Answer
a) $a=8.68m/s^2$
b) $\alpha=71.7rad/s^2$
c) $\Delta y=9.77m$
Work Step by Step
(a) We can find the linear acceleration as follows:
$\Sigma F_y=-T+mg=ma$
$\implies (-\frac{1}{2}Ma)+mg=ma$
$\implies mg=(m+\frac{1}{2}M)$
This simplifies to:
$a=(\frac{m}{m+\frac{1}{2}M})g$
We plug in the known values to obtain:
$a=[\frac{2.85}{2.85+\frac{1}{2}(0.742)}](9.81)$
$a=8.68m/s^2$
(b) We can find the required angular acceleration as
$\alpha=\frac{a}{R}$
We plug in the known values to obtain:
$\alpha=\frac{8.68}{0.121}$
$\alpha=71.7rad/s^2$
(c) We can find the required distance as
$\Delta y=0+\frac{1}{2}(8.68)(1.50)^2$
$\Delta y=9.77m$
