Question

Holaa!! Question ❤

An uniform iron rod of length L and mass M is placed up on x - axis at a distance x from the origin, Find out the center of mass of the rod :)

Thenku ^^"


​

Answer 1

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Gievn that :}}}}}}$

• The length of the iron rod is considered to be L
• The mass of the rod is considered to be M
• The distance of it from the origin is considered to be x

${\maltese \; \; {\underline{\underline{\textsf{\textbf{To Find :}}}}}}$

• The centre of mass of the iron rod

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Assumptions :}}}}}}$

✸ We know that the length of the rod is considered to be L and the mass is said to be M so , the unitary length of the rod will be such that ;

$\longrightarrow \sf L = \dfrac{Mass }{ Unitary \; Length}$

$\longrightarrow \sf Unitary \; Length = \dfrac{Mass}{Length}$

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Required Solution:}}}}}}$

We , know that the formula to find the centre of mass in a n number of particle system is ,

$\bigstar \; {\underline{\boxed{\tt{\vec{r}com = \dfrac{m_1 \vec{r}_1 +m_2\vec{r}_2 + . \; . \; . \; m_n \vec{r}_n }{m_1 + m_2 + . \; .\; . \; m_n} }}}}$

Where m1 , m2 so on mn are the masses of the particles and r1 , r2 so on rn are the position vectors of the particles , or the formula can be written in such way too :

$\qquad \bullet \; {\footnotesize{\underline{\boxed{\bf{ \dfrac{\sum \limits_{i = 1}^{n} m_i r_i}{M} }}}}}$  

Since it is difficult to find out the position vectors and the masses of the particles let's integrate to find the centre of mass

$\qquad \maltese \; {\footnotesize{\underline{\boxed{\bf{ \vec{r} com =\dfrac{ \int \limits_{0}^{l} dm \; x}{ \int \limits _{0}^{l} dm} }}}}}$

In this case let us consider to position vector of center of mass as xcom

$\vec{r}com = \vec{x}com$

Now let us consider the mass of each element to be

$\sf dm = \dfrac{M}{L} dx$

Now let's find out the centre of mass of the particles using formula ;

$\longrightarrow {\bf{ \vec{x} com =\dfrac{ \int \limits_{0}^{l} dm \; x}{ \int \limits _{0}^{l} dm} }}$

Since , we know that the rod lies on x - axis so the co - ordinates of the y - z plane will be (0,0 )

★ Integrating we get ;

$\longrightarrow {\sf{ \vec{x} com =\dfrac{ \int \limits_{0}^{l} dm \; x}{ \int \limits _{0}^{l} dm} }} \\ \\ \\ \longrightarrow {\sf{ \vec{x} com =\dfrac{ \int \limits_{0}^{l} \dfrac{M}{L} \; dx \; x}{ \int \limits _{0}^{l} \dfrac{M}{L} dx } }} \\ \\ \\ \longrightarrow {\sf{ \vec{x} com =\dfrac{ \dfrac{M}{L} \int \limits_{0}^{l} \; dx \; x}{ \dfrac{M}{L} \int \limits _{0}^{l} dx } }}$

• Canceling M/L on both the numerator and denominator ;

$\longrightarrow {\sf{ \vec{x} com =\dfrac{ \int \limits_{0}^{l} \; dx \; x}{ \int \limits _{0}^{l} dx } }}$

Now let's integrate the numerator and the denominator by using formula ;

$\qquad \maltese \; {\footnotesize{\underline{\boxed{\bf\int \limits_{}^{} x^n dx = \dfrac{x^n+ 1}{n + 1} + c }}}}$

Here C is the integration constant and in the above given case we have the limits too, Instead writing + C we enclose the result with the limits

$\longrightarrow {\bf{ \vec{x} com = \dfrac{\bigg[\dfrac{x^{1 + 1}}{1+ 1}\bigg]^l_0 }{\bigg[\dfrac{x^{0 + 1}}{1}\bigg]^l_0}}}$

$\longrightarrow {\bf{ \vec{x} com = \dfrac{\bigg[\dfrac{x^{2}}{2}\bigg]^l_0 }{\bigg[\dfrac{x^{1}}{1}\bigg]^l_0}}}$

$\longrightarrow {\bf{ \vec{x} com = \dfrac{\bigg[\dfrac{x^{2}}{2}\bigg]^l_0 }{\bigg[x \bigg]^l_0}}}$  

$\longrightarrow {\bf{ \vec{x} com = \dfrac{\dfrac{1}{2} \bigg[l^2 - 0^2\bigg] }{\bigg[ l - 0\bigg]}}}$

$\longrightarrow {\bf{ \vec{x} com = \dfrac{1}{2} * \dfrac{l^2 }{ l }}}$

$\longrightarrow {\bf{ \vec{x} com = {\purple{\underline{\boxed{\bf \dfrac{l}{2} }}}}\bigstar} }$

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Therefore :}}}}}}$

• The centre of mass of the particle would lie at a distance of l/2 or the centre of the rod

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

* Note : If any error found in viewing from the app kindly see the answer from the web

See this question at : https://brainly.in/question/43794495

Answer 2

Ques-

An uniform iron rod of length L and mass M is placed up on x - axis at a distance x from the origin, Find out the center of mass of the rod

Answer -

Centre of mass of a uniform object lies at its geometric centre. So, centre of mass of the rod lies at a distance of L/2 from the the end.