Two children each with mass m = 29.5 kg, sit on opposite ends of a narrow board

Question

Two children each with mass m = 29.5 kg, sit on opposite ends of a narrow board with length L = 5.5 m, width W = 0.20 m, and mass M = 6.0 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 of the board?

b) What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 1.43 rad/s?

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

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

Explanation / Answer

Given:
mass of each child, m = 29.5 kg
length of thin narrow board, L = 5.5 m
width, a = .2 m
mass of narrow board, M = 6.0 kg

a) The moment of inertia for a narrow thin board is I = 1/3* M * a^2 (if it wasn't a narrow board it would be I = 1/3 * M * (a^2 + b^2), where b would be the length.

Since we are calculating I through the center, and each child weighs the same, we will use I = 2m*(L/2)^2

So I = 1/3*Ma^2 + 2*m*(L/2)^2

Plug in the values and we get I = 446.2 kg * m^2

b) angular momentum is, L = * I , so

L = 1.43 rad/s * (446.2 kg * m^2)

L = 638.1

c) Conservation of angular moment states,

I1 * 1 = I2 * 2 or w2 = (I1 * w1) / I2

So it becomes our answer from part b divided by the new I, which we calculate using

I = 1/3*Ma^2 + 2*m*(L/4)^2 = 111.63, so w2 = 5.72 rad/s

d) To calcular the change in kinetic energy we use, KE2 - KE1

KE = 1/2 I * w^2

KE1 = 1/2 * 446.2 * 1.43^2 = 456.2

KE2 = 1/2 * 111.63 * 5.72^2 = 1826.2

Change in KE = 1826.2 - 456.2 = 1370 J

I tried to be as detailed as possible, please rate. PM me for any questions at all. Thank you

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