Answer
$2.84rad/s$
Work Step by Step
We can find the final angular speed as follows:
$\omega_f=\frac{\frac{1}{2}Mr^2\omega_i+mvr}{\frac{1}{2}Mr^2+mr^2}$
This simplifies to:
$\omega_f=\frac{Mr\omega_i+2mv}{Mr+2mr}$
We plug in the known values to obtain:
$\omega_f=\frac{(155)(2.63)(4.03)+2(59.4)(3.41)}{(155)(2.63)+2(59.4)(2.63)}$
$\omega_f=2.84rad/s$
