Answer
a) $\omega=1.1\times 10^2\frac{rad}{s}$
b) $h_i=0.56m$
Work Step by Step
(a) We know that
$mgh_i=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
$\implies mgh_i=\frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2$
$mgh_i=\frac{7}{10}mv^2$
This can be simplified as:
$v=\sqrt{\frac{10}{7}gh_i}$
As $\omega =v\frac{v}{r}$
$\implies \omega=\frac{1}{r}\sqrt{\frac{10gh_i}{7}}$
We plug in the known values to obtain:
$\omega=\frac{1}{0.029}\sqrt{\frac{10(9.81)(0.78)}{7}}$
$\omega=1.1\times 10^2\frac{rad}{s}$
(b) As we know that
$mgh_i=\frac{1}{2}mv^2$
This can be rearranged as:
$h_i=\frac{v^2}{g}$
$\implies h_i=\frac{(\frac{10}{7}ghi)}{2g}=\frac{5}{7}(0.78)=0.56m$
