Figure shows two blocks, each of mass $m$ , suspended from the ends of a rigid weightless rod of length $l_{1} + l_{2}$ with $l_{1} = 20 c m$ and $l_{2} = 80 c m$ . The rod is held in horizontal position shown in figure and then released. Calculate the accelerations of the two blocks as they starts to move.

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$a_{1} = 2.70 m / s^{2}, a_{2} = 3.12 m / s^{2}$

$a_{1} = 1.30 m / s^{2}, a_{2} = 5.92 m / s^{2}$

$a_{1} = 3.73 m / s^{2}, a_{2} = 6.00 m / s^{2}$

$a_{1} = 1.73 m / s^{2}, a_{2} = 6.92 m / s^{2}$

answer is C.

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Detailed Solution

$N e t t o r q u e, τ_{n e t} = m g l_{2} - m g l_{1} I f α i s t h e a n g u l a r a c c e l e r a t i o n s, t h e n α = \frac{τ_{n e t}}{l} = \frac{m g (l_{2} - l_{1})}{m {l_{1}}^{2} + m {l_{2}}^{2}} = \frac{g (l_{2} - l_{1})}{{l_{1}}^{2} + {l_{2}}^{2}} = \frac{9.8 (0.8 - 0.2)}{0 . 2^{2} + 0 . 8^{2}} = 8.65 r a d / s^{2} ∴ a_{1} = α l_{1} = 8.65 \times 0.2 = 1.73 m / s^{2} a_{2} = α l_{2} = 8.65 \times 0.8 = 6.92 m / s^{2}$