91Ó°ÊÓ

In the given question, Energy is to be stored in a 70.0-kg flywheel in the shape of a uniform solid disk with a radius R = 1.20 m.

The maximum allowed radial acceleration of a point on its rim is 3500 m/s².

We know the definition of radical acceleration \( = r{\omega ^2}\).

Also, we know that the moment of the inertia,

\(I = m{r^2}\).

Radical kinetic energy,

\(K = \frac{1}{2} \times {I^2} \times {\omega ^2}\).

Angular velocity through the radical acceleration.

\(3500\frac{m}{{{s^2}}} = r{\omega ^2}\)

\(3500\frac{m}{{{s^2}}} = 1.2m \times {\omega ^2}\)

\({\omega ^2} = \frac{{3500\frac{m}{{{s^2}}}}}{{1.2m}}\)

\({\omega ^2} = 2916.66\)

\(\omega = \sqrt {2916.66} \)

\(\omega = 54\)

We know that ,

\(I = m{r^2}\)

\( = 70 \times {\left( {1.20} \right)^2}\)

\( = 70 \times 1.44\)

\( = 100.8kg.{m^2}\)

We know that ,

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

\( = \frac{1}{2} \times {\left( {100.8} \right)^2} \times {54^2}\)

\( = 0.5 \times 10160.64 \times 2916\)

\( = 14813280J\)

Hence , the maximum kinetic energy that can be stored in the flywheel \(14813280J\).