## Physics (10th Edition)

Published by Wiley

# Chapter 9 - Rotational Dynamics - Problems - Page 249: 78

#### Answer

The rolling wheel: $KE=54J$ and $h_{max}=2.76m$ The sliding wheel: $KE=36J$ and $h_{max}=1.84m$

#### Work Step by Step

a) The rolling wheel's total kinetic energy: $$KE=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$$ The wheel has a rolling motion, so $\omega=\frac{v}{R}$. Its moment of inertia has the formula $I=\frac{1}{2}mR^2$ Therefore, its rotational KE is $\frac{1}{2}I\omega^2=\frac{1}{4}mR^2\frac{v^2}{R^2}=\frac{1}{4}mv^2$ $$KE=\frac{3}{4}mv^2=\frac{3}{4}(2kg)(6m/s)^2=54J$$ a) The sliding wheel's total kinetic energy: $$KE=\frac{1}{2}mv^2=\frac{1}{2}(2kg)(6m/s)^2=36J$$ b) According to the principle of conservation of mechanical energy, $$E_f=E_0$$ $$KE_f+PE_f=KE_0+PE_0$$ For both wheels, they start off from the ground, so their initial potential energy $PE_0=0$. As they reach their maximum height, their speed is zero, $KE_f=0$. Therefore, $$PE_f=KE_0$$ $$mgh_f=KE_0$$ $$h_{max}=\frac{KE_0}{mg}=\frac{KE_0}{19.6}$$ a) The rolling wheel has $KE_0=54J$, so $h_{max}=2.76m$ b) The sliding wheel has $KE_0=36J$, so $h_{max}=1.84m$

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