#### Answer

$v = 2.4~m/s$

#### Work Step by Step

We can find the rotational inertia of the disk.
$I = \frac{1}{2}MR^2$
$I = \frac{1}{2}(0.10~kg)(0.040~m)^2$
$I = 8.0\times 10^{-5}~kg~m^2$
We can use the rotational kinetic energy to find the angular velocity.
$KE_{rot} = 0.15~J$
$\frac{1}{2}I~\omega^2 = 0.15~J$
$\omega^2 = \frac{(2)(0.15~J)}{I}$
$\omega = \sqrt{\frac{(2)(0.15~J)}{I}}$
$\omega = \sqrt{\frac{(2)(0.15~J)}{8.0\times 10^{-5}~kg~m^2}}$
$\omega = 61.2~rad/s$
We can find the speed of a point on the rim.
$v = \omega ~R$
$v = (61.2~rad/s)(0.040~m)$
$v = 2.4~m/s$