To find the center of mass of the rod, we can use the formula for the center of mass of a non-uniform rod with a variable mass density.
Given:
• The length of the rod = L
• The linear mass density λ=λ0x\lambda = \lambda_0 x, where xx is the distance from the origin along the x-axis.
• We are to find the distance of the center of mass from the origin.
Step 1: Mass element
At a small distance dxdx from the origin, the mass of the element is:
dm=λdx=λ0xdxdm = \lambda dx = \lambda_0 x dx
where xx is the position along the rod.
Step 2: Total mass of the rod
To find the total mass of the rod, integrate the mass element over the length of the rod:
M=∫0Lλ0x dxM = \int_0^L \lambda_0 x \, dx
Solving this:
M=λ0∫0Lx dx=λ0[x22]0L=λ0L22M = \lambda_0 \int_0^L x \, dx = \lambda_0 \left[\frac{x^2}{2}\right]_0^L = \frac{\lambda_0 L^2}{2}
Step 3: Center of mass formula
The center of mass xcmx_{\text{cm}} is given by the formula:
xcm=1M∫0Lx dmx_{\text{cm}} = \frac{1}{M} \int_0^L x \, dm
Substitute dm=λ0xdxdm = \lambda_0 x dx into the formula:
xcm=1M∫0Lxλ0x dx=1M∫0Lλ0x2 dxx_{\text{cm}} = \frac{1}{M} \int_0^L x \lambda_0 x \, dx = \frac{1}{M} \int_0^L \lambda_0 x^2 \, dx
Now, integrate:
xcm=1M⋅λ0∫0Lx2 dx=1M⋅λ0[x33]0L=1M⋅λ0⋅L33x_{\text{cm}} = \frac{1}{M} \cdot \lambda_0 \int_0^L x^2 \, dx = \frac{1}{M} \cdot \lambda_0 \left[\frac{x^3}{3}\right]_0^L = \frac{1}{M} \cdot \lambda_0 \cdot \frac{L^3}{3}
Step 4: Substitute for M
Substitute M=λ0L22M = \frac{\lambda_0 L^2}{2} into the equation for xcmx_{\text{cm}}:
xcm=λ0L33λ0L22=2L3x_{\text{cm}} = \frac{\frac{\lambda_0 L^3}{3}}{\frac{\lambda_0 L^2}{2}} = \frac{2L}{3}
Final Answer:
The distance of the center of mass from the origin is 2L3\frac{2L}{3}.
So, the correct answer is (B) 2L3\frac{2L}{3}.
