## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 317: 76

#### Answer

We can rank the flywheels in order of their angular momentum, from smallest to largest: $A = E = F \lt B \lt C \lt D$

#### Work Step by Step

We can find the angular momentum for each flywheel: Flywheel A: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(10~kg)(0.20~m)^2~(30~rad/s)$ $L = 6.0~kg~m^2/s$ Flywheel B: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(20~kg)(0.20~m)^2~(30~rad/s)$ $L = 12.0~kg~m^2/s$ Flywheel C: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(20~kg)(0.40~m)^2~(15~rad/s)$ $L = 24.0~kg~m^2/s$ Flywheel D: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(20~kg)(0.40~m)^2~(30~rad/s)$ $L = 48.0~kg~m^2/s$ Flywheel E: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(20~kg)(0.10~m)^2~(60~rad/s)$ $L = 6.0~kg~m^2/s$ Flywheel F: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(5~kg)(0.20~m)^2~(60~rad/s)$ $L = 6.0~kg~m^2/s$ We can rank the flywheels in order of their angular momentum, from smallest to largest: $A = E = F \lt B \lt C \lt D$

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