Question:

An object consists of two uniform thin rods, each of mass M and length L, connected together at their ends to form a symmetric 'X' shape (as shown in the figure). The object is rotated about an axis perpendicular to the plane of the X and passing through the intersection of the rods.

Calculate the moment of inertia of this object about the axis of rotation.

(A) ML²/2 (B) ML²/4 (C) 3ML²/2 (D) 7ML²/4 (E) 2ML²

Solution:

To calculate the moment of inertia of the object, we need to consider the contribution from each rod separately and then sum them up.

Let's consider the moment of inertia for one of the rods first. We can use the formula for the moment of inertia of a thin rod rotated about its center: I = (1/12)ML². However, since we are rotating the rod about an axis perpendicular to its plane and passing through one of its ends, we need to use the parallel-axis theorem to shift the moment of inertia to this new axis.

The parallel-axis theorem states that I = I_cm + Md², where I_cm is the moment of inertia about the center of mass of the object, M is the mass of the object, and d is the distance between the axis of rotation and the center of mass of the object.

For a rod of length L, the distance between the axis of rotation and the center of mass is L/2. Thus, the moment of inertia for one rod about the axis of rotation is:

I₁ = (1/12)ML² + M(L/2)² = (1/12)ML² + M(L²/4) = (1/12 + 1/4) ML² = (1/3)ML².

Since there are two identical rods in the object, the moment of inertia for both rods will be 2(I₁) = 2((1/3)ML²) = (2/3)ML².

Finally, we can sum up the individual contributions to find the total moment of inertia of the object:

I_total = (2/3)ML².

Therefore, the correct answer is (C) 3ML²/2.