Question

A rod has length 50cm, with non-uniform mass per length density, p = pox, where p. is a constant and x is the distance from the left end of the rod. The left end of the rod is pivoted and the right end is tied to a string, and the rod is kept horizontal. What is the string's tension?

Answer 1

1/12 * ρ₀ * g

Answer 2

The rod has length $L = 50$ cm = 0.5 m.
The mass per unit length density is given by $\rho = \rho_0 x$, where $x$ is the distance from the left end (pivot).

First, we find the total mass $M$ of the rod by integrating the density over the length:
$M = \int_0^L \rho \,dx = \int_0^L \rho_0 x \,dx = \rho_0 \left[ \frac{x^2}{2} \right]_0^L = \rho_0 \frac{L^2}{2}$.

Next, we find the position of the center of mass $x_{CM}$ from the left end:
$x_{CM} = \frac{\int_0^L x \,dm}{\int_0^L dm} = \frac{\int_0^L x (\rho_0 x \,dx)}{M} = \frac{\int_0^L \rho_0 x^2 \,dx}{M} = \frac{\rho_0 \left[ \frac{x^3}{3} \right]_0^L}{M} = \frac{\rho_0 \frac{L^3}{3}}{M}$.
Substituting $M = \rho_0 \frac{L^2}{2}$:
$x_{CM} = \frac{\rho_0 \frac{L^3}{3}}{\rho_0 \frac{L^2}{2}} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}$.

The rod is in horizontal equilibrium, pivoted at the left end and supported by a string at the right end. The forces acting on the rod are:

1. The weight of the rod, $W = Mg$, acting vertically downwards at the center of mass $x_{CM} = \frac{2L}{3}$ from the pivot.
2. The tension $T$ in the string, acting vertically upwards at the right end, which is at a distance $L$ from the pivot.
3. The reaction force at the pivot (which does not create torque about the pivot).

For rotational equilibrium about the pivot, the net torque must be zero. Taking counter-clockwise torque as positive:
$\sum \tau_{pivot} = 0$
The torque due to tension is $T \times L$ (counter-clockwise).
The torque due to weight is $W \times x_{CM}$ (clockwise).
$T \times L - W \times x_{CM} = 0$
$T L = W x_{CM}$
$T L = (Mg) \left( \frac{2L}{3} \right)$
$T = Mg \frac{2}{3}$.

Substitute the expression for $M = \rho_0 \frac{L^2}{2}$:
$T = \left( \rho_0 \frac{L^2}{2} \right) g \frac{2}{3} = \rho_0 \frac{L^2}{3} g$.

Now, substitute the given length $L = 50$ cm = 0.5 m:
$T = \rho_0 \frac{(0.5 \text{ m})^2}{3} g = \rho_0 \frac{0.25 \text{ m}^2}{3} g = \rho_0 \frac{1/4}{3} g = \frac{1}{12} \rho_0 g$.

The tension in the string is $\frac{1}{12} \rho_0 g$.