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Two masses m and 3m are suspended from a light frictionless pulley with the help of a massless string.if the system is set free find the acceleration of the centre of mass

Asked by Debdulal | 27 Sep, 2019, 02:40: PM
Figure shows the coordinate system for finding the acceleration of centre of mass of system
that consists of two blocks of mass m and 3m respectively.

Let d is the diameter of pulley or distance between strings on both side

x-coordinate of centre of mass ( in metre ) = [ m×(d/2) + 3m×(d/2) ] /(m+3m) = d/4

magnitude of acceleration of blocks ( in m/s2 ) = [( 3m - m )/(m+3m) ] g = g/2  m/s2

direction of acceleration is +y direction for block of mass m and -y direction for block of mass 3m

Hence acceleration of block of mass ( in m/s2 ) = [ m×a - 3m×(-a) ](m+3m) = (-a/2 ) = -g/4

Answered by Thiyagarajan K | 28 Sep, 2019, 09:52: AM
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