## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 311: 9

#### Answer

The fraction of the total kinetic energy that is the rotational kinetic energy of the wheels is $0.0194$

#### Work Step by Step

We can find an expression for the total rotational kinetic energy of the two wheels: $KE_{rot} = 2\times \frac{1}{2}I\omega^2$ $KE_{rot} = (I)(\frac{v}{r})^2$ $KE_{rot} = (0.080~kg~m^2)(\frac{v}{0.32~m})^2$ $KE_{rot} = (0.78125~kg)~v^2$ We can find an expression for the translational kinetic energy: $KE_{tr} = \frac{1}{2}mv^2$ $KE_{tr} = \frac{1}{2}(79~kg)~v^2$ $KE_{tr} = (39.5~kg)~v^2$ We can find the fraction of the total kinetic energy that is the rotational kinetic energy of the wheels: $\frac{KE_{rot}}{KE} = \frac{KE_{rot}}{KE_{tr}+KE_{rot}}$ $\frac{KE_{rot}}{KE} = \frac{(0.78125~kg)~v^2}{(39.5~kg)~v^2+(0.78125~kg)~v^2}$ $\frac{KE_{rot}}{KE} = 0.0194$ The fraction of the total kinetic energy that is the rotational kinetic energy of the wheels is $0.0194$.

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