91Ó°ÊÓ

1. Mass of small cylinder, m_s = 1.25kg
2. Mass of large cylinder, m_l = 1.25kg
3. Angular velocity of small cylinder, ${\mathrm{Ó\neg }}_s=235rad/s$
4. Angular velocity of small cylinder, ${\mathrm{Ó\neg }}_l=235rad/s$
5. Radius of small cylinder, $R_s=0.25m$
6. Radius of large cylinder,$R_l=0.25m$

We can find the moment of inertia of each cylinder using the given data. Using this value of rotational inertia in a rotational kinetic energy formula, we will find rotational kinetic energy of smaller and larger cylinder by using the formula given below.

To calculate rotational kinetic energy of the small cylinder, we need to find its moment of inertia.

Moment of inertia of small solid cylinder

$\begin{array}{rcl}{l}_s& =& \frac{1}{2}{m}_s{R}_s^2\\ & =& -\frac{1}{2}×1.25kg×{\left(0.25m^2\right)}^2\\ & =& 0.039kg.m^2\end{array}$

Using this in rotational kinetic energy formula

$\begin{array}{rcl}{K}_s& =& \frac{1}{2}{l}_s{\mathrm{Ó\neg }}_s^2\\ & =& \frac{1}{2}×0.039kg/m^2×{\left(235rad/s\right)}^2\\ & =& 1076J\\ & =& 1.08kJ\end{array}$

Rotational kinetic energy of small cylinder is 1.08kJ.

To calculate rotational kinetic energy of the large cylinder, we need to find its moment of inertia.

Moment of inertia of large solid cylinder

$\begin{array}{rcl}{l}_l& =& \frac{1}{2}{m}_l{R}_l^2\\ & =& \frac{1}{2}×1.25kg×{\left(0.75m^2\right)}^2\\ & =& 0.3515kg.m^2\end{array}$

Using this in rotational kinetic energy formula

$\begin{array}{rcl}{K}_l& =& \frac{1}{2}{l}_l{\mathrm{Ó\neg }}_l^2\\ & =& \frac{1}{2}×0.3515kg.m^2×{\left(235rad/s\right)}^2\\ & =& 9705.79J\\ & =& 9.71kJ\end{array}$

Rotational kinetic energy of large cylinder is 9.71kJ.