## University Physics with Modern Physics (14th Edition)

(a) $I_2 = f^5~I_1$ (b) $KE_2 = 6.4\times 10^8~J$
(a) $I_2 = \sum m_2r_2^2$ $I_2 = \sum (f^3~m_1)~(f~r_1)^2$ $I_2 = f^5~\sum (m_1)~(r_1)^2$ $I_2 = f^5~I_1$ The original moment of inertia will be multiplied by $f^5$. (b) $KE_1 = \frac{1}{2}I_1\omega^2$ $KE_2 = \frac{1}{2}I_2\omega^2$ $KE_2 = \frac{1}{2}(f^5~I_1)\omega^2$ $KE_2 = f^5~\frac{1}{2}I_1\omega^2$ $KE_2 = f^5~KE_1$ $KE_2 = (48)^5~(2.5~J)$ $KE_2 = 6.4\times 10^8~J$