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November 27, 2014

November 27, 2014

Posted by **Katie** on Friday, March 16, 2012 at 10:33pm.

What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 2.89 rad/s?

While the system is rotating, the children pull themselves toward the center of the board until they are half as far from the center as before. What is the resulting angular speed?

What is the change in kinetic energy of the system as a result of the children changing their positions?

- physics -
**Elena**, Saturday, March 17, 2012 at 12:25pmMoment of inertia of the board respectively the pivot point is

Io = m(L^2 + W^2)/12 = 9.4(4.5^2+0.3^2)/12 = 15.93 kg•m^2,

Moment of inertia of the system (board+2 point mass-ñhildren) respectively

the pivot point is

I = Io + 2mR^2 = 15.93 + 2•11•(4.5/2)^2 = 127.305 kg•m^2

An angular momentum

L = I•ω =127.305•2.89 = 367.9 kg•m^2/s.

In the second position: the separation of each point mass (child)

from the pivot point is r = L/4, then,

I1 = Io + 2mr^2 = 15.93+ 2•11•(4.5/4)^2 = 43.77 kg•m^2.

According to the law of conservation of

the angular momentum of the system

L = Iω = I1•ω1.

ω1= Iω/ I1 = 367.9/43.77=8.4 rad/s.

ΔKE = KE2- KE1 = I1(ω1)^2/2- I(ω)^2/2 =

= 43.77•(8.4)^2/2 - 127.305•(2.89)^2/2 =

= 1548-532=1016 J.

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