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Question

(C) Let $T$ be the tension in the string and $a$ be the acceleration of the mass $m$.For the falling mass $m$,the equation of motion is: $mg - T = ma$

Vedclass · Accepted Answer

$\frac{2}{3}g$

Vedclass · Answer

(C) Let $T$ be the tension in the string and $a$ be the acceleration of the mass $m$.For the falling mass $m$,the equation of motion is: $mg - T = ma$  => $T = m(g - a)$  --- $(1)$For the rotating pulley,the torque $	au = I\alpha = TR$.Given that the pulley is a uniform disk,its moment of inertia $I = \frac{1}{2}MR^2$.Since the string does not slip,the angular acceleration $\alpha = \frac{a}{R}$.Substituting these into the torque equation: $(\frac{1}{2}MR^2)(\frac{a}{R}) = TR$ => $\frac{1}{2}Ma = T$ --- $(2)$Equating $(1)$ and $(2)$: $m(g - a) = \frac{1}{2}Ma$.If we assume the mass of the pulley $M$ is equal to the mass $m$ (as implied by the context of such standard problems),then $m(g - a) = \frac{1}{2}ma$.Dividing by $m$: $g - a = \frac{a}{2}$ => $g = \frac{3a}{2}$.Therefore,the acceleration $a = \frac{2}{3}g$.

$A$ mass $m$ is suspended from a string wrapped around a frictionless pulley. The pulley has a mass $M$ and radius $R$. If the pulley is a uniform circular disk and the string does not slip on the pulley,what is the acceleration of the mass?

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