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:
\[ \lambda(y) = 2 + y \text{ kg/m} \]
This means that the mass per unit length of the rod changes as we move along the y-axis.

### Step 2: Set up the expression for the center of mass
The center of mass \( Y_{cm} \) for a continuous mass distribution along the y-axis can be calculated using the formula:
\[ Y_{cm} = \frac{\int y \, dm}{\int dm} \]
where \( dm \) is the differential mass element.

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

### Step 4: Substitute \( dm \) into the center of mass formula
Now we can substitute \( dm \) into the center of mass formula:
\[ Y_{cm} = \frac{\int_0^3 y \, (2 + y) \, dy}{\int_0^3 (2 + y) \, dy} \]

### Step 5: Calculate the numerator
First, we calculate the numerator:
\[ \int_0^3 y(2 + y) \, dy = \int_0^3 (2y + y^2) \, dy \]
Calculating this integral:
\[ = \left[ y^2 + \frac{y^3}{3} \right]_0^3 = \left[ 3^2 + \frac{3^3}{3} \right] - \left[ 0 + 0 \right] = 9 + 9 = 18 \]

### Step 6: Calculate the denominator
Now, we calculate the denominator:
\[ \int_0^3 (2 + y) \, dy = \left[ 2y + \frac{y^2}{2} \right]_0^3 = \left[ 2(3) + \frac{3^2}{2} \right] - \left[ 0 + 0 \right] = 6 + \frac{9}{2} = 6 + 4.5 = 10.5 \]

### Step 7: Calculate \( Y_{cm} \)
Now we can substitute the results back into the center of mass formula:
\[ Y_{cm} = \frac{18}{10.5} = \frac{36}{21} = \frac{12}{7} \approx 1.714 \text{ m} \]

### Conclusion
Thus, the position of the center of mass of the rod from \( y = 0 \) is approximately:
\[ Y_{cm} \approx 1.7 \text{ m} \]

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