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2 particles bearing mass in the ratio n:1 are interconnected by a light inextensible string that passes over a smooth pulley.If the system is released,then the acceleration of the centre of mass of the system is? please solve the above question with proper explaination.
draw the fbd of the system , write eqn of motion and obtain the acceleration of the bodies let us assumw it comes out as a now acceleration of the centre of mass will be given by = (m1a-m2a)/(m1+m2) negative sign because of the oppositedly directed accn of the two bodies
draw the fbd of the system , write eqn of motion and obtain the acceleration of the bodies
let us assumw it comes out as a
now acceleration of the centre of mass will be given by
= (m1a-m2a)/(m1+m2)
negative sign because of the oppositedly directed accn of the two bodies
ans is [n-1/n+1]^2*g in downwad direction. if im right then like it , i will give the solution
solution find acc. of both the blocks and put those in Acm=m1A1+m2aA2/m1+m2 with signs!!!!!!!! [very important]. thanxxx
solution
find acc. of both the blocks and put those in Acm=m1A1+m2aA2/m1+m2 with signs!!!!!!!! [very important].
thanxxx
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