An object consists of 5 masses, each of mass m, connected by massless rods. The two short rods have length L and the two long rods have length 2L. What is the moment of inertia about an axis perpendicular to the diagram (out of the page) and through the mass labeled x (the middle one)?
To find the moment of inertia of the object about an axis perpendicular to the diagram and through the mass labeled x, we need to consider the contributions of each mass individually.
The moment of inertia of an individual point mass rotating about an axis is given by the formula: I = mr², where I is the moment of inertia, m is the mass of the point, and r is the distance of the point from the axis of rotation.
Let's consider the different masses in the object:
1. Masses connected by the short rods:
- The moment of inertia of the two masses closest to the mass labeled x can be calculated using the formula I = m₁r₁² + m₂r₂², where m₁ and m₂ are the masses and r₁ and r₂ are the distances of each mass from the axis of rotation.
- Since the two short rods have length L, we can see that the distances of the masses closest to x from the axis are L and L, respectively.
- Therefore, the moment of inertia contributed by the two masses connected by the short rods is 2mL².
2. Masses connected by the long rods:
- The moment of inertia of the two masses on the outer ends connected by the long rods can also be calculated using the same formula.
- In this case, since the long rods have length 2L, the distances of the outer masses from the axis of rotation are 3L and 3L.
- Therefore, the moment of inertia contributed by the two masses connected by the long rods is 2m(3L)².
3. The middle mass labeled x:
- Finally, the moment of inertia of the mass labeled x itself can be calculated using the formula I = mx², where x is the distance of the mass from the axis of rotation.
- As the axis of rotation passes through this mass, x = 0, resulting in a moment of inertia of 0.
To find the total moment of inertia about the given axis, we sum up the contributions from each mass:
Total moment of inertia = 2mL² + 2m(3L)² + 0
= 2mL² + 18mL²
= 20mL²
Therefore, the moment of inertia about an axis perpendicular to the diagram and through the mass labeled x is 20mL².