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

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0.01 rad $s^{- 2}$

1.0 rad $s^{- 2}$

100 rad $s^{- 2}$

10 rad $s^{- 2}$

answer is B.

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Detailed Solution

Net torque, t = 30g × 2 - 20g × 2 = 20g, I = 30 × $2^{2}$ + 20 × $2^{2}$ = 120 + 80 = 200 kg $m^{2}$ . $α = \frac{τ}{I} = \frac{20 g}{200} = \frac{g}{10} = \frac{10}{10} = 1.0 r a d s^{- 2}$ .

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