Question

Yoni rod of uniform thickness is placed along x axis with one end of origin if length of rod is l and its linear mass density is proportional to X then find the distance of centre of mass from origin​

Answer 1

Answer:

Given:

Rod of uniform thickness and linear mass density is placed along x axis.

Let linear mass density be λ.

To find:

Distance of Centre of Mass from origin.

Calculation:

As per question ;

$\lambda \: \propto \: x$

Introducing a constant :

$\lambda \: = k x$

Now , trying to find out the centre of mass :

$\displaystyle \bar x = \dfrac{ \int \: x \: dm}{ \int \: dm}$

We will be considering a small mass element at a distance of x from origin (0,0)

$\displaystyle \implies\bar x = \dfrac{ \int \: x \: ( \lambda \: \times dx)}{ \int \: (\lambda \: \times dx)}$

Putting value of λ :

$\displaystyle \implies\bar x = \dfrac{ \int \: x \: ( kx \: \times dx)}{ \int \: (kx \: \times dx)}$

$\displaystyle \implies\bar x = \dfrac{ \int \: ( {x}^{2} \: \times dx)}{ \int \: (x \: \times dx)}$

$\implies \bar x = \dfrac{ \bigg \{\dfrac{ {x}^{3} }{3} \bigg \} }{ \bigg \{\dfrac{ {x}^{2} }{2} \bigg \}}$

Putting the limits :

$\implies \bar x = \dfrac{ \bigg \{\dfrac{ {x}^{3} }{3} \bigg \}_{0}^{l} }{ \bigg \{\dfrac{ {x}^{2} }{2} \bigg \}_{0}^{l}}$

$\implies \bar x = \dfrac{2}{3} l$

So final answer :

$\boxed{ \huge{ \red{ \bold{\bar x = \dfrac{2}{3} l}}}}$

Answer 2

SoLuTiOn :

⏭ Given:

✏ A rod of uniform thickness is placed along x-axis with one end of origin.

✏ Length of rod = L

✏ Linear mass density of rod is directly proportional to x(distance from origin).

⏭ To Find:

✏ Distance of centre of mass from origin.

⏭ Formula:

✏ Formula of centre of mass for uniform mass distribution is given by

$\orange{ \bigstar} \: \underline{ \boxed{ \bold{ \sf{ \purple{x{ \tiny{cm}} = \frac{ \int{x \: dm}}{ \int{dm}} }}}}} \: \orange{ \bigstar}$

⏭ Calculation:

$\mapsto \sf \: As \: per \: given \: question \\ \\ \sf \: \lambda \propto \: x \: \implies \: \lambda = kx \\ \\ \mapsto \sf \: k = proportionality \: constant$

• Centre of mass of rod :

$\implies \sf \: x{ \tiny{cm}} = \frac{ \int{x \: dm}}{ \int{dm}} \\ \\ \implies \sf \: we \: know \: that \: \red{dm = \lambda \: dx} \\ \\ \implies \sf \: x{ \tiny{cm}} = \frac{ \int{x \: ( \lambda \: dx)}}{ \int( \lambda \: dx)} = \frac{ \int \: x \: (kx \: dx)}{ \int(kx \: dx)} \\ \\ \implies \sf \: x{ \tiny{cm}} = \frac{( \frac{ {x}^{3} }{3} )}{ (\frac{ {x}^{2} }{2}) } =\lim_{0 \to L} {\frac{2x}{3} } \\ \\ \implies \sf \: \underline{ \boxed{ \bold{ \sf{ \pink{x{ \tiny{cm}} = \frac{2L}{3}}}}}} \: \orange{ \bigstar}$