Answer
$T_2=4992N$ and $\alpha=6.34rad/s^2$
Work Step by Step
From figure c), according to Newton's 2nd law of motion, $$T_2-mg=ma_y$$
Because the cable pulling the crate is rolled around the smaller pulley, $a_y=l_2\alpha=0.2\alpha$. Therefore, $$T_2-mg=0.2m\alpha$$
The crate's mass $m=451kg$, $g=9.8m/s^2$, so $$T_2-4419.8=90.2\alpha$$ $$T_2-90.2\alpha=4419.8 (1)$$
Both $T_1$ and $T_2$ produce torques with lever arm $l_1$ and $l_2$ respectively. We have $$\sum\tau=T_1l_1-T_2l_2=I\alpha$$ $$T_2l_2+I\alpha=T_1l_1$$
We have $T_1=2150N$, $l_1=0.6m, l_2=0.2m, I=46kg.m^2$ $$0.2T_2+46\alpha=1290 (2)$$
Solving (1) and (2), we get $T_2=4992N$ and $\alpha=6.34rad/s^2$
