A thin bar of length L has a mass per unit length `lambda`, that increases linerarly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is `lambda_(0)`, then the distance of the centre of mass from the lighter end is

Options

A. `(L)/(3)+(lambda_(0)L^(2))/(8M)`
B. `(L)/(3)+(lambda_(0)L^(2))/(4M)`
C. `(L)/(2)-(lambda_(0)L^(2))/(4M)`
D. `(2L)/(3)-(lambda_(0)L^(2))/(6M)`

Answer

To find the distance of the center of mass from the lighter end of a thin bar of length \( L \) with a mass per unit length \( \lambda \) that increases linearly, we can follow these steps:

### Step 1: Define the mass per unit length
The mass per unit length \( \lambda(x) \) at a distance \( x \) from the lighter end can be expressed as:
\[ \lambda(x) = \lambda_0 + ax \]
where \( \lambda_0 \) is the mass per unit length at the lighter end (when \( x = 0 \)), and \( a \) is a constant that represents the rate of increase of mass per unit length.

### Step 2: Find the total mass
The total mass \( M \) of the bar can be calculated by integrating the mass per unit length over the length of the bar:
\[ M = \int_0^L \lambda(x) \, dx = \int_0^L (\lambda_0 + ax) \, dx \]
Calculating this integral:
\[ M = \int_0^L \lambda_0 \, dx + \int_0^L ax \, dx = \lambda_0 L + \frac{aL^2}{2} \]

### Step 3: Express \( a \) in terms of \( M \)
From the equation for total mass, we can express \( a \):
\[ M = \lambda_0 L + \frac{aL^2}{2} \implies a = \frac{2(M - \lambda_0 L)}{L^2} \]

### Step 4: Set up the center of mass equation
The center of mass \( x_{cm} \) can be found using the formula:
\[ x_{cm} = \frac{1}{M} \int_0^L x \cdot \lambda(x) \, dx \]
Substituting \( \lambda(x) \):
\[ x_{cm} = \frac{1}{M} \int_0^L x (\lambda_0 + ax) \, dx \]
Calculating this integral:
\[ x_{cm} = \frac{1}{M} \left( \int_0^L x \lambda_0 \, dx + \int_0^L ax^2 \, dx \right) \]
\[ = \frac{1}{M} \left( \lambda_0 \frac{L^2}{2} + a \frac{L^3}{3} \right) \]

### Step 5: Substitute \( a \) and simplify
Substituting \( a \) into the expression for \( x_{cm} \):
\[ x_{cm} = \frac{1}{M} \left( \lambda_0 \frac{L^2}{2} + \frac{2(M - \lambda_0 L)}{L^2} \cdot \frac{L^3}{3} \right) \]
\[ = \frac{1}{M} \left( \lambda_0 \frac{L^2}{2} + \frac{2L(M - \lambda_0 L)}{3} \right) \]

### Step 6: Final expression for center of mass
After simplifying, we find:
\[ x_{cm} = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M} \]

### Final Answer
Thus, the distance of the center of mass from the lighter end is:
\[ x_{cm} = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M} \]