## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 8 - Rotational Motion - Problems: 55

#### Answer

$1.63 \times 10^4 J$

#### Work Step by Step

The energy required equals the ride’s final rotational kinetic energy. The final rotational speed is 0.8976 rad/s. The rotational inertia of the solid disk is $\frac{1}{2}MR^2=40500 kg \cdot m^2$. $$KE_{rot}=\frac{1}{2}I \omega^2 \approx 1.63 \times 10^4 J$$

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