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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 9 / Problem 21E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: A wheel of diameter 40.0 cm starts from rest and

Chapter 9, Problem 21E

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QUESTION:

Problem 21E

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s2. Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) arad = ω2r and (b) arad = v2/r.

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QUESTION:

Problem 21E

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s2. Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) arad = ω2r and (b) arad = v2/r.

ANSWER:

Solution 21E

Step 1 of 2:

To calculate the the angular velocity at the end of two revolutions

$a)\ {\omega{}}_{z}^{2}={\omega{}}_{oz}^{2}+2{\alpha{}}_{z}(\theta{}-{\theta{}}_{0})$

${\omega{}}_{z}=\sqrt{2{\alpha{}}_{z}(\theta{}-{\theta{}}_{0})}$

$=\sqrt{2(3\ rad/{s}^{2})(4\pi{}\ rad)}$

$=8.68\ rad/s$

${a}_{rad}=\frac{{v}^{2}}{r}\ \ \ \ \ and\ \ \ \ \ we\ know\ that\ \ v=r\omega{}$

$so\ {a}_{rad}=\frac{{(r\omega{})}^{2}}{r}$

$=r{\omega{}}^{2}$

$=(0.2m)(8.68\ rad/s{)}^{2}$

$=15.1\ m/{s}^{2}$

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