Answer
$$\begin{aligned} v &=1.4 \mathrm{m} / \mathrm{s} \end{aligned}$$
Work Step by Step
From Table $10-2$ , the rotational inertia of the spherical shell is $2 M R^{2} / 3,$ so the kinetic $\text { energy (after the object has descended distance } h)$ is
$$
K=\frac{1}{2}\left[\frac{2}{3} M R^{2}\right] \omega_{\text {sphere }}^{2}+\frac{1}{2} I \omega_{\text {pulley }}^{2}+\frac{1}{2} m v^{2}
$$
since it started from remain, then this energy shoud be equal (in the absence of friction) to the potential energy $m g h$ with which the system started. We substitute $v / r$ for the pulley's angular speed and $v / R$ for that of sphere and solve for $v .$
$$\begin{aligned} v &=\sqrt{\frac{m g h}{\frac{1}{2} m+\frac{1}{2}+\frac{M}{3}}}=\sqrt{\frac{2 g h}{1+\left(I / m r^{2}\right)+(2 M / 3 m)}} \\ \\ &=\sqrt{\frac{2(9.8)(0.82)}{\left.2(0.60)(0.050)^{2}\right)+2(4.5) / 3(0.60)}}=1.4 \mathrm{m} / \mathrm{s} \end{aligned}$$
