Question

Rotational kinetic energy of a system with 4 rods of length L making a square and the rotational axis is passing through the diagonal of the square formed. each rod of mass M

Answer 1

$\frac{1}{3} ML^2 \omega^2$

Answer 2

The system consists of four identical rods, each of mass M and length L, forming a square. The axis of rotation passes through a diagonal of the square. Let the square be ABCD, and let the axis of rotation be the diagonal AC.

We need to calculate the moment of inertia of the system about the axis AC. The total moment of inertia is the sum of the moments of inertia of the four rods: AB, BC, CD, and DA.

$I_{system} = I_{AB} + I_{BC} + I_{CD} + I_{DA}$

Consider a rod of mass M and length L. The moment of inertia about an axis passing through one end and making an angle $\theta$ with the rod is given by $I = \frac{1}{3} ML^2 \sin^2 \theta$.

Let's consider each rod:

1. Rod AB: The axis AC passes through end A of the rod AB. The angle between the rod AB and the diagonal AC is 45 degrees (since it's a square, the diagonal bisects the angle at the vertex).
   So, for rod AB, $\theta = 45^\circ$.
   $I_{AB} = \frac{1}{3} ML^2 \sin^2(45^\circ) = \frac{1}{3} ML^2 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{3} ML^2 \times \frac{1}{2} = \frac{1}{6} ML^2$.

2. Rod BC: The axis AC passes through end C of the rod BC. The angle between the rod BC and the diagonal AC is 45 degrees.
   So, for rod BC, $\theta = 45^\circ$.
   $I_{BC} = \frac{1}{3} ML^2 \sin^2(45^\circ) = \frac{1}{3} ML^2 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{3} ML^2 \times \frac{1}{2} = \frac{1}{6} ML^2$.

3. Rod CD: The axis AC passes through end C of the rod CD. The angle between the rod CD and the diagonal AC is 45 degrees.
   So, for rod CD, $\theta = 45^\circ$.
   $I_{CD} = \frac{1}{3} ML^2 \sin^2(45^\circ) = \frac{1}{3} ML^2 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{3} ML^2 \times \frac{1}{2} = \frac{1}{6} ML^2$.

4. Rod DA: The axis AC passes through end A of the rod DA. The angle between the rod DA and the diagonal AC is 45 degrees.
   So, for rod DA, $\theta = 45^\circ$.
   $I_{DA} = \frac{1}{3} ML^2 \sin^2(45^\circ) = \frac{1}{3} ML^2 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{3} ML^2 \times \frac{1}{2} = \frac{1}{6} ML^2$.

The total moment of inertia of the system about the diagonal AC is the sum of the moments of inertia of the four rods:

$I_{system} = I_{AB} + I_{BC} + I_{CD} + I_{DA} = \frac{1}{6} ML^2 + \frac{1}{6} ML^2 + \frac{1}{6} ML^2 + \frac{1}{6} ML^2 = 4 \times \frac{1}{6} ML^2 = \frac{2}{3} ML^2$.

The rotational kinetic energy of the system is given by $KE_{rot} = \frac{1}{2} I_{system} \omega^2$, where $\omega$ is the angular velocity of rotation.

$KE_{rot} = \frac{1}{2} \left(\frac{2}{3} ML^2\right) \omega^2 = \frac{1}{3} ML^2 \omega^2$.