91Ó°ÊÓ

We know that \(K = \frac{1}{2}I{\omega ^2}\)

interpreting equation\(I = {m_1}{r_1} + {m_2}{r_2} + \_\_\_\_\_\).

Over a sum of particles that up a body.

In the above expression, each term will have the mass multiplied by \({f^3}\)and \({f^2}\)and the distance multiplied by f.

\({f^3}\left( {{f^2}} \right) = {f^2}\)

b.

Given in the question, \(\frac{1}{{48}}\) -scalemodel has rotational kinetic energy of 2.5 J.

We know that \({I_{scaled}} = {\left( {\frac{1}{{48}}} \right)^5} \times I\),

Since \(K = \frac{1}{2}I{\omega ^2}\) and

\({I_{sacled}} = {\left( {\frac{1}{{48}}} \right)^5}I\)then

\({K_{sacled}} = {\left( {\frac{1}{{48}}} \right)^5}K\)__________(1)

\({K_{sacled}} = 2.5J\)

Put the \({K_{sacled}}\) value in equation (1)

\(2.5J = {\left( {\frac{1}{{48}}} \right)^5}K\)

\(2.5J \times {\left( {48} \right)^5} = K\)

\(K = 6.37 \times {10^8}J\)

Hence, the material rotating at the same angular velocity is\(6.37 \times {10^8}J\).