A mass m supported by massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
${g$ = acceleration due to gravity $}$

Question

TestHub · Accepted Answer

Equation of motion for the falling mass $m$ is:
$m g - T = ma$

Equation of rotational motion for the hollow cylinder is:
$τ = I α$
$T \cdot R = I α$

The moment of inertia of a uniform hollow cylinder about its central axis is $I = m R^{2}$ .

For no slipping of the string, the linear acceleration $a$ and angular acceleration $α$ are related by $a = R α$ , which gives $α = \frac{a}{R}$ .

Substituting $I$ and $α$ into the torque equation:
$T \cdot R = (m R^{2}) (\frac{a}{R})$
$T = ma$

Substituting $T = ma$ into the first equation:
$m g - ma = ma$
$m g = 2 ma$
$a = \frac{g}{2}$

Answer: $\frac{g}{2}$

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