A man stands on a platform that is rolating (without friction)with an angular speed of...

We have a rolating platform with angular speed of ${\displaystyle {\omega }_i=1.20re\frac{v}{s}}$, a man with stretched arms stands on this
platform and he holds a brick in each hand. It is given that the rolation inertia of the system about the central
vertical axis of the platform is ${\displaystyle I_i=6.0kg\cdot m^2}$, then the man moves the bricks such that the rolational inertia
decreases to ${\displaystyle I_i=6.0kg\cdot m^2}$
(a) First we need to find the final angular momentum of the system, since there is no external force, then the
angular momentum is conserved, that is,
${\displaystyle L_i=L_f}$
but the angular momentum is ${\displaystyle L=I\omega }$, so we can write,
${\displaystyle I_i{\omega }_i=I_f{\omega }_f}$
solve for the final angular momentum and then substitute with the givens to get,
${\displaystyle {\omega }_f=\left(\frac{I_i}{I_i}{\omega }_i\right)=\left(\frac{6.0kg\cdot m^2}{2.0kg\cdot m^2}\right)\left(1.20re\frac{v}{s}\right)}$
${\displaystyle =3.60re\frac{v}{s}}$
${\displaystyle {\omega }_f=3.60re\frac{v}{s}}$
(b) The kinetic energy in terms of the rolational inertia and the angular speed is given by
${\displaystyle K=\frac{1}{2}I{\omega }^2}$
using this equation we can write the initial and final kinetical energy as
${\displaystyle K_i=\frac{1}{2}{I}_i{\omega }_i^2}$ ${\displaystyle K_F=\frac{1}{2}{I}_F{\omega }_F^2}$
their ratio is, therefore,
${\displaystyle \frac{K_f}{K_i}=\frac{I_f{\omega }_f^2}{I_i{\omega }_i^2}=\frac{I_f{\omega }_f^2}{I_i{\omega }_i^2}}$
${\displaystyle =\left(\frac{6.0kg\cdot m^2}{2.0kg\cdot m^2}\right){\left(\frac{3.60re\frac{v}{s}}{1.20re\frac{v}{s}}\right)}^2}$
${\displaystyle =3.0}$
${\displaystyle \frac{K_f}{K_i}=30}$
(c) The kinetic energy came from the man's internal energy, where the man did work on the brick(using his internal
energy - muscles) to decrease the rolational inertia ( by pulling them closer).

${\displaystyle {\omega }_1=1.2re\frac{v}{s}}$
${\displaystyle I_1=6.0kg{m}^2}$
${\displaystyle {\omega }_2=?}$
${\displaystyle I_2=2.0kg{m}^2}$
for conservation of angular momentum,
Initial angular momentum=final angular momentum
${\displaystyle L_1=L_2}$
${\displaystyle I_1{\omega }_1=I_2{\omega }_2}$
solving for ${\displaystyle {\omega }_2,}$
${\displaystyle {\omega }_2=\frac{I_1\star {\omega }_1}{I_2}}$
${\displaystyle {\omega }_2=\frac{6.0\star 1.2}{2.0}}$
${\displaystyle {\omega }_2=3.6re\frac{v}{s}}$
the resulting angular speed of the platform is ${\displaystyle 3.6re\frac{v}{s}}$.
(b)the ratio of the new kinetic energy to the system kinetic energy
let the new kinetic energy ${\displaystyle =K_2}$
original kinetic enegry ${\displaystyle =K_1}$
Kinetic energy of s rotational body ${\displaystyle {=}^1\mathrm{\angle }2I_2{\omega }^2}$
${\displaystyle \frac{K_2}{K_1}{=}^1\mathrm{\angle }2I_1{\omega }_1^2{=}^1\mathrm{\angle }2I_2{\omega }^2}$
${\displaystyle \frac{K_2}{K_1}=I_1{\omega }_1^2=I_2{\omega }_2^2}$
${\displaystyle \frac{K_2}{K_1}=\frac{2.0\star {3.6}^2}{6.0\star {1.2}^2}}$
${\displaystyle \frac{K_2}{K_1}=\frac{25.92}{8.64}}$
${\displaystyle K_2{K}_1=3}$
The ratio of the new kinetic wnergy to the old kinetic energy is 3.

Solution:
$L_F-L_I$
$I_f{\omega }_f=I_i{\omega }_i$
The initial angular velocity is,
${\omega }_i=1.2\frac{rev}{s}\times \frac{2\pi rad}{rev}$
$=7.54rad/s$
The inal angular velocity is,
$$\begin{gathered} I_f{\omega }_f=I_i{\omega }_i \\ {\omega }_f=\frac{I_i}{I_f}{\omega }_i \\ =\frac{6kg\cdot m^2}{2kg\cdot m^2}\times 7.54rad/s \\ =22.62rad/s \end{gathered}$$
The change in totational kinetic energy is,
$$\begin{gathered} \mathrm{△}\omega =\frac{1}{2}{I}_f{\omega }_f^2-\frac{1}{2}{I}_i{\omega }_i^2 \\ =\frac{1}{2}(I_f{\omega }_f^2-I_i{\omega }_i^2) \\ =0.5[(2)(22.62{)}^2-(6)(7.5{)}^2] \\ =341J \end{gathered}$$
Hence, the change in roational kinetical energy is 341 J.