Answer
The rotational kinetic energy is $~~1.5~J$
Work Step by Step
In part (c), we found that the linear speed as it reaches the end of the string is $~~v = 0.56~m/s$
We can find the angular speed:
$\omega = \frac{v}{r}$
$\omega = \frac{0.56~m/s}{3.2\times 10^{-3}~m}$
$\omega = 175~rad/s$
We can convert the rotational inertia to units of $kg~m^2$:
$I = (950~g~cm^2)(\frac{1~kg}{1000~g})(\frac{1~m}{10^4~cm^2}) = 9.5\times 10^{-5}~kg~m^2$
We can find the rotational kinetic energy:
$K_{rot} = \frac{1}{2}I\omega^2$
$K_{rot} = \frac{1}{2}(9.5\times 10^{-5}~kg~m^2)(175~rad/s)^2$
$K_{rot} = 1.5~J$
The rotational kinetic energy is $~~1.5~J$
