Posted by Katie on .
Two children, each with mass m = 11.0 kg, sit on opposite ends of a narrow board with length L = 4.5 m, width W = 0.30 m, and mass M = 9.4 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction. What is the rotational inertia of the board plus the children about a vertical axis through the center of the board?
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?
Moment 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.