For the given figure,find the acceleration of | vedclass

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(A) Let $m_1 = 1\, kg$ and $m_2 = 0.5\, kg$. The pulley has mass $M = 2\, kg$ and radius $R = 0.1\, m$.For the $m_1$ block moving downwards with accel

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$2$

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(A) Let $m_1 = 1\, kg$ and $m_2 = 0.5\, kg$. The pulley has mass $M = 2\, kg$ and radius $R = 0.1\, m$.For the $m_1$ block moving downwards with acceleration $a$:$m_1 g - T_1 = m_1 a \implies 10 - T_1 = a$ $...(I)$For the $m_2$ block moving upwards with acceleration $a$:$T_2 - m_2 g = m_2 a \implies T_2 - 5 = 0.5 a$ $...(II)$For the rotation of the pulley:$(T_1 - T_2) R = I \alpha = (\frac{1}{2} M R^2) (\frac{a}{R}) = \frac{1}{2} M R a$$T_1 - T_2 = \frac{1}{2} M a = \frac{1}{2} (2) a = a$ $...(III)$Adding equations $(I), (II),$ and $(III)$:$(10 - T_1) + (T_2 - 5) + (T_1 - T_2) = a + 0.5 a + a$$5 = 2.5 a$$a = \frac{5}{2.5} = 2\, m/s^2$.

For the given figure,find the acceleration of the $1\, kg$ block. The string is massless,the mass of the pulley is $M = 2\, kg$,and the diameter of the pulley is $0.2\, m$. (in $m/s^2$)

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