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

tebollahb

tebollahb

Answered

2022-01-12

A man stands on a platform that is rolating (without friction)with an angular speed of 1.2revs; 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 6.0kgm2. If by moving the bricks the man decreases the rolational inertia of the
system to 2.0kgm2, 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?

Answer & Explanation

Timothy Wolff

Timothy Wolff

Expert

2022-01-13Added 26 answers

We have a rolating platform with angular speed of ωi=1.20revs, 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 Ii=6.0kgm2, then the man moves the bricks such that the rolational inertia
decreases to Ii=6.0kgm2
(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,
Li=Lf
but the angular momentum is L=Iω, so we can write,
Iiωi=Ifωf
solve for the final angular momentum and then substitute with the givens to get,
ωf=(IiIiωi)=(6.0kgm22.0kgm2)(1.20revs)
=3.60revs
ωf=3.60revs
(b) The kinetic energy in terms of the rolational inertia and the angular speed is given by
K=12Iω2
using this equation we can write the initial and final kinetical energy as
Ki=12Iiωi2 KF=12IFωF2
their ratio is, therefore,
KfKi=Ifωf2Iiωi2=Ifωf2Iiωi2
=(6.0kgm22.0kgm2)(3.60revs1.20revs)2
=3.0
KfKi=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

Lindsey Gamble

Expert

2022-01-14Added 38 answers

ω1=1.2revs
I1=6.0kgm2
ω2=?
I2=2.0kgm2
for conservation of angular momentum,
Initial angular momentum=final angular momentum
L1=L2
I1ω1=I2ω2
solving for ω2,
ω2=I1ω1I2
ω2=6.01.22.0
ω2=3.6revs
the resulting angular speed of the platform is 3.6revs.
(b)the ratio of the new kinetic energy to the system kinetic energy
let the new kinetic energy =K2
original kinetic enegry =K1
Kinetic energy of s rotational body =12I2ω2
K2K1=12I1ω12=12I2ω2
K2K1=I1ω12=I2ω22
K2K1=2.03.626.01.22
K2K1=25.928.64
K2K1=3
The ratio of the new kinetic wnergy to the old kinetic energy is 3.
star233

star233

Expert

2022-01-15Added 238 answers

Solution:
LFLI
Ifωf=Iiωi
The initial angular velocity is,
ωi=1.2revs×2πradrev
=7.54 rad/s
The inal angular velocity is,
Ifωf=Iiωiωf=IiIfωi=6kgm22kgm2×7.54 rad/s=22.62 rad/s
The change in totational kinetic energy is,
ω=12Ifωf212Iiωi2=12(Ifωf2Iiωi2)=0.5[(2)(22.62)2(6)(7.5)2]=341 J
Hence, the change in roational kinetical energy is 341 J.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?