## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 316: 63

#### Answer

The total kinetic energy of the hollow cylinder is $Mv^2$ The total kinetic energy of the solid sphere is $\frac{7}{10}Mv^2$ The total kinetic energy of the solid cylinder is $\frac{3}{4}Mv^2$ We can rank the objects by their kinetic energies, from smallest to largest: $solid~sphere \lt solid ~cylinder \lt hollow~cylinder$

#### Work Step by Step

We can find the total kinetic energy of the hollow cylinder: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(MR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}Mv^2$ $KE = Mv^2$ We can find the total kinetic energy of the solid sphere: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$ $KE = \frac{7}{10}Mv^2$ We can find the total kinetic energy of the solid cylinder: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$ $KE = \frac{3}{4}Mv^2$ We can rank the objects by their kinetic energies, from smallest to largest: $solid~sphere \lt solid ~cylinder \lt hollow~cylinder$

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