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 6.0kg⋅m2. If by moving the bricks the man decreases the rolational inertia of the system to 2.0kg⋅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?

Question

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 6.0kg⋅m2. If by moving the bricks the man decreases the rolational inertia of the  system to 2.0kg⋅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 · Accepted Answer

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 centralvertical axis of the platform is Ii=6.0kg⋅m2, then the man moves the bricks such that the rolational inertiadecreases to Ii=6.0kg⋅m2(a) First we need to find the final angular momentum of the system, since there is no external force, then theangular momentum is conserved, that is,Li=Lfbut the angular momentum is L=Iω, so we can write,Iiωi=Ifωfsolve for the final angular momentum and then substitute with the givens to get,ωf=(IiIiωi)=(6.0kg⋅m22.0kg⋅m2)(1.20revs)=3.60revsωf=3.60revs(b) The kinetic energy in terms of the rolational inertia and the angular speed is given byK=12Iω2using this equation we can write the initial and final kinetical energy asKi=12Iiωi2 KF=12IFωF2their ratio is, therefore,KfKi=Ifωf2Iiωi2=Ifωf2Iiωi2=(6.0kg⋅m22.0kg⋅m2)(3.60revs1.20revs)2=3.0KfKi=30(c) The kinetic energy came from the man&#039;s internal energy, where the man did work on the brick(using his internalenergy - muscles) to decrease the rolational inertia ( by pulling them closer).

Lindsey Gamble · Accepted Answer

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=I1⋆ω1I2  ω2=6.0⋆1.22.0  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 =1∠2I2ω2  K2K1=1∠2I1ω12=1∠2I2ω2  K2K1=I1ω12=I2ω22  K2K1=2.0⋆3.626.0⋆1.22  K2K1=25.928.64  K2K1=3  The ratio of the new kinetic wnergy to the old kinetic energy is 3.

star233 · Accepted Answer

Solution: LF−LI Ifωf=Iiωi The initial angular velocity is, ωi=1.2revs×2πradrev =7.54 rad/s The inal angular velocity is, Ifωf=Iiωiωf=IiIfωi=6kg⋅m22kg⋅m2×7.54 rad/s=22.62 rad/s The change in totational kinetic energy is, △ω=12Ifωf2−12Iiωi2=12(Ifωf2−Iiωi2)=0.5[(2)(22.62)2−(6)(7.5)2]=341 J Hence, the change in roational kinetical energy is 341 J.

A man stands on a platform that is rolating (without friction)with an

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