Question

A particle attached to a light wire of length l is
projected horizontally with a velocity of √(5gl).
Calculate its net acceleration when wire becomes
horizontal
(1) Zero
(2) √10 g
(3) g
(4) 3g​

Answer 1

Answer:

First, we need to apply Conservation of Mechanical Energy between the Starting point and the Point of consideration.

$KE1 + PE1 = KE2 + PE2$

$= > \dfrac{1}{2} m( { \sqrt{5gl}) }^{2} + 0 = mgl + \dfrac{1}{2}m {v}^{2}$

$= > 5gl = 2gl + {v}^{2}$

$= > {v}^{2} = 3gl$

So centripetal force ( directed inwards)

$a_{c} \: = \dfrac{ {v}^{2} }{l} = \dfrac{3gl}{l} = 3g$

Acceleration due to gravity ( acting downwards)

$a_{g} \: = g$

These 2 acceleration vectors are perpendicular to one another :

Net acceleration be "a net".

$a_{net} \: = \sqrt{ {g}^{2} + {(3g)}^{2} } = g \sqrt{10}$

So final answer (Option 2)

$\boxed{ \huge{ \red{ \sf{a_{net} \: =g \sqrt{10}}}}}$

Answer 2

Explanation:

Refer to the attachment!!!!!!