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Acceleration of block $m_1$ is (given moment of inertia of pulley is I and string does not slip over the pulley )

 

\[\begin {array} {1 1} (a)\;\frac{(m_1-m_2)g}{(m_1+m_2)} & \quad (b)\;\frac{(m_1-m_2)g}{(m_1+m_2+I/l^2)}\\ (c)\;\frac{(m_1-m_2+I/R^2)}{(m_1+m_2)} & \quad  (d)\;\frac{(m_1-m_2+I/R^2)}{(m_1+m_2)} \end {array}\]

1 Answer

$m_1g - T_1=m_1a$
$T_2-m_2g =m_2 a$
$(T_1-T_2) R= I \alpha$
$\qquad= \large\frac{Ia}{R}$
Since the string does not slip over the pulley the acceleration of the string and the tangential accelerated a point on the sum of the pulley is the same .
Solving the above three equation :
$a= \large\frac{m_1 g-m_2 g}{(m_1+m_2+ I/R^2)}$
answered Nov 27, 2013 by meena.p
 

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