Answer
(a) $KE = 140~J$
(b) 33% of the kinetic energy is rotational kinetic energy.
Work Step by Step
(a) We can find the angular velocity.
$\omega = (0.50~rev/s)(2\pi~rad/rev)$
$\omega = \pi~rad/s$
We can find the translational velocity.
$v = \omega~R$
$v = (\pi~rad/s)(0.50~m)$
$v = (0.50~\pi)~m/s$
We can find the kinetic energy.
$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
$KE = \frac{1}{2}mv^2+\frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$KE = \frac{3}{4}mv^2$
$KE = \frac{3}{4}(75~kg)(0.50~pi~m/s)^2$
$KE = 140~J$
(b) The rotational energy is $\frac{1}{4}mv^2$. We can find the percent of the kinetic energy that is rotational kinetic energy.
$\frac{KE_{rot}}{KE}\times 100\% =\frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2}\times 100\% = 33\%$
33% of the kinetic energy is rotational kinetic energy.
