Calculate the linear acceleration of the blocks of mass $m_{A} = 10 k g, m_{B} = 8 k g$ mass of disc shapped pulley $m_{P} = 2 k g$ $(9 = 10 m s^{- 2})$

\frac{20}{19}

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\frac{30}{19}

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\frac{40}{19}

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\frac{25}{19}

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Solution

The correct option is A $\frac{20}{19}$
Lets consider $R =$ radius of disc
Given,
$m_{A} = 10 k g$
$m_{B} = 8 k g$
$m_{P} = 2 k g$
$g = 10 m / s^{2}$
From free body diagram,
$m_{A} g - T_{1} = m_{A} a$
$100 - T_{1} = 10 a$
$T_{1} = 100 - 10 a$ . . . . . . . ..(1)
$T_{2} - m_{B} g = m_{B} a$
$T_{2} - 80 = 8 a$
$T_{2} = 80 + 8 a$ . . . . . . . (2)
Total torgue on pully is given by
$(T_{1} - T_{2}) R = I α = \frac{m_{P} R^{2}}{2} . \frac{a}{R}$
$T_{1} - T_{2} = \frac{m_{p} a}{2} = \frac{2 a}{2} = a$ . . . . . (3)
Substitute the value of $T_{1}, T_{2}$ from equation (1) and (2) respectively in equation (3),
$100 - 10 a - (80 + 8 a) = a$
$100 - 10 a - 80 - 8 a = a$
$20 - 18 a = a$
$20 = 19 a$
$a = \frac{20}{19} m / s^{2}$
The correct option is A.

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