If the linear mass density of a rod of length 3 m (lying from` y=0` to` y=3`m) varies as `lambda`=(2+y)kg/m,then position of the centre of mass of the rod from y=0 is nearly at

Options

A. 2m
B. 1.7m
C. 2.5m
D. 2.7m

Answer

To find the position of the center of mass of a rod with a varying linear mass density, we can follow these steps:

### Step 1: Define the linear mass density
The linear mass density of the rod is given as:
\[ \lambda(y) = 2 + y \quad \text{(kg/m)} \]
where \( y \) is the position along the length of the rod.

### Step 2: Set up the expressions for the center of mass
The center of mass \( y_{cm} \) of a continuous mass distribution is given by the formula:
\[ y_{cm} = \frac{\int_0^L y \, dm}{\int_0^L dm} \]
where \( L \) is the length of the rod.

### Step 3: Express \( dm \) in terms of \( dy \)
The mass element \( dm \) can be expressed in terms of the linear mass density:
\[ dm = \lambda(y) \, dy = (2 + y) \, dy \]

### Step 4: Calculate the numerator and denominator
1. **Numerator**:
\[ \int_0^3 y \, dm = \int_0^3 y \cdot (2 + y) \, dy \]
Expanding this:
\[ = \int_0^3 (2y + y^2) \, dy \]
Now, integrate term by term:
\[ = \left[ y^2 + \frac{y^3}{3} \right]_0^3 \]
Evaluating at the limits:
\[ = \left[ 3^2 + \frac{3^3}{3} \right] - \left[ 0 + 0 \right] = 9 + 9 = 18 \]

2. **Denominator**:
\[ \int_0^3 dm = \int_0^3 (2 + y) \, dy \]
Integrating:
\[ = \left[ 2y + \frac{y^2}{2} \right]_0^3 \]
Evaluating at the limits:
\[ = \left[ 2(3) + \frac{3^2}{2} \right] - \left[ 0 + 0 \right] = 6 + 4.5 = 10.5 \]

### Step 5: Calculate the center of mass
Now substituting the values into the center of mass formula:
\[ y_{cm} = \frac{18}{10.5} = \frac{36}{21} \approx 1.714 \, \text{m} \]

### Final Answer
The position of the center of mass of the rod from \( y=0 \) is approximately \( 1.7 \, \text{m} \).

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