Answer

Verified

81k+ views

**Hint:**We need to consider the area of cross section of the rod of length L and negligible width. Mass is the product of density and volume where volume is given by the length of an element.

**Formula Used:**

The formulae used in the solution are given here.

The x-coordinate of mass is given by ${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$.

**Complete Step by Step Solution:**Centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particles appears to be concentrated.

It has been given that the linear density of rod of length L and placed along the x-axis with the lighter end at the origin, is given by $\lambda = B{x^2}$ where $B$ is a constant.

Density is a measure of mass per volume. The average density of an object equals its total mass divided by its total volume.

Consider an element of length $dx$ of the rod at a distance $x$ from the left end. Since the rod is a linear rod, it has negligible thickness and the area of cross section is $a$.

Area of the rod $a = dx$.

Thus, the mass of the element, $dm$ is given by $dm = \lambda dx = \left( {B{x^2}} \right)dx$

The x-coordinate of mass is given by ${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$.

It can also be written as, ${X_{CM}} = \dfrac{{\int {dmx} }}{{\int {dm} }} = \dfrac{{\int\limits_0^L {x\left( {B{x^2}} \right)adx} }}{{\int\limits_0^L {\left( {B{x^2}} \right)adx} }}$

Simplifying the equation, we get, $\dfrac{{B\dfrac{{\left[ {{x^4}} \right]_0^L}}{4}}}{{B\dfrac{{\left[ {{x^3}} \right]_0^L}}{3}}} = \dfrac{{B{L^4}}}{4} \times \dfrac{3}{{B{L^3}}}$

Thus, the x-coordinates of the centre of mass are: ${X_{CM}} = \dfrac{{3L}}{4}$.

The y-coordinate is equal to zero, because the rod is linear and has no vertical component.

**Hence the correct answer is Option A.**

**Note:**Motion of the point of centre of mass is identical to the motion of a single particle whose mass is equal to the sum of all individual particles of the system and the resultant of all the forces exerted on all the particles of the system by surrounding bodies (or) action of a field of force is exerted directly to that particle. This point is called the centre of mass of the system of particles. The concept of centre of mass (COM) is useful in analysing the complicated motion of the system of objects, particularly When two and more objects collide or an object explodes into fragments.

Recently Updated Pages

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

Which of the following is not a redox reaction A CaCO3 class 11 chemistry JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The value of intlimits02pi max left sin xcos x right class 12 maths JEE_Main

Man A sitting in a car moving with a speed of 54 kmhr class 11 physics JEE_Main