Question

Q3.
A mass 'm' is supported by a 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?
R
V m
Oa) 2g/3
Ob) g/2
Oc) 5g/6
d) 9​

Answer 1

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle{14}}\put(7,0){\line(0,-1){20}}\qbezier(-7.1,0)(-6.8,6)(0,7)\qbezier(0,7)(6.3,6)(7,0)\put(4.5,-25){\framebox(5,5){\sf{m}}}\put(-1.6,9){\sf{m}}\put(0,0){\vector(1,0){7}}\put(2,1){\footnotesize\textsf{R}}\put(7,-25){\vector(0,-1){10}}\put(4.5,-39){\sf{mg}}\multiput(0,0)(3,-23){2}{\put(10,5){\vector(0,-1){10}}\put(12,-0.5){\sf{a}}}\end{picture}$

Let the linear acceleration of the string be $\sf{a}$ acting downwards.

Then we can see the point of contact of the string with the cylinder has the same linear acceleration $\sf{a.}$

But the point of contact of the string with the block goes down with the block due to its weight. Hence this point has the acceleration, nothing but $\sf{g.}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle{14}}\put(7,0){\line(0,-1){20}}\qbezier(-7.1,0)(-6.8,6)(0,7)\qbezier(0,7)(6.3,6)(7,0)\put(4.5,-25){\framebox(5,5){\sf{m}}}\put(-1.6,9){\sf{m}}\put(0,0){\vector(1,0){7}}\put(2,1){\footnotesize\textsf{R}}\put(7,-25){\vector(0,-1){10}}\put(4.5,-39){\sf{mg}}\multiput(0,0)(3,-10){2}{\put(10,5){\vector(0,-1){10}}\put(12,-0.5){\sf{a}}}\put(6,-10){\line(1,0){2}}\multiput(7,0)(0,-20){2}{\circle*{1}}\end{picture}$

Thus by constraint equation applied on the string, we have,

$\longrightarrow\sf{-a+g=a}$

$\longrightarrow\sf{2a=g}$

$\longrightarrow\underline{\underline{\sf{a=\dfrac{g}{2}}}}$

Hence (b) is the answer.