Answer
(a) the total moment of inertia:
$I = \frac{{365}}{2}$ $g\cdot c{m^2}$
(b) the angular velocity is $\omega \simeq 122.53$ rad/s.
Work Step by Step
(a) Referring to Figure 19, we set the $z$-axis as the axis of symmetry for the yo-yo. So then, we can use the result from Exercise 45, that is ${I_z} = \frac{1}{2}M{R^2}$ to compute the total moment inertia.
$\begin{array}{*{20}{c}}
{{\rm{Component}}}&{{\rm{Mass}}}&{{\rm{Radius}}}&{{I_z}}\\
{{\rm{Disk{\ }1}}}&{20}&3&{\frac{1}{2}\left( {20} \right)\left( 9 \right) = 90}\\
{{\rm{Axle}}}&5&1&{\frac{1}{2}\left( 5 \right)\left( 1 \right) = \frac{5}{2}}\\
{{\rm{Disk{\ }2}}}&{20}&3&{\frac{1}{2}\left( {20} \right)\left( 9 \right) = 90}
\end{array}$
Since $I$ is the sum of the moments of the three components of the yo-yo, we get the total moment of inertia:
$I = 90 + \frac{5}{2} + 90 = \frac{{365}}{2}$ $g\cdot c{m^2}$
(b) Since the total mass of the yo-yo is $M=45$ g, so the potential energy lost is
$PE = Mgh = \left( {45} \right)\left( {980} \right)100$
$PE = 4.41 \times {10^6}$ $g\cdot c{m^2}/{s^2}$
Since the yo-yo as a whole rotates with angular velocity $\omega $; using $r=3$, the translational velocity of the yo-yo as it is falling down the string is
$v = \omega r = 3\omega $
So, the translational kinetic energy is
${\rm{TKE}} = \frac{1}{2}M{v^2} = \frac{1}{2}\left( {45} \right)\left( 9 \right){\omega ^2} = \frac{{405}}{2}{\omega ^2}$
Using $I = \frac{{365}}{2}$ from part (a), we compute the rotational kinetic energy:
${\rm{RKE}} = \frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{{365}}{2}} \right){\omega ^2} = \frac{{365}}{4}{\omega ^2}$
Since the potential energy is the sum of the rotational kinetic energy and the translational kinetic energy, we get
$PE = RKE + TKE$
$4.41 \times {10^6} = \frac{{365}}{4}{\omega ^2} + \frac{{405}}{2}{\omega ^2}$
$4.41 \times {10^6} = \frac{{1175}}{4}{\omega ^2}$
$\omega \simeq 122.53$
So, the angular velocity is $\omega \simeq 122.53$ rad/s.
