We have a rolating platform with angular speed of \(\displaystyle\omega_{{i}}={1.20}{r}{e}\frac{{v}}{{s}}\), 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 \(\displaystyle{I}_{{i}}={6.0}{k}{g}\cdot{m}^{{2}}\), then the man moves the bricks such that the rolational inertia

decreases to \(\displaystyle{I}_{{i}}={6.0}{k}{g}\cdot{m}^{{2}}\)

(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,

\(\displaystyle\omega_{{f}}={\left({\frac{{{I}_{{i}}}}{{{I}_{{i}}}}}\omega_{{i}}\right)}={\left({\frac{{{6.0}{k}{g}\cdot{m}^{{2}}}}{{{2.0}{k}{g}\cdot{m}^{{2}}}}}\right)}{\left({1.20}{r}{e}\frac{{v}}{{s}}\right)}\)

\(\displaystyle={3.60}{r}{e}\frac{{v}}{{s}}\)

\(\displaystyle\omega_{{f}}={3.60}{r}{e}\frac{{v}}{{s}}\)

(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,

\(\displaystyle{\frac{{{K}_{{f}}}}{{{K}_{{i}}}}}={\frac{{{I}_{{f}}{\omega_{{f}}^{{2}}}}}{{{I}_{{i}}{\omega_{{i}}^{{2}}}}}}={\frac{{{I}_{{f}}{\omega_{{f}}^{{2}}}}}{{{I}_{{i}}{\omega_{{i}}^{{2}}}}}}\)

\(\displaystyle={\left({\frac{{{6.0}{k}{g}\cdot{m}^{{2}}}}{{{2.0}{k}{g}\cdot{m}^{{2}}}}}\right)}{\left({\frac{{{3.60}{r}{e}\frac{{v}}{{s}}}}{{{1.20}{r}{e}\frac{{v}}{{s}}}}}\right)}^{{2}}\)

\(\displaystyle={3.0}\)

\(\displaystyle{\frac{{{K}_{{f}}}}{{{K}_{{i}}}}}={30}\)

(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).

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 \(\displaystyle{I}_{{i}}={6.0}{k}{g}\cdot{m}^{{2}}\), then the man moves the bricks such that the rolational inertia

decreases to \(\displaystyle{I}_{{i}}={6.0}{k}{g}\cdot{m}^{{2}}\)

(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,

\(\displaystyle\omega_{{f}}={\left({\frac{{{I}_{{i}}}}{{{I}_{{i}}}}}\omega_{{i}}\right)}={\left({\frac{{{6.0}{k}{g}\cdot{m}^{{2}}}}{{{2.0}{k}{g}\cdot{m}^{{2}}}}}\right)}{\left({1.20}{r}{e}\frac{{v}}{{s}}\right)}\)

\(\displaystyle={3.60}{r}{e}\frac{{v}}{{s}}\)

\(\displaystyle\omega_{{f}}={3.60}{r}{e}\frac{{v}}{{s}}\)

(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,

\(\displaystyle{\frac{{{K}_{{f}}}}{{{K}_{{i}}}}}={\frac{{{I}_{{f}}{\omega_{{f}}^{{2}}}}}{{{I}_{{i}}{\omega_{{i}}^{{2}}}}}}={\frac{{{I}_{{f}}{\omega_{{f}}^{{2}}}}}{{{I}_{{i}}{\omega_{{i}}^{{2}}}}}}\)

\(\displaystyle={\left({\frac{{{6.0}{k}{g}\cdot{m}^{{2}}}}{{{2.0}{k}{g}\cdot{m}^{{2}}}}}\right)}{\left({\frac{{{3.60}{r}{e}\frac{{v}}{{s}}}}{{{1.20}{r}{e}\frac{{v}}{{s}}}}}\right)}^{{2}}\)

\(\displaystyle={3.0}\)

\(\displaystyle{\frac{{{K}_{{f}}}}{{{K}_{{i}}}}}={30}\)

(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).