Question

help me how to solve the question please​

Answer 1

QUESTION:

A bob of mass m is rotating in a vertical plane, if velocity of the bob at the lower most position is V = √(3gl). The angle rotated by the bob before string become loose is.

ANSWER:

• Option 2) π/2 + arcsin(1/3)

GIVEN:

• A bob of mass m is rotating in a vertical plane,

• Vlocity of the bob at the lower most position is V = √(3gl).

TO FIND:

• The angle rotated by the bob before string becomes loose.

EXPLANATION:

$\boxed{ \bold{ \large{ \gray{T = \dfrac{m}{r}(u^2+gl-3gl)}}}}$

Let l = l' where the tension is zero( string becomes loose)

u = √(3gl)

T = 0

$\sf 0 = \dfrac{m}{r}(( \sqrt{3gl} )^2+gl-3gl')$

$\sf 3gl+gl-3gl' = 0$

$\sf 4gl - 3gl' = 0$

$\sf 4gl = 3gl'$

$\sf 4l = 3l'$

$\sf l' = \dfrac{4}{3} l$

$\boxed{ \bold{ \large{ \gray{T - mgcos\ \theta = \dfrac{mv^2}{r}}}}}$

T = 0

r = l

$\boxed{ \bold{ \large{ \gray{v = \sqrt{u^2- 2gh}}}}}$

$\sf 0 - mgcos\ \theta = \dfrac{mv^2}{l}$

$\sf - mgcos\ \theta = \dfrac{m( \sqrt{ {u}^{2} - 2gh} {)}^{2} }{l}$

u = √(3gl)

h = l' = 4/3 l

$\sf - gcos\ \theta = \dfrac{( \sqrt{3gl} {)}^{2} - 2g \left( \dfrac{4}{3}l \right)}{l}$

$\sf - gcos\ \theta = \dfrac{3gl- \left( \dfrac{8g}{3}l \right)}{l}$

$\sf - gcos\ \theta =gl \left( \dfrac{ \left( \dfrac{9 - 8}{3} \right)}{l} \right)$

$\sf -cos\ \theta = \dfrac{ \left( \dfrac{1}{3} \right)}{1}$

$\sf cos\ \theta = - \dfrac{1}{3}$

$\sf \theta = {cos}^{ - 1} \left( - \dfrac{1}{3} \right)$

$\boxed{ \bold{ \large{ \gray{Cos ^{-1} (-x) = \pi - Cos ^{-1} x}}}}$

$\sf \theta = \pi - {cos}^{ - 1} \left( \dfrac{1}{3} \right)$

$\boxed{ \bold{ \large{ \gray{\dfrac{ \pi}{2} - {sin}^{ - 1} (x)= {cos}^{ - 1} (x)}}}}$

$\sf \theta = \pi - \dfrac{ \pi}{2} + {sin}^{ - 1} ( \dfrac{1}{3} )$

$\sf \theta = \dfrac{ \pi}{2} + {sin}^{ - 1} ( \dfrac{1}{3} )$

Hence the angle rotated by the bob before string becomes loose = π/2 + arcsin(1/3).