A cylinder is suspended by two strings wrapped around the cylinder near each end, the free ends of the string being attached to hooks on the ceiling, such that the length of the cylinder is horizontal. From the position of rest, the cylinder is allowed to roll down as suspension strings unwind. Calculate: (i) the downward linear acceleration of the cylinder and (ii) tension in the strings. Mass of cylinder is 12kg.
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Let the downward linear acceleration of the cylinder be a. If M be the mass of the cylinder, then$\text{Mg}-2\text{T}=\text{Ma}$
$\text{or }\text{T}=\frac{1}{2}\text{m}(\text{g}-\text{a})\dots(1)$
Now, Torque = Moment of inertia × angular acceleration
$\text{i.e.,}2\text{Tr}=\text{I}\alpha$
(when $\alpha$ = liner acceleration/ r)
$\text{or }2\text{Tr}=\frac{1}{2}\text{mr}^2\times\Big(\frac{\text{a}}{\text{r}}\Big)$
$\text{or }\text{T}=\frac{\text{ma}}{4}\dots(2)$
From eqns (1) and (2), we get
$\text{m}\frac{\text{a}}{4}=\frac{1}{2}\text{m}(\text{g}-\text{a})$
Solving eqns (3), we get $\text{a}=\Big(\frac{2}{3}\Big)\text{g}$
Substituting the value of a in eq. (2), we get
$\text{T}=\frac{\text{m}\Big(\frac{2}{3}\Big)\text{g}}{4}$
$=\frac{\text{m}\times2\text{g}}{12}=\frac{12\times2\text{g}}{12}=2\text{kgf}.$
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