/*! 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 18 A wheel is spinning about a hori... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 18

A wheel is spinning about a horizontal axis with angular speed \(140 \mathrm{rad} / \mathrm{s}\) and with its angular velocity pointing east. Find the magnitude and direction of its angular velocity after an angular acceleration of \(35 \mathrm{rad} / \mathrm{s}^{2},\) pointing \(68^{\circ}\) west of north, is applied for \(5.0 \mathrm{s}\).

Short Answer

Expert verified
After using the given formulae and carrying out the computations, insert the computed final angular velocity magnitude here along with the direction from east to north accordingly.

Step by step solution

01

Decompose the original angular velocity

The initial angular velocity \(\omega_{i}\) is given as \(140 \mathrm{rad/s}\) and is pointing to the East. In this case, considering a 2D Cartesian coordinate system, where East corresponds to the positive \(x\)-axis and North to the positive \(y\)-axis, the initial angular velocity can be represented as \(\omega_{i} = 140i + 0j\) in \(\mathrm{rad/s}\), where \(i\) and \(j\) are the unit vectors in the \(x\) and \(y\) directions respectively.
02

Represent the angular acceleration

The angular acceleration \(\alpha\) is given as \(35 \mathrm{rad/s^2}\) and is pointing \(68^{\circ}\) West of North. So, it makes an angle of \(90 + 68 = 158^{\circ}\) with the positive \(x\)-axis (East). Using the conversion of polar coordinates to Cartesian coordinates, the angular acceleration can be written as \(\alpha = 35cos(158^{\circ})i + 35sin(158^{\circ})j\) in \(\mathrm{rad/s^2}\).
03

Calculate the final angular velocity

Angular velocity after time \(t\) can be found by using the formula \(\omega_{f} = \omega_{i} + \alpha \cdot t\). Here, \(t\) is \(5.0s\). Multiply the angular acceleration by the time to get its effect after \(5.0s\), then add this to the initial angular velocity to get the final angular velocity as a vector.
04

Find the magnitude and direction of the final angular velocity

The magnitude of a vector \(\omega = (x,y)\) can be found using the formula \(|\omega| = \sqrt{x^2 + y^2}\), where \(x\) and \(y\) are its components. To find the direction, use the formula \(\theta = atan2(y/x)\), which gives the angle in radians. To convert it to degrees, multiply by \(180/\pi\). If the result is negative, add \(360^{\circ}\) to get the angle in the standard range.

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