Problem 12.25

12.25. a)
We need to find the moment of inertia of the system. Let's use the parallel axis theorem.
I = I_cm + M d²
$I = \frac{1}{12} ML^{2} + Md^{2}$
$I = \frac{13}{12}M$
now, the period is given from
$T = 2\pi \sqrt{\frac{I}{mgd}}$
$T = 2\pi \sqrt{\frac{\frac{13 M}{12}}{Mg(1m)}}$
T= 2.09 sec

b)
For the simple pendulum, $T = 2\pi \sqrt{\frac{\ell}{g}}$
T = 2.01 sec
So, if we compare them, we see that they are
$\frac{.08}{2.01} ~=~ 4.08\%$ different.