Answer
(a) $a_i = \frac{F_i}{m_i}$
(b) $\tau_i = r_i~F_i$
(c) $a_i = \alpha~r_i$
(d) $\sum_{i=1}^{N}~\tau_i = I~\alpha$
Work Step by Step
(a) $F_i = m_i~a_i$
$a_i = \frac{F_i}{m_i}$
(b) We can find the torque acting on the particle $m_i$:
$\tau_i = r_i~F_i$
(c) $a_i = \alpha~r_i$
(d) We can find the sum of all the torques:
$\sum_{i=1}^{N}~\tau_i = \sum_{i=1}^{N}~r_i~F_i$
$\sum_{i=1}^{N}~\tau_i = \sum_{i=1}^{N}~r_i~(m_i~a_i)$
$\sum_{i=1}^{N}~\tau_i = \sum_{i=1}^{N}~r_i~(m_i~\alpha~r_i)$
$\sum_{i=1}^{N}~\tau_i = \sum_{i=1}^{N}~(m_i~r_i^2)~\alpha$
$\sum_{i=1}^{N}~\tau_i = (\sum_{i=1}^{N}~(m_i~r_i^2))~\alpha$
$\sum_{i=1}^{N}~\tau_i = I~\alpha$
