## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 319: 98

#### Answer

We can list the shapes in the order they arrive at the finish line: frictionless cube, solid sphere, solid cylinder, hollow sphere, hollow cylinder

#### Work Step by Step

Let $h$ be the vertical height of the starting position. We can use conservation of energy to find the speed $v$ of each object at the bottom of the slope. solid sphere: $U_g = KE_{tr}+KE_{rot}$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$ $Mgh = \frac{7}{10}Mv^2$ $v = \sqrt{\frac{10~gh}{7}}$ hollow sphere: $U_g = KE_{tr}+KE_{rot}$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{3}MR^2)~(\frac{v}{R})^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{3}Mv^2$ $Mgh = \frac{5}{6}Mv^2$ $v = \sqrt{\frac{6~gh}{5}}$ solid cylinder: $U_g = KE_{tr}+KE_{rot}$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)~(\frac{v}{R})^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$ $Mgh = \frac{3}{4}Mv^2$ $v = \sqrt{\frac{4~gh}{3}}$ hollow cylinder: $U_g = KE_{tr}+KE_{rot}$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(MR^2)~(\frac{v}{R})^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}Mv^2$ $Mgh = Mv^2$ $v = \sqrt{gh}$ frictionless cube: $U_g = KE_{tr}$ $Mgh = \frac{1}{2}Mv^2$ $Mgh = \frac{1}{2}Mv^2$ $v = \sqrt{2gh}$ If a shape has a larger speed at the bottom of the ramp, then it arrives to the finish line earlier since it has a larger acceleration. We can list the shapes in the order they arrive at the finish line: frictionless cube, solid sphere, solid cylinder, hollow sphere, hollow cylinder

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