Answer
(a) $I = 1.93~kg~m/s$
(b) $I = 6.53~kg~m/s$
(c) I = 0
(d) $I = 1.15~kg~m/s$
Work Step by Step
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) $I = mr^2 = (4.60~kg)(0.500~m)^2$
$I = 1.15~kg~m/s$
