The linear density of a non`-`uniform rod of length `2m` is given by `lamda(x)=a(1+bx^(2))` where `a` and `b` are constants and `0lt=xlt=2`. The centre of mass of the rod will be at. `X=`

Options

A. `(3+6b)/(6+8b)`
B. `(6+2b)/(3+4b)`
C. `(3+6b)/(3+4b)`
D. `(1+2b)/(3+4b)`

Explanation

To find the center of mass of the non-uniform rod with a given linear density, we can follow these steps:

### Step 1: Define the linear density function
The linear density of the rod is given by:
\[ \lambda(x) = a(1 + bx^2) \]
where \( a \) and \( b \) are constants, and \( x \) varies from \( 0 \) to \( 2 \).

### Step 2: Express the mass element
The mass of a small element \( dx \) of the rod at position \( x \) can be expressed as:
\[ dm = \lambda(x) \, dx = a(1 + bx^2) \, dx \]

### Step 3: Set up the center of mass formula
The center of mass \( X_{cm} \) of the rod can be calculated using the formula:
\[ X_{cm} = \frac{1}{M} \int_0^L x \, dm \]
where \( M \) is the total mass of the rod and \( L \) is the length of the rod (which is \( 2 \) m).

### Step 4: Calculate the total mass \( M \)
To find \( M \), we integrate \( dm \) over the length of the rod:
\[ M = \int_0^2 dm = \int_0^2 a(1 + bx^2) \, dx \]
Calculating this integral:
\[ M = a \left[ x + \frac{bx^3}{3} \right]_0^2 = a \left[ 2 + \frac{b(2)^3}{3} \right] = a \left[ 2 + \frac{8b}{3} \right] = 2a + \frac{8ab}{3} \]

### Step 5: Calculate the integral for the numerator
Now we calculate the integral for the numerator:
\[ \int_0^2 x \, dm = \int_0^2 x \cdot a(1 + bx^2) \, dx = a \int_0^2 (x + bx^3) \, dx \]
Calculating this integral:
\[ \int_0^2 (x + bx^3) \, dx = \left[ \frac{x^2}{2} + \frac{bx^4}{4} \right]_0^2 = \left[ \frac{(2)^2}{2} + \frac{b(2)^4}{4} \right] = \left[ 2 + 4b \right] \]
Thus,
\[ \int_0^2 x \, dm = a(2 + 4b) \]

### Step 6: Substitute into the center of mass formula
Now we substitute \( M \) and the integral into the center of mass formula:
\[ X_{cm} = \frac{1}{M} \int_0^2 x \, dm = \frac{a(2 + 4b)}{2a + \frac{8ab}{3}} \]

### Step 7: Simplify the expression
To simplify, we can multiply the numerator and denominator by \( 3 \):
\[ X_{cm} = \frac{3a(2 + 4b)}{6a + 8ab} = \frac{6a + 12ab}{6a + 8ab} \]

### Step 8: Final expression for the center of mass
Thus, the center of mass \( X_{cm} \) is given by:
\[ X_{cm} = \frac{6 + 12b}{6 + 8b} \]