#### Answer

$I=I_{CM}+M(\frac{1}{2})^{2}=\frac{1}{12}M(1)^{2}+\frac{1}{4}M=\frac{1}{3}M$
$T=\frac{Mg}{2}$
Thus, angular acceleration =$\frac{T}{I}=\frac{3g}{2}$
a). Tangential acceleration of the 100 cm position = $ 1\times \frac{3g}{2}=1.5g$
Yes, the result is surprising as it is more than g.
b). $a_{t}=g$
$g=r \times \frac{3g}{2}$
Thus, $r=\frac{200}{3} cm=66.67cm$.
Hence, tangential acceleration is g at position 66.67 cm.