A uniform disc with mass M = 4,kg and radius | vedclass

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(D) For the falling block of mass $m$,the equation of motion is: $mg - T = ma$  ...........$(1)$For the rotating disc of mass $M$,the torque equation

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$10$

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(D) For the falling block of mass $m$,the equation of motion is: $mg - T = ma$  ...........$(1)$For the rotating disc of mass $M$,the torque equation is: $TR = I\alpha = (\frac{1}{2}MR^2)\alpha$  ...........$(2)$Since the cord does not slip,the linear acceleration $a$ of the block is related to the angular acceleration $\alpha$ of the disc by: $a = R\alpha$  or $\alpha = \frac{a}{R}$  ...........$(3)$Substituting $(3)$ into $(2)$: $TR = \frac{1}{2}MR^2(\frac{a}{R}) \implies T = \frac{1}{2}Ma$Given $M = 4\,kg$ and $m = 2\,kg$,we have $T = \frac{1}{2}(4)a = 2a$Substitute $T = 2a$ into $(1)$: $mg - 2a = ma \implies (2)(10) - 2a = 2a$$20 = 4a \implies a = 5\,m/s^2$Now,calculate the tension $T$: $T = 2a = 2(5) = 10\,N$.

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