A thin rod of length 10 feet as amass distribution such that its mass varies linearly from 2 pounds at one end to 6 pounds at the other. Determine the position of the centre of the mass from one end?

To determine the position of the center of mass of the rod, we need to find the point along the rod where the total mass can be considered to be concentrated.

Let's denote the position of the center of mass as x feet from one end of the rod. We can set up an equation using the concept of center of mass:

Total mass * position of center of mass = sum of individual masses * their positions

(2 + 6) * x = 2 * 0 + 6 * 10

8x = 60

x = 7.5 feet

Therefore, the center of mass of the rod is located 7.5 feet from one end.

Let \(x\) be the distance from the lighter end of the rod to the center of mass. Then the linear mass density function is given by

$$\rho(x) = 2 + \frac{4}{10}x = 0.2 + 0.4x \text{ pound/ft}$$

The total mass of the rod is

$$m = \int_0^{10} \rho(x) \ dx = \int_0^{10}(0.2 +0.4x ) \ dx = 0.2x + 0.2x^2 \bigg|_{0}^{10} = 2 + 20 = 22\text{ pounds}$$

The moment about the lighter end of the rod is

$$M = \int_0^{10} x\rho(x) \ dx = \int_0^{10} x(0.2+0.4x) \ dx = 0.2\frac{x^2}{2} +0.4\frac{x^3}{3} \bigg|_{0}^{10}$$

$$M = 10+40 = 50\text{ pound-ft}$$

Therefore, the position of the center of mass from the lighter end is

$$x = \frac{M}{m}=\frac{50}{22} = 2.27 \text{ feet}$$