# Chapter 11 - Section 11.4 - Conservation of Angular Momentum - Example - Page 194: 11.2

$2.5 \times 10^5$ revolutions per day

#### Work Step by Step

We know angular momentum is conserved, so we find: $\frac{2}{5}MR_1^2\omega_1=\frac{2}{5}MR_2^2\omega_2^2 \\ \omega_2 = \frac{R_1^2}{R_2^2}\times \omega_1 = \frac{(2\times 10^7)^2}{(6 \times 10^3)^2} \times \frac{1}{45} = 2.5 \times 10^5$

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