A thin rod of length 'L' is lying along the x -axis with its ends at x=0 and x=L . Its linear density (mass/length) varies with x as k( x \over L )^n, where n can be zero or any positive number. If the position X_CM of the center of mass of the rod is plotted against 'n', which of the following graphs best approximates the dependence of X_CM on n ?

Center of Mass PYQ — JEE Main 2008 · Shift 0 · Q78 with Solution

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Correct answer: A

Given linear mass density

The rod lies from $x=0$ to $x=L$ , and its linear density is

\lambda(x)=k\left(\frac{x}{L}\right)^n,

where $n=0,1,2,\dots$

We need the center of mass:

X_{CM}=\frac{\int x\,dm}{\int dm}.

Since $dm=\lambda(x)\,dx$ ,

X_{CM}=\frac{\int_0^L x\lambda(x)\,dx}{\int_0^L \lambda(x)\,dx}.

Compute the denominator (total mass)

M=\int_0^L k\left(\frac{x}{L}\right)^n dx =\frac{k}{L^n}\int_0^L x^n\,dx

Using

\int_0^L x^n dx=\frac{L^{n+1}}{n+1},

we get

M=\frac{k}{L^n}\cdot \frac{L^{n+1}}{n+1}=\frac{kL}{n+1}.

Compute the numerator

\int_0^L x\lambda(x)\,dx =\int_0^L x\,k\left(\frac{x}{L}\right)^n dx =\frac{k}{L^n}\int_0^L x^{n+1}dx

Now,

\int_0^L x^{n+1}dx=\frac{L^{n+2}}{n+2}.

So,

\int_0^L x\lambda(x)dx =\frac{k}{L^n}\cdot \frac{L^{n+2}}{n+2} =\frac{kL^2}{n+2}.

Find $X_{CM}$

X_{CM}=\frac{\frac{kL^2}{n+2}}{\frac{kL}{n+1}} =L\frac{n+1}{n+2}.

So the center of mass is

\boxed{X_{CM}=L\frac{n+1}{n+2}}.

Study how it depends on $n$

Let

f(n)=\frac{X_{CM}}{L}=\frac{n+1}{n+2}=1-\frac{1}{n+2}.

This tells us:

At $n=0$ , $X_{CM}=L\frac{1}{2}=\frac{L}{2}.$
As $n$ increases, $X_{CM}$ increases.
As $n\to\infty$ , $X_{CM}\to L.$ So it approaches $L$ asymptotically.

Now check curvature:

f'(n)=\frac{1}{(n+2)^2}>0,

so the graph is increasing.

f''(n)=-\frac{2}{(n+2)^3}<0,

so the graph is concave downward.

Thus the graph:

starts at $L/2$ when $n=0$ ,
rises with $n$ ,
gradually flattens,
approaches $L$ asymptotically from below.

Match with options

The correct graph must show an increasing, concave-down curve starting from $L/2$ at $n=0$ and tending to $L$ for large $n$ .

Hence the correct option is:

\boxed{A}

Comparison with stored answer

Stored correct answer: $A$

Our derived answer: $A$

They agree.

More from Center of Mass

View written solutionFree

Correct answer: A

Given linear mass density

The rod lies from $x=0$ to $x=L$ , and its linear density is

\lambda(x)=k\left(\frac{x}{L}\right)^n,

where $n=0,1,2,\dots$

We need the center of mass:

X_{CM}=\frac{\int x\,dm}{\int dm}.

Since $dm=\lambda(x)\,dx$ ,

X_{CM}=\frac{\int_0^L x\lambda(x)\,dx}{\int_0^L \lambda(x)\,dx}.

Compute the denominator (total mass)

M=\int_0^L k\left(\frac{x}{L}\right)^n dx =\frac{k}{L^n}\int_0^L x^n\,dx

Using

\int_0^L x^n dx=\frac{L^{n+1}}{n+1},

we get

M=\frac{k}{L^n}\cdot \frac{L^{n+1}}{n+1}=\frac{kL}{n+1}.

Compute the numerator

\int_0^L x\lambda(x)\,dx =\int_0^L x\,k\left(\frac{x}{L}\right)^n dx =\frac{k}{L^n}\int_0^L x^{n+1}dx

Now,

\int_0^L x^{n+1}dx=\frac{L^{n+2}}{n+2}.

So,

\int_0^L x\lambda(x)dx =\frac{k}{L^n}\cdot \frac{L^{n+2}}{n+2} =\frac{kL^2}{n+2}.

Find $X_{CM}$

X_{CM}=\frac{\frac{kL^2}{n+2}}{\frac{kL}{n+1}} =L\frac{n+1}{n+2}.

So the center of mass is

\boxed{X_{CM}=L\frac{n+1}{n+2}}.

Study how it depends on $n$

Let

f(n)=\frac{X_{CM}}{L}=\frac{n+1}{n+2}=1-\frac{1}{n+2}.

This tells us:

At $n=0$ , $X_{CM}=L\frac{1}{2}=\frac{L}{2}.$
As $n$ increases, $X_{CM}$ increases.
As $n\to\infty$ , $X_{CM}\to L.$ So it approaches $L$ asymptotically.

Now check curvature:

f'(n)=\frac{1}{(n+2)^2}>0,

so the graph is increasing.

f''(n)=-\frac{2}{(n+2)^3}<0,

so the graph is concave downward.

Thus the graph:

starts at $L/2$ when $n=0$ ,
rises with $n$ ,
gradually flattens,
approaches $L$ asymptotically from below.

Match with options

The correct graph must show an increasing, concave-down curve starting from $L/2$ at $n=0$ and tending to $L$ for large $n$ .

Hence the correct option is:

\boxed{A}

Comparison with stored answer

Stored correct answer: $A$

Our derived answer: $A$

They agree.

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