Answer
(a) $81\frac{rad}{s^2}$
(b) $0.14m$
Work Step by Step
(a) We know that
$I=\frac{1}{2}MR^2$
We plug in the known values to obtain:
$I=\frac{1}{2}(0.99)(0.055)^2=0.0015Kg.m^2$
Now $\tau=rF$
$\tau=(0.055)(2.2)=0.12N.m$
We can find the angular acceleration as
$\alpha=\frac{\tau}{I}$
We plug in the known values to obtain:
$\alpha=\frac{0.121}{0.0015}=81\frac{rad}{s^2}$
(b) As $s=r\theta$
$s=r(\frac{1}{2}\alpha t^2)$
We plug in the known values to obtain:
$s=(0.055)(\frac{1}{2}(81)(0.25)^2)$
$s=0.14m$
