Question

3 thin uniform rods each of length L & mass M are joined to form an equilateral triangle .find the moment of inertia of the system about an axis along 1 side of the triangle?

Answer 1

Problem:

3 thin uniform rods each of length L & mass M are joined to form an equilateral triangle .find the moment of inertia of the system about an axis along 1 side of the triangle?

Solution:

According to the definition of momentum of inertia:

$$I = \sum_{i} m_{i} r_{i}^{2}$$

In this case:

$$\begin{array}{l} I = 2 * \int_{0}^{L \frac{\sqrt{3}}{2}} r^{2} \, dm = \int_{0}^{L \frac{\sqrt{3}}{2}} r^{2} \frac{M}{L} \, dl = \int_{0}^{L \frac{\sqrt{3}}{2}} r^{2} \frac{M}{L} \frac{dr}{\cos \frac{\pi}{6}} = \frac{2M}{\sqrt{3}L} \int_{0}^{L \frac{\sqrt{3}}{2}} r^{2} \, dr \\ = \frac{1}{2} M L^{2} \end{array}$$

Here integral gives the momentum of inertia of only one rod. Thus, it is multiplied by 2.

Answer: $I = \frac{1}{2} M L^{2}$.