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Chapter 11: Problem 42

A thin rod of length \(\ell\) and mass \(M\) rotates about a vertical axis through its center with angular velocity \(\omega .\) The rod makes an angle \(\phi\) with the rotation axis. Determine the magnitude and direction of \(\overrightarrow{\mathbf{L}}\).

Short Answer

Expert verified
Magnitude is \( \frac{1}{12} M\ell^2 \omega \), direction is along the axis inclined at \( \phi \).

Step by step solution

01

Understand the Problem

We have a thin rod with rotational characteristics, specifically mass, length, and angular velocity. The rod rotates about its center at an angle. We need to find its angular momentum vector, which involves direction and magnitude.
02

Define the Angular Momentum Formula

Angular momentum for a rigid body rotating about a fixed axis is given by \[ \overrightarrow{\mathbf{L}} = \mathbf{I} \cdot \overrightarrow{\mathbf{\omega}} \]where \( \mathbf{I} \) is the moment of inertia and \( \overrightarrow{\mathbf{\omega}} \) is the angular velocity vector.
03

Determine Moment of Inertia for the Rod

The moment of inertia for a rod of length \( \ell \) rotating about its center is\[ \mathbf{I} = \frac{1}{12} M\ell^2 \] This measures how the mass is distributed relative to the axis of rotation.
04

Calculate the Angular Velocity Vector

The rod rotates with angular velocity \( \omega \), and it makes an angle \( \phi \) with the vertical axis. Thus, the angular velocity vector considering its direction is \( \overrightarrow{\mathbf{\omega}} = \omega \hat{n} \), where \( \hat{n} \) is the unit vector representing the axis of rotation.
05

Determine the Components of Angular Momentum

Substitute the values into the angular momentum formula:\[ \overrightarrow{\mathbf{L}} = \frac{1}{12} M\ell^2 \omega \hat{n} \]This shows the magnitude is \( \frac{1}{12} M\ell^2 \omega \) in the direction of \( \hat{n} \).
06

Direction of Angular Momentum

Since the rod makes an angle \( \phi \) with the axis, the direction of the angular momentum vector \( \overrightarrow{\mathbf{L}} \) is along the axis at an angle \( \phi \). This is perpendicular to the plane containing the rotation axis and the rod.

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Key Concepts

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

Moment of Inertia
To understand how a body rotates, understanding its moment of inertia is essential. Moment of inertia, denoted as \( I \), is like a rotational equivalent of mass. It tells us how much resistance an object poses against changes in its rotational motion.
For a thin rod rotating about its center, the moment of inertia is calculated by:
  • \( I = \frac{1}{12} M\ell^2 \)
  • \( M \) is the mass of the rod.
  • \( \ell \) is the length of the rod.
This formula reveals how the distribution of the rod's mass about the rotational axis influences its resistance to rotational change. The further the mass is from the axis, the greater the moment of inertia. Set against other rotating objects, the rod's moment of inertia will be different depending on its shape and rotational axis position.
In simpler terms, for a thin rod, the longer and heavier it is, the more difficult it becomes to rotate it quickly.
Angular Velocity
Angular velocity is a measure of how fast something spins around an axis. Represented as \( \omega \), it's analogous to linear velocity, but for rotational motion. It provides us information about the speed and direction of the rotation.
In scenarios involving rotational motion, you'll often deal with an angular velocity vector. This vector includes but isn't limited to:
  • Magnitude: how fast the rotation occurs (e.g., radians per second).
  • Direction: defined along the axis of rotation, given by a unit vector \( \hat{n} \).

The angular velocity is crucial as it dictates not only how quickly something spins but also affects quantities like angular momentum. In our exercise, the angular velocity is given as \( \omega \) and is oriented in the direction of the unit vector \( \hat{n} \), due to the rod's alignment during its rotation. The speed isn't just about the number but also understanding that it has components in multiple directions in three-dimensional space.
Rigid Body Rotation
Rigid body rotation refers to the movement of rotating objects that do not deform. Each part of the body maintains a constant distance from every other part during motion. This simplification makes calculations feasible for most real-world applications.
For rigid bodies like our rotating rod, understanding its rotation relies on parameters such as:
  • Moment of inertia: defines how the mass is spread out relative to the axis.
  • Angular velocity: tells us about the rotational speed and direction.

Relying on these, angular momentum \( \overrightarrow{\mathbf{L}} = \mathbf{I} \cdot \overrightarrow{\mathbf{\omega}} \) becomes a primary concern, fusing both inertia and velocity into a single concept. This translates to practical insights like how energy and forces interact during motion. Consequently, rigid body rotation underscores many physical applications, from simple paddles in water to more complex machinery in engineering.

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Most popular questions from this chapter

(II) A particle is located at \(\vec{\mathbf{r}}=(4.0 \mathbf{i}+3.5 \hat{\mathbf{j}}+6.0 \hat{\mathbf{k}}) \mathrm{m}\) . A force \(\vec{\mathbf{F}}=(9.0 \hat{\mathbf{j}}-4.0 \hat{\mathbf{k}}) \mathrm{N}\) acts on it. What is the torque, calculated about the origin? components of the linear acceleration are: \(\mathbf{a}_{\mathrm{tan}}=\overline{\boldsymbol{\alpha}} \times \mathbf{r}\) and \(\mathbf{a}_{\mathrm{R}}=\vec{\epsilon} \times \mathbf{v}\)

Competitive ice skaters commonly perform single, double, and triple axel jumps in which they rotate \(1 \frac{1}{2}, 2 \frac{1}{2},\) and \(3 \frac{1}{2}\) revolutions, respectively, about a vertical axis while airborne. For all these jumps, a typical skater remains airborne for about 0.70 s. Suppose a skater leaves the ground in an "open" position (e.g., arms outstretched) with moment of inertia \(I_{0}\) and rotational frequency \(f_{0}=1.2 \mathrm{rev} / \mathrm{s},\) maintaining this position for \(0.10 \mathrm{~s}\). The skater then assumes a "closed" position (arms brought closer) with moment of inertia I, acquiring a rotational frequency \(f,\) which is maintained for 0.50 s. Finally, the skater immediately returns to the "open" position for 0.10 s until landing (see Fig. \(11-49\) ). (a) Why is angular momentum conserved during the skater's jump? Neglect air resistance. (b) Determine the minimum rotational frequency \(f\) during the flight's middle section for the skater to successfully complete a single and a triple axel. (c) Show that, according to this model, a skater must be able to reduce his or her moment of inertia in midflight by a factor of about 2 and 5 in order to complete a single and triple axel, respectively.

A potter's wheel is rotating around a vertical axis through its center at a frequency of \(1.5 \mathrm{rev} / \mathrm{s} .\) The wheel can be considered a uniform disk of mass \(5.0 \mathrm{~kg}\) and diameter \(0.40 \mathrm{~m} .\) The potter then throws a \(2.6-\mathrm{kg}\) chunk of clay, approximately shaped as a flat disk of radius \(8.0 \mathrm{~cm},\) onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it?

(1I) An Atwood machine (Fig. 16) consists of two masses, \(m_{\mathrm{A}}=7.0 \mathrm{kg}\) and \(m_{\mathrm{B}}=8.2 \mathrm{kg}\) . connected by a cord that passes over a pulley free to rotate about a fixed axis. The pulley is a solid cylinder of radius \(R_{0}=0.40 \mathrm{m}\) and mass 0.80 \(\mathrm{kg}\) . (a) Determine the acceleration \(a\) of each mass. (b) What percentage of error in \(a\) would be made if the moment of inertia of the pulley were ignored? Ignore friction in the pulley bearings.

A particle of mass \(m\) uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius \(R\) : $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta+\hat{\mathbf{j}} R \sin \theta $$ with \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2},\) where the constants \(\omega_{0}\) and \(\alpha\) are the initial angular velocity and angular acceleration, respectively. Determine the object's tangential acceleration \(\overrightarrow{\mathbf{a}}_{\tan }\) and determine the torque acting on the object using \((a) \vec{\tau}=\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\) (b) \(\vec{\tau}=I \vec{\alpha}\)
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