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

(a) $I = 1.93~kg~m/s$ (b) $I = 6.53~kg~m/s$ (c) I = 0 (d) $I = 1.15~kg~m/s$
Let $m_b$ be the mass of each ball and let $m_u$ be the mass of the uniform bar. (a) $I = \frac{1}{12}m_uL^2+(2\times m_br^2)$ $I = \frac{1}{12}(4.00~kg)(2.00~m)^2+2\times (0.300~kg)(1.00~m)^2$ $I = 1.93~kg~m/s$ (b) $I = \frac{1}{3}m_uL^2+0 + m_br^2$ $I = \frac{1}{3}(4.00~kg)(2.00~m)^2+(0.300~kg)(2.00~m)^2$ $I = 6.53~kg~m/s$ (c) Since the distance of all the mass from the axis of rotation is zero, $I = 0$. (d) $I = mr^2 = (4.60~kg)(0.500~m)^2$ $I = 1.15~kg~m/s$