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# A wheel is turning about an axis through its center with ISBN: 9780321675460 31

## Solution for problem 38E Chapter 9

University Physics | 13th Edition

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Problem 38E

A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

Step-by-Step Solution:

Solution 38 E Step 1: Data given Velocity = d = 8.20 rev Time t = 12 s Kinetic energy 36.0 J We need to find moment of inertia We shall use s = u + 1/2 (at ) Here initial velocity is = 0 Hence we have 2 8.20 rev × 2 = 1/2 × a × (12 s) Rearrange to find acceleration a we get 8.20 rev×2 a = (1/2)×(12 s) a = 0.71 rev/s 2 Step 2 : We need to find velocity it is obtained using 2 = 2ad 2 2 = 2 × 0.71 rev/s × 8.20 rev × 2 2 =73.16 rad/s

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##### ISBN: 9780321675460

The full step-by-step solution to problem: 38E from chapter: 9 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 38E from 9 chapter was answered, more than 498 students have viewed the full step-by-step answer. This full solution covers the following key subjects: wheel, axis, its, center, kinetic. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?” is broken down into a number of easy to follow steps, and 59 words.

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