Two children, each with mass m = 18.5 kg , sit on opposite ends of a narrow boar

Question

Two children, each with mass m = 18.5 kg, sit on opposite ends of a narrow board with length L = 4.1 m, width W = 0.17 m, and mass M = 5.8 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction.

(a) What is the rotational inertia of the board plus the children about a vertical axis through the center of the board?

(b) What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 2.24 rad/s?

(c) 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?

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

Explanation / Answer

The rotational inertia is the I of the beam, plus the MR2 of each child

I = 1/12ML^2 + MR^2+ MR^2

I = 1/12 * 2.9 * 4.1^2 + 18.5 * 2.05^2 + 18.5 * 2.05^2

I = 159.55 kgm2

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The angular momentum is found by L = IW

L = (159.55* 2.24) = 357.39 kgm^2/s
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Angular momentum is conserved,

so L = IW

I changes to (1/12*5.8 * 4.1^2 + (18.5*0.0425^2 ) + (18.5* 0.0425^2)

New I = 8.191 kgm2

So, 357.39 = 8.191 * W

W = 43.63 rad/s

Change in KE = 0.5 I Wf^2 - 0.5IWi^2

KE = 0.5* 8.191* 43.63^2 - (0.5 * 159.55* 2.24^2)

KE chagne = 7395.82 Joules

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