Answer
(a) $K=0.012J$ (b) $48rad/s$
Work Step by Step
(a) The kinetic energy of a rotating object is equal to $$K=\frac{1}{2}I\omega^2$$ To find the moment of inertia, use the formula for the moment of inertia of a disk, which is $$I=\frac{1}{2}MR^2$$ Substituting known values of $m=12g=0.012kg$ and $R=6.0cm=0.060m$ yields a moment of inertia of $$I=\frac{1}{2}(0.012kg)(0.060m)^2=2.2\times 10^{-5} kg \times m^2$$ Substituting known values of $I=2.2\times 10^{-5} kg \times m^2$ and $\omega=34rad/s$ yields a kinetic energy of $$K=\frac{1}{2}(2.2\times 10^{-5} kg \times m^2)(34rad/s)^2=0.012J$$ (b) To double the kinetic energy, the new angular speed must be $$\omega'^2=2\omega^2$$ This means that $$\omega'=\sqrt{2} \omega$$ This means that the new angular speed must be $$\omega'=\sqrt{2}(34rad/s)=48rad/s$$
