Answer
Using the parallel axis theorem and items $(e)$ and $(h)$ in Table $10-2,$ the rotational
inertia is
$$
I=\frac{1}{12} m L^{2}+m(L / 2)^{2}+\frac{1}{2} m R^{2}+m(R+L)^{2}=10.83 m R^{2}
$$
where $L=2 R$ has been used. If we take the base of the rod to be at the coordinate origin $(x=0, y=0)$ then the center of mass is at
$$
y=\frac{m L / 2+m(L+R)}{m+m}=2 R
$$
Comparing(differentiate) the place shown in the textbook figure to its upside down (inverted)
place shows that the change in center of mass position (in absolute value) is $|\Delta y|=4 R .$ The corresponding loss in gravitational potential energy is converted into kinetic energy. Thus,
$$
K=(2 m) g(4 R) \Rightarrow \omega=9.82 \mathrm{rad} / \mathrm{s}
$$
where Eq. $10-34$ has been used.
Work Step by Step
Using the parallel axis theorem and items $(e)$ and $(h)$ in Table $10-2,$ the rotational
inertia is
$$
I=\frac{1}{12} m L^{2}+m(L / 2)^{2}+\frac{1}{2} m R^{2}+m(R+L)^{2}=10.83 m R^{2}
$$
where $L=2 R$ has been used. If we take the base of the rod to be at the coordinate origin $(x=0, y=0)$ then the center of mass is at
$$
y=\frac{m L / 2+m(L+R)}{m+m}=2 R
$$
Comparing(differentiate) the place shown in the textbook figure to its upside down (inverted)
place shows that the change in center of mass position (in absolute value) is $|\Delta y|=4 R .$ The corresponding loss in gravitational potential energy is converted into kinetic energy. Thus,
$$
K=(2 m) g(4 R) \Rightarrow \omega=9.82 \mathrm{rad} / \mathrm{s}
$$
where Eq. $10-34$ has been used.
