Find centre of mass of given rod of linear mass density `lambda=(a+b(x/l)^2)` , x is distance from one of its end. Length of the rod is `l`.

Options

A. `4/3((a+b)/(2a+3b))L`
B. `3/2((a+b)/(2a+b))L`
C. `3/2((2a+b)/(3a+b))L`
D. `3/4((2a+b)/(3a+b))L.`

Answer

To find the center of mass of a rod with a given linear mass density \(\lambda = a + b\left(\frac{x}{l}\right)^2\), where \(x\) is the distance from one end of the rod and \(l\) is the length of the rod, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Linear Mass Density**:
The linear mass density of the rod is given by:
\[ \lambda(x) = a + b\left(\frac{x}{l}\right)^2 \]

2. **Express the Differential Mass Element**:
The mass element \(dm\) can be expressed in terms of the linear mass density:
\[ dm = \lambda(x) \, dx = \left(a + b\left(\frac{x}{l}\right)^2\right) dx \]

3. **Total Mass of the Rod**:
The total mass \(M\) of the rod can be calculated by integrating \(dm\) from \(0\) to \(l\):
\[ M = \int_0^l dm = \int_0^l \left(a + b\left(\frac{x}{l}\right)^2\right) dx \]

4. **Calculate the Total Mass**:
We can split the integral:
\[ M = \int_0^l a \, dx + \int_0^l b\left(\frac{x}{l}\right)^2 \, dx \]
The first integral:
\[ \int_0^l a \, dx = a \cdot l \]
The second integral:
\[ \int_0^l b\left(\frac{x}{l}\right)^2 \, dx = b\left(\frac{1}{l^2}\right) \int_0^l x^2 \, dx = b\left(\frac{1}{l^2}\right) \cdot \frac{l^3}{3} = \frac{bl}{3} \]
Thus, the total mass \(M\) is:
\[ M = al + \frac{bl}{3} = l\left(a + \frac{b}{3}\right) \]

5. **Calculate the Center of Mass**:
The center of mass \(x_{cm}\) is given by:
\[ x_{cm} = \frac{1}{M} \int_0^l x \, dm \]
Substituting \(dm\):
\[ x_{cm} = \frac{1}{M} \int_0^l x \left(a + b\left(\frac{x}{l}\right)^2\right) dx \]
This can be split into two integrals:
\[ x_{cm} = \frac{1}{M} \left( \int_0^l ax \, dx + \int_0^l b\left(\frac{x^3}{l^2}\right) dx \right) \]

6. **Calculate Each Integral**:
The first integral:
\[ \int_0^l ax \, dx = a \cdot \frac{l^2}{2} \]
The second integral:
\[ \int_0^l b\left(\frac{x^3}{l^2}\right) dx = \frac{b}{l^2} \cdot \frac{l^4}{4} = \frac{bl^2}{4} \]
Thus:
\[ \int_0^l x \, dm = a \frac{l^2}{2} + \frac{bl^2}{4} \]

7. **Combine and Simplify**:
Now substituting back into the center of mass formula:
\[ x_{cm} = \frac{1}{M} \left( a \frac{l^2}{2} + \frac{bl^2}{4} \right) \]
Substitute \(M\):
\[ x_{cm} = \frac{1}{l\left(a + \frac{b}{3}\right)} \left( a \frac{l^2}{2} + \frac{bl^2}{4} \right) \]
Simplifying gives:
\[ x_{cm} = \frac{l}{a + \frac{b}{3}} \left( \frac{a}{2} + \frac{b}{4} \right) \]

8. **Final Expression**:
After simplifying the above expression, we find:
\[ x_{cm} = \frac{3l}{4} \cdot \frac{2a + b}{3a + b} \]

### Final Answer:
The center of mass of the rod is:
\[ x_{cm} = \frac{3l(2a + b)}{4(3a + b)} \]