tebollahb

2022-01-12

A man stands on a platform that is rolating (without friction)with an angular speed of $1.2re\frac{v}{s}$; his arms outstreched
and he holds a brick in each hand. The rolational inertia of the system consisting of the man, bricks, and platform about
the central vertical axis of the platform is $6.0kg\cdot m2$. If by moving the bricks the man decreases the rolational inertia of the
system to $2.0kg\cdot m2$, what are(a) the resulting angular speed of the platform ans (b) the ratio of the new kinetic energyof
the system to the original kinetic energy?
c) What source provided the added kinetic energy?

Timothy Wolff

Expert

We have a rolating platform with angular speed of ${\omega }_{i}=1.20re\frac{v}{s}$, 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 ${I}_{i}=6.0kg\cdot {m}^{2}$, then the man moves the bricks such that the rolational inertia
decreases to ${I}_{i}=6.0kg\cdot {m}^{2}$
(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,
${L}_{i}={L}_{f}$
but the angular momentum is $L=I\omega$, so we can write,
${I}_{i}{\omega }_{i}={I}_{f}{\omega }_{f}$
solve for the final angular momentum and then substitute with the givens to get,
${\omega }_{f}=\left(\frac{{I}_{i}}{{I}_{i}}{\omega }_{i}\right)=\left(\frac{6.0kg\cdot {m}^{2}}{2.0kg\cdot {m}^{2}}\right)\left(1.20re\frac{v}{s}\right)$
$=3.60re\frac{v}{s}$
${\omega }_{f}=3.60re\frac{v}{s}$
(b) The kinetic energy in terms of the rolational inertia and the angular speed is given by
$K=\frac{1}{2}I{\omega }^{2}$
using this equation we can write the initial and final kinetical energy as
${K}_{i}=\frac{1}{2}{I}_{i}{\omega }_{i}^{2}$ ${K}_{F}=\frac{1}{2}{I}_{F}{\omega }_{F}^{2}$
their ratio is, therefore,
$\frac{{K}_{f}}{{K}_{i}}=\frac{{I}_{f}{\omega }_{f}^{2}}{{I}_{i}{\omega }_{i}^{2}}=\frac{{I}_{f}{\omega }_{f}^{2}}{{I}_{i}{\omega }_{i}^{2}}$
$=\left(\frac{6.0kg\cdot {m}^{2}}{2.0kg\cdot {m}^{2}}\right){\left(\frac{3.60re\frac{v}{s}}{1.20re\frac{v}{s}}\right)}^{2}$
$=3.0$
$\frac{{K}_{f}}{{K}_{i}}=30$
(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).

Lindsey Gamble

Expert

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

star233

Expert

Solution:
${L}_{F}-{L}_{I}$
${I}_{f}{\omega }_{f}={I}_{i}{\omega }_{i}$
The initial angular velocity is,
${\omega }_{i}=1.2\frac{rev}{s}×\frac{2\pi rad}{rev}$

The inal angular velocity is,

The change in totational kinetic energy is,

Hence, the change in roational kinetical energy is 341 J.

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