Three children are sitting on a see-saw in such a way that it is balanced. A $20 k g$ and a $30 k g$ boy are on opposite sides at a distance of $2 m$ from the pivot. If the third boy jumps off, thereby destroying balance, then the initial angular acceleration of the board is: (Neglect weight of board)

0.01 r a d s^{- 2}

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1.0 r a d s^{- 2}

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10 r a d s^{- 2}

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100 r a d s^{- 2}

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Solution

The correct option is A $1.0 r a d s^{- 2}$
We have the net torque as $τ = 30 g \times 2 g - 20 g \times 2 = 20 g$
and moment of inertia
$I = 30 \times 2^{2} + 20 \times 2^{2} = 120 + 80 = 200 k g m^{2}$
Thus we get angular acceleration as
$α = \frac{τ}{I} = \frac{20 g}{200} = \frac{g}{100} = \frac{10}{10} = 1.0 r a d s^{- 2}$

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Three children are sitting on a see-saw in such a way that it is balanced. A 20kg and a 30kg boy are on opposite sides at a distance of 2m from the pivot. If the third boy jumps off, thereby destroying balance, then the initial angular acceleration of the board is: (Neglect weight of board)

The correct option is A 1.0rads−2We have the net torque as τ=30g×2g−20g×2=20gand moment of inertiaI=30×22+20×22=120+80=200kgm2Thus we get angular acceleration asα=τI=20g200=g100=1010=1.0rads−2