show that the moment of inertia of a uniform rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass is 1/12(ML^2)
To derive the moment of inertia of a uniform rod about an axis perpendicular to the rod and passing through its center of mass, we can use the parallel axis theorem.
1. Begin by considering an infinitesimally small element of the rod, dx, located at a distance x from the center of mass. The mass of this element can be written as dm = (M/L)dx, where M represents the total mass of the rod.
2. The moment of inertia of this small element about the axis passing through the center of mass is given by dI = dm * (x^2).
3. Now, we need to integrate this expression over the entire length of the rod, from -L/2 to L/2, to obtain the total moment of inertia. Therefore, the moment of inertia can be written as:
I = ∫(-L/2 to L/2) dm * (x^2)
4. Substituting dm = (M/L)dx into the equation, we get:
I = ∫(-L/2 to L/2) (M/L)dx * (x^2)
5. Simplifying the expression and integrating, we have:
I = (M/L) * ∫(-L/2 to L/2) x^2 dx
6. Evaluating the integral, we get:
I = (M/L) * [(x^3)/3] | from -L/2 to L/2
At x = L/2: [(L/2)^3)/3]
At x = -L/2: [(-L/2)^3)/3]
I = (M/L) * [(L^3)/24 + (L^3)/24]
= (M/L) * (2L^3)/24
= (M/L) * (L^3)/12
= (M/12) * L^2
7. Since the total mass of the rod is equal to M and the length of the rod is equal to L, the moment of inertia can be expressed as:
I = (M/12) * L^2
= 1/12 * (M * L^2)
Therefore, we have shown that the moment of inertia of a uniform rod about an axis perpendicular to the rod and passing through its center of mass is equal to 1/12 * (M * L^2).