Answer
$1.63 \times 10^4 J$
Work Step by Step
The energy required equals the ride’s final rotational kinetic energy.
The final rotational speed is 0.8976 rad/s.
The rotational inertia of the solid disk is $\frac{1}{2}MR^2=40500 kg \cdot m^2$.
$$KE_{rot}=\frac{1}{2}I \omega^2 \approx 1.63 \times 10^4 J$$
