Question

if linear density of a rod of length 3 M varies as Lambda is equals to 2 + X then the position of the centre of gravity of the road is

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 ^_^

Answer 2

The position of the centre of gravity of the given rod is 12/7 m.

The linear density of a rod of length 3m varies as λ = 2 + x.

We have to find the position of the centre of gravity of the road.

Concepts :

• Centre of gravity of a uniform rigid body is also centre of mass.
• And the centre of mass for an element of rigid body is given by,

$x_{cm}=\frac{\int{dm\times x}}{\int{dm}}$

• Linear mass density is rate of change of mass per unit length.

$\implies \lambda=\frac{dm}{dx}=(2+x)\\\\\implies dm=(2+x)dx$

Now centre of mass of rigid body is,

$x_{cm}=\frac{\int\limits^3_0{(2+x)xdx}}{\int\limits^3_0{(2+x)dx}}\\\\\\=\frac{\left[x^2+\frac{x^3}{3}\right]^3_0}{\left[2x+\frac{x^2}{2}\right]^3_0}\\\\\\=\frac{12}{7}$

As we already mentioned that centre of mass for a uniform rigid body is also the centre of gravity.

Hence centre of gravity of the rod is 12/7 m.

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