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 with a mass per unit length that increases linearly, we can follow these steps:

### Step 1: Define the mass per unit length
The mass per unit length \( \lambda(x) \) of the bar increases linearly from \( \lambda_0 \) at \( x = 0 \) to some value at \( x = L \). We can express this linear relationship as:
\[ \lambda(x) = ax + \lambda_0 \]
where \( a \) is a constant that we need to determine.

### Step 2: Determine 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:
\[ M = \int_0^L \lambda(x) \, dx = \int_0^L (ax + \lambda_0) \, dx \]
Calculating this integral:
\[ M = \int_0^L ax \, dx + \int_0^L \lambda_0 \, dx = \left[ \frac{a x^2}{2} \right]_0^L + \left[ \lambda_0 x \right]_0^L = \frac{a L^2}{2} + \lambda_0 L \]

### Step 3: Solve for \( a \)
Rearranging the equation for \( M \):
\[ M = \frac{a L^2}{2} + \lambda_0 L \]
From this, we can solve for \( a \):
\[ a = \frac{2(M - \lambda_0 L)}{L^2} \]

### Step 4: Find the center of mass
The x-coordinate of the center of mass \( x_{cm} \) is given by:
\[ x_{cm} = \frac{1}{M} \int_0^L x \cdot \lambda(x) \, dx = \frac{1}{M} \int_0^L x (ax + \lambda_0) \, dx \]
Calculating this integral:
\[ x_{cm} = \frac{1}{M} \left( \int_0^L ax^2 \, dx + \int_0^L \lambda_0 x \, dx \right) = \frac{1}{M} \left( \left[ \frac{a x^3}{3} \right]_0^L + \left[ \frac{\lambda_0 x^2}{2} \right]_0^L \right) \]
\[ = \frac{1}{M} \left( \frac{a L^3}{3} + \frac{\lambda_0 L^2}{2} \right) \]

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

### Step 6: Simplify the expression
Now we simplify:
\[ x_{cm} = \frac{1}{M} \left( \frac{2L}{3}M - \frac{2\lambda_0 L^2}{3} + \frac{\lambda_0 L^2}{2} \right) \]
Finding a common denominator for the last two terms:
\[ = \frac{1}{M} \left( \frac{2L}{3}M + \left( -\frac{4\lambda_0 L^2}{6} + \frac{3\lambda_0 L^2}{6} \right) \right) \]
\[ = \frac{1}{M} \left( \frac{2L}{3}M - \frac{\lambda_0 L^2}{6} \right) \]

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