Question

Q2. A chain of length / lies on a smooth inclined plane with
half of its length hanging vertically (part AB). From this
position the hanging part AB starts falling vertically
downwards. Find the velocity of the chain when it is on
the verge of leaving the inclined surface.
​

Answer 1

Answer:

$\boxed{\huge{\pink{ \dfrac{1}{2} \sqrt{g \: l}}}}$

Explanation:

$\bold{\underline{\green{Given}}}\longrightarrow \\ a \: chain \: of \: length \: l \: lies \: on \: a \: plane \\ as \: shown \: in \: figure \: and \: chain \: starts \: falling \: downwards$

$\bold{\underline{\red{To \: find}}}\longrightarrow \\ velocity \: of \: chain \: when \: it \: is \: on \: verge \\ to \: leave \: the \: plane$

$\bold{\underline{\pink{Concept \: used}}}\longrightarrow \\ force \: = mass \times acceleration$

$\bold{\underline{\blue{Solution}}}\longrightarrow \\ let \: mass \: of \: whole \: chain \: be \: m \: \\ and \: tension \: in \: chain \: be \: T \: \\ and \: acceleration \: be \: a$

$now \: for \: motion\:of \: hanging \: part \: of \: chain \\ mass \: of \: hangig \: part = \frac{m}{2} \\ now \\ \dfrac{m}{2} g \: - T = \dfrac{m}{2} a...............(1) \\ now \: for \: motion \: of \: part \: that \: is \: on \: plane \\ T \: - \: \dfrac{m}{2}g \: sin30 ^{0} = \dfrac{m}{2} a \\ T \: - \dfrac{m}{2} g \: \dfrac{1}{2} = \dfrac{m}{2} a \\ T \: - \: \frac{m}{4} g \: = \dfrac{m}{2} a.............(2)$

$adding \: equations \: 1 \: and \: 2$

$\dfrac{m}{2} g - T + T - \dfrac{m}{4} g = \dfrac{m}{2} a + \dfrac{m}{2} a$

$( \dfrac{m}{2} - \dfrac{m}{4} )g = ( \dfrac{m}{2} + \dfrac{m}{2} )a$

$\dfrac{m}{4} g = ma$

$a = \dfrac{g}{4}$

$now \: for \: leaving \: the \: plane \: part \: of \\ chain \: which \: is \: on \: plane \: cover \: \\ distance \: \dfrac{l}{2} \\ now \\ u = 0 \:and \: s = \dfrac{l}{2} and \: a = \dfrac{g}{4} \\ now \\ {v}^{2} = {u}^{2} + 2as \\ = > {v}^{2} = (0) ^{2} + 2 \: ( \dfrac{g}{4} ) \: ( \dfrac{l}{2} ) \\ = > {v}^{2} = \dfrac{1}{4} gl \\ = > v \: = \sqrt{ \dfrac{1}{4}gl } \\ = > v = \dfrac{1}{2} \sqrt{gl}$