understanding the problem, formulating equations, and performing calculations carefully. In this exercise, we dealt with calculating the moment of inertia for a bicycle wheel, a common task in physics related to rotational dynamics.

  • **Understanding the System**: First, we assess the components of the bicycle wheel — the rim and the spokes. Each part contributes differently to the moment of inertia.
  • **Setting Up Equations**: We use formulas that correspond to the shape and distribution of mass in each part. For instance, for the rim, we use the formula for a thin ring, and for the spoke, we treat it like a rod rotating about an end.
  • **Performing Calculations**: With known masses and dimensions, we perform calculations, ensuring units are consistent and correctly applied in formulas.

This systematic approach ensures the problem is broken down into manageable calculations, allowing for accurate results. Developing a methodical process for tackling such problems is vital in physics and improves capability to solve complex tasks.
Rotational Dynamics
Rotational dynamics is the branch of physics that deals with the motion of objects rotating about an axis. Unlike translational motion, it involves quantities like angular velocity and moment of inertia.

The moment of inertia is a measure of how difficult it is to change an object's rotational speed. It can be thought of as the rotational analog of mass in linear dynamics. For the bicycle wheel, the rim and spokes each have their own moments of inertia.

Understanding **rotational inertia** also helps in comprehending energy conservation principles in rotating systems. The kinetic energy in rotational motion can be described using the formula: \[ KE_{rot} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

Rotational dynamics becomes essential in understanding how forces applied to the wheel result in acceleration and sustained motion. The insights gained from analyzing rotational dynamics and moments of inertia are pivotal in designing and optimizing systems involving wheels, rotors, and even celestial bodies.

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Most popular questions from this chapter

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about \(9.3 \times 10^{19} \mathrm{~J}\).

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (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 \(\mu_{k}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

In San Francisco a very simple technique is used to turn around a cable car when it reaches the end of its route. The car rolls onto a turntable, which can rotate about a vertical axis through its center. Then, two people push perpendicularly on the car, one at each end, as shown in the drawing. The turntable is rotated one-half of a revolution to turn the car around. If the length of the car is \(9.20 \mathrm{~m}\) and each person pushes with a \(185-\mathrm{N}\) force, what is the magnitude of the net torque applied to the car?

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. This distance is called the radius of gyration of the rod.

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 \(-2.4 \mathrm{rad} /\) 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?

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