/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 The solid half-circular cylinder... [FREE SOLUTION] | 91Ó°ÊÓ

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The solid half-circular cylinder of mass \(m\) revolves about the \(z\) -axis with an angular velocity \(\omega\) as shown. Determine its angular momentum H with respect to the \(x-y-z\) axes.

Short Answer

Expert verified
\( H = \frac{1}{4}mr^2\omega \)

Step by step solution

01

Identify the Geometry and Motion

The solid half-circular cylinder is symmetrical about the z-axis and revolves around it at an angular velocity \( \omega \). In polar coordinates, the geometry is easier to describe.
02

Define the Moment of Inertia

For a half-cylinder revolving about the z-axis, we use its moment of inertia about that axis. The moment of inertia \( I_z \) for a full cylinder is \( \frac{1}{2}mr^2 \). Therefore, for a half-cylinder, \( I_z = \frac{1}{4}mr^2 \).
03

Apply the Angular Momentum Formula

Angular momentum \( H \, \) with respect to the z-axis can be calculated using the formula: \[ H = I_z \cdot \omega \] where \( I_z \) is the moment of inertia and \( \omega \) is the angular velocity.
04

Substitute and Calculate

Substituting the moment of inertia into the angular momentum formula: \[ H = \left( \frac{1}{4}mr^2 \right)\cdot \omega = \frac{1}{4}mr^2\omega \]

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Key Concepts

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

Moment of Inertia
Moment of inertia is a fundamental concept when studying rotational dynamics. It can be thought of as the rotational equivalent of mass in linear motion and measures an object's resistance to changes in its rotation. The moment of inertia
  • depends on the object's shape and mass distribution,
  • is calculated about the axis of rotation,
  • is expressed in units of mass times square distance (e.g., kg·m²).
The moment of inertia for different objects varies greatly. For instance, cylindrical objects, like the half-circular cylinder in our exercise, have distinct moment of inertia expressions. For the full cylinder revolving about its central axis, the formula is \(\frac{1}{2}mr^2\). Since our object is a half-cylinder, its moment of inertia is half that, \(\frac{1}{4}mr^2\). Knowing the correct formula allows us to help rotate objects about their axes efficiently.
Rotational Dynamics
Understanding rotational dynamics involves examining how forces affect objects in rotational motion. Several factors play into the dynamics of objects in rotation, and the angular momentum is a vital part of it. Angular momentum \(H\) can be compared to linear momentum for objects in translational motion. It depends significantly on:
  • The moment of inertia \(I\) of the object,
  • The angular velocity \(\omega\), which is the speed of rotation along with direction.
The formula \(H = I_x \cdot \omega\) quantifies this angular momentum. This relationship highlights how both the mass distribution and the spinning speed influence rotational dynamics. Applying this to our half-cylinder, we substitute the moment of inertia \(\frac{1}{4}mr^2\), yielding \(H = \frac{1}{4}mr^2 \omega\). Grasping these details in rotational dynamics helps us predict and control movement efficiently.
Polar Coordinates
Polar coordinates are an alternate mathematical system used when dealing with rotation. Compared to Cartesian coordinates, polar coordinates are ideal when describing systems involving circles or spiral paths. The two components of this system consist of:
  • A radial distance \(r\) from a reference point (origin),
  • An angle \(\theta\) from a reference direction (typically the x-axis).
In our specific exercise, the polar coordinates simplify the description of the half-cylinder revolving symmetrically about the \(z\)-axis. When an object rotates, associating these motions with radial distances and angles makes calculations more intuitive, especially large-scale or circular movements. This system aids in breaking down complex motion equations into simpler components, aligning conveniently with the symmetrical nature of rotational systems.

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Most popular questions from this chapter

The rotor \(B\) spins about its inclined axis \(O A\) at the angular speed \(N_{1}=200\) rev/min, where \(\beta=30^{\circ}\) Simultaneously, the assembly rotates about the vertical \(z\) -axis at the rate \(N_{2} .\) If the total angular velocity of the rotor has a magnitude of 40 rad/s, determine \(N_{2}\)

The spacecraft shown is symmetrical about its \(z\) -axis and has a radius of gyration of \(720 \mathrm{mm}\) about this axis. The radii of gyration about the \(x\) - and \(y\) -axes through the mass center are both equal to \(540 \mathrm{mm}\) When moving in space, the \(z\) -axis is observed to generate a cone with a total vertex angle of \(4^{\circ}\) as it precesses about the axis of total angular momentum. If the spacecraft has a spin velocity \(\dot{\phi}\) about its \(z\) -axis of \(1.5 \mathrm{rad} / \mathrm{s}\), compute the period \(\tau\) of each full precession. Is the spin vector in the positive or negative \(z\) -direction?

The pendulum oscillates about the \(x\) -axis according to \(\theta=\frac{\pi}{6} \sin 3 \pi t\) radians, where \(t\) is the time in seconds. Simultaneously, the shaft \(O A\) revolves about the vertical \(z\) -axis at the constant rate \(\omega_{z}=2 \pi \mathrm{rad} / \mathrm{sec}\) Determine the velocity \(\mathbf{v}\) and acceleration a of the center \(B\) of the pendulum as well as its angular acceleration \(\alpha\) for the instant when \(t=0\)

The rotor and shaft are mounted in a clevis which can rotate about the \(z\) -axis with an angular velocity \(\Omega .\) With \(\Omega=0\) and \(\theta\) constant, the rotor has an angular velocity \(\omega_{0}=-4 \mathbf{j}-3 \mathbf{k}\) rad/s. Find the velocity \(\mathbf{v}_{A}\) of point \(A\) on the rim if its position vector at this instant is \(\mathbf{r}=0.5 \mathbf{i}+1.2 \mathbf{j}+1.1 \mathbf{k} \mathrm{m} .\) What is the rim speed \(v_{B}\) of any point \(B ?\)

Determine the bending moment \(\mathbf{M}\) at the tangency point \(A\) in the semicircular rod of radius \(r\) and mass \(m\) as it rotates about the tangent axis with a constant and large angular velocity \(\omega .\) Neglect the moment \(m g r\) produced by the weight of the rod.

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