Answer
(a) $49.41rad/s^2$
(b) $0.0849m$
Work Step by Step
(a) We know that
$\alpha=\frac{2(rF-\tau)}{Mr^2}$
We plug in the known values to obtain:
$\alpha=\frac{2(0.055m\times 2.2N-0.047N.m)}{(0.99Kg)(0.055m)^2}$
$\alpha=49.41rad/s^2$
(b) We know that
$x=r\Delta \theta$
$\implies x=\frac{1}{2}r\alpha t^2$ (because $\Delta \theta=\frac{1}{2}\alpha t^2$)
We plug in the known values to obtain:
$x=\frac{1}{2}(0.055m)(49.41rad/s^2)(0.25s)^2$
$x=0.0849m$
