Answer
(a) $0.078Kg.\frac{m^2}{s}$
(b) $31\frac{rad}{s}$
Work Step by Step
(a) The angular momentum is given as
$\Delta L=\tau \Delta t$
We plug in the known values to obtain:
$\Delta L=(0.12)(0.65)=0.078Kg.\frac{m^2}{s}$
(b) The angular speed is given as
$\omega=\frac{L}{I}$
We plug in the known values to obtain:
$\omega=\frac{0.078}{2.5\times 10^{-3}}$
$\omega=31\frac{rad}{s}$
