91Ó°ÊÓ

We have

Mass of cylinder$\left(m\right)=75kg$.

Diameter of cylinder $\left(D\right)=1.0m$.

Radius of cylinder$\left(R\right)=D/2=0.50m$.

Angular acceleration $\left(\mathrm{Ó\neg }\right)=0.50rev/s$.

We have the total kinetic energy which is the sum of kinetic energy of translation of CM and the rotational kinetic energy, given by,

$K_{tot}=K_{tr}+K_{rot}   ⋯⋯\left(A\right)$, where,

$$\begin{gathered} K_{tr}=\frac{1}{2}m{v}^2  ⋯⋯\left(1\right) \\ {K}_{rot}=\frac{1}{2}I{\mathrm{Ó\neg }}^2⋯⋯\left(2\right) \end{gathered}$$

The velocity of center of mass can be expressed as:

$v=R\mathrm{Ó\neg }$

Thus, equation (1) becomes,

$K_{tr}=\frac{1}{2}m{R}^2{\mathrm{Ó\neg }}^2  ⋯⋯\left(3\right)$.

Now, the moment of inertia of the cylinder about the axis passing through its center is,

$I=\frac{1}{2}m{R}^2$.

Thus, equation (2) becomes,

$K_{rot}=\frac{1}{2}\frac{1}{2}m{R}^2{\mathrm{Ó\neg }}^2=\frac{1}{4}m{R}^2{\mathrm{Ó\neg }}^2   ⋯⋯\left(4\right)$.

Putting (3) and ( 4 ) in equation $\left(A\right)$, we get,

$$\begin{gathered} K_{tot}=\frac{1}{2}m{R}^2{\mathrm{Ó\neg }}^2+\frac{1}{4}m{R}^2{\mathrm{Ó\neg }}^2 \\ ⇒K_{tot}=\frac{3}{4}m{R}^2{\mathrm{Ó\neg }}^2 \end{gathered}$$

Thus, substituting the values in above equation, we get,

$$\begin{gathered} K_{tot}=\frac{3}{4}\left(75\right){\left(0.5\right)}^2{\left(0.50·2\mathrm{π}\right)}^2 \\ {K}_{tot}=140 J \end{gathered}$$

since$1rev=2\mathrm{Ï€}rad$.

Hence, the total kinetic energy is, $140J$.

The ratio of rotational to total kinetic energy is,

$$\begin{gathered} \frac{K_{rot}}{K_{tot}}=\frac{\frac{1}{4}m{R}^2{\mathrm{Ó\neg }}^2  }{\frac{3}{4}m{R}^2{\mathrm{Ó\neg }}^2} \\ ⇒\frac{K_{rot}}{K_{tot}}=\frac{1}{3} \end{gathered}$$

Hence, the percent of the rotational kinetic energy is,$33\%$.