a)Two kids sit on the ends of a uniform board that is 6.62 meters long and 0.304

Question

a)Two kids sit on the ends of a uniform board that is 6.62 meters long and 0.304 meters wide. The board rotates horizontally about a frictionless pivot through its center. If the board has a mass of 6.77 kilograms and each kid has a mass of 8.95 kilograms, find the rotational inertia for the board and kids system.

b)With the kids at opposite ends of the board, the board is initially rotating at 2.79 radians per second. If the kids both scoot in towards the center of the board, what will the new angular speed be when they are both halfway towards the center?

Explanation / Answer

The length of the boarrd is L = 6.62 m
The width of the board is w = 0.304 m
The mass of the board is M = 6.77 kg
The mass of each kid is m = 8.95 kg

a)
The moment of inertia of the board is Ib = 1/12 M(L^2 + w^2)

Ib = 1/12 *6.77 * (6.62^2 + 0.304 ^2)

Ib = 24.77 kgm^2

The moments of inertia of the each kid is Ik = m(L/2)^2

Ik = 8.95*(3.31)^2

Ik = 98.06 kgm^2

The total rotational inertia of the board and kids system is

I = Ib + 2Ik

I = 220.89 kgm^2

b)
The initial angular momentum 1 = 2.79 rad/s

The initial kinetic energy of the system is KEi = 0.5*I*1^2

When the kids are half of their way from the centre

The moment of inertia of the system will be I' = 24.77 + 2*8.95*(1.655)^2

I' = 73.8 k/m^2

Let 2 be the final angular velocity, then

Final kinetic energy of the system is KEf = 0.5*I'*2^2

According to the conservation of energy

KEi = KEf

2/1 = (I/I')

2 = 4.83 rad/s

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