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A smooth pulley A of mass M_(0)is lying on a smooth table . A light string passes round the pulley and has masses M_(1) "and" M_(2)attached to its ends , the two portions of the string being perpendicular to the edge of the table so that the masses hang vertically . Calculate the acceleration of the pulley . |
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Answer» SOLUTION :Let the length of portions of string on the table be `l_(0),l_(1) "and" l_(2) ` as shown in figure . Let mass `M_(0)` move to the right by x on the table ,` M_(1) `goes down by `y_(1) "and" M_(2) ` goes up by `y_(2)` Then `l_(0)` becomes `(l_(0)-x)` l_(1)` becomes `(l_(1)+y_(1))` and `l_(2) ` becomes `(l_(2)-y_(2))`. As the length of string remains unchanged `2l_(0)+l_(1)+l_(2)=2(l_(0)-x)+l_(1)+y_(1)+l_(2)-y_(2)` or `2x=y_(1)-y_(2)` Diff. twice w.r. to t, we get `2(d^(2)x)/(dt^(2))=(d^(2)y_(1))/(dt^(2)-(d^(2)y_(2))/(dt^(2))` If a_(0),a_(1)"and"a_(2)` are ACCELERATION of  `M_(0),M_(1)` and `M_(2)` respectively , then `2a_(0)=a_(1)-a_(2)`....(1) For motion of `M_(0),2T=M_(0)a_(0)`....(2) For motion of mass `M_(1)` `M_(1)g-T=M_(1)a_(1)`....(3) For motion of mass `M_(2)`, `T-M_(2)g=M_(2)a_(2)`....(4) Substituting values of `a_(0),a_(1)` and `a_(2)` from (2) ,(3) and (4) in equation (1) , we get `2((2T)/M_(0))=(g-T/M_(1)-(T/M_(2)-g),` `(4T)/M_(0)=2g-T(1/M_(1)+1/M_(2)) `i.e., (4/M_(0)+1/M_(1)+1/M_(2))T=2g` This gives `T=(2M_(0)M_(1)M_(2)g)/(4M_(1)M_(2)+M_(0)(M_(1)+M_(2))`....(5) `:. ` Aceeleration of PULLEY A from (2) `a_(0) =(2T)/M_(0)=(4M_(1)M_(2)g)/(4M_(1)M_(2)+M_(0)(M_(1)+M_(2))`....(6) |
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