Question

Find the centre of gravity of a rod when the density at any point of the rod varies as the distance from one end of the rod.​

Answer 1

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 ^_^

sorry for before distrubances