If linear density of a rod of length 3cm varies as lemda=2+x,then the positionof the centre of gravity of the rod
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Heya........!!!!
It is Given that => λ = x + 2 .
=> Considering a small part dx of the Rod and then it's mass will be dm .
=> Then it's mass dm => λ × dx = ( x + 2 )dx .
=> let Xc denotes the centre of Gravity
Then through Integrating , , we get ,,
=>> Xc = ( ∫ xdm ÷ ∫ dm )
=> Xc = ∫ l xdx ÷ ∫ l dx (( here l = lamda ))
=> Putting the value of lamba we get ;
➡Xc = ∫ x(x + 2 ) dx ÷ ∫ ( x + 2 ) dx
((( Note** = We are integrating with limit of Lenght ( L ) to 0 )))
On solving that Integral and Put L = 3m as it is given in the question .
♦♦Finally we get =>> Xc = 12/7 m
Hope It Helps u ^_^
4.4
It is Given that => λ = x + 2 .
=> Considering a small part dx of the Rod and then it's mass will be dm .
=> Then it's mass dm => λ × dx = ( x + 2 )dx .
=> let Xc denotes the centre of Gravity
Then through Integrating , , we get ,,
=>> Xc = ( ∫ xdm ÷ ∫ dm )
=> Xc = ∫ l xdx ÷ ∫ l dx (( here l = lamda ))
=> Putting the value of lamba we get ;
➡Xc = ∫ x(x + 2 ) dx ÷ ∫ ( x + 2 ) dx
((( Note** = We are integrating with limit of Lenght ( L ) to 0 )))
On solving that Integral and Put L = 3m as it is given in the question .
♦♦Finally we get =>> Xc = 12/7 m
Hope It Helps u ^_^
4.4
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