AP Physics 1 Exam Question:

A uniform rod of length L and mass M is suspended horizontally from one end. The rod is displaced by an angle θ and then released. Determine the moment of inertia of the rod about its pivot point.

(a) Derive an expression for the moment of inertia of the rod about its pivot point, assuming the rod is a uniform rectangular shape.

(b) Calculate the moment of inertia of the rod about its pivot point, given that the length L = 2 m and the mass M = 3 kg.

Answer:

(a) To derive the expression for the moment of inertia of the rod about its pivot point, we can consider the rod as a collection of infinitesimally thin rectangular elements. Let's assume the thickness of each rectangular element is Δx.

The moment of inertia of each rectangular element with respect to the pivot point is given by:

dI = dm * x^2

Where dI is the moment of inertia of the infinitesimally small element, dm is the mass of the element, and x is the perpendicular distance of the element from the pivot point.

Since the rod is uniform, the mass distribution is linear. Therefore, the mass of each element can be expressed as:

dm = (M/L) * Δx

Substituting this into the equation for dI, we have:

dI = ((M/L) * Δx) * x^2

To find the total moment of inertia (I) of the rod, we need to integrate this expression over the entire length L of the rod:

I = ∫ dI

I = ∫ ((M/L) * Δx) * x^2 from x = 0 to x = L

I = (M/L) ∫ x^2 d(x) from x = 0 to x = L

I = (M/L) * (x^3/3) from x = 0 to x = L

I = (M/L) * [ (L^3/3) - (0^3/3) ]

I = (M/L) * (L^3/3)

I = M * L^2 / 3

Therefore, the expression for the moment of inertia (I) of the rod about its pivot point is given by I = M * L^2 / 3.

(b) Given that L = 2 m and M = 3 kg:

I = (3 kg) * (2 m)^2 / 3

I = 12 kg·m^2 / 3

I = 4 kg·m^2

The moment of inertia of the rod about its pivot point is 4 kg·m^2.

Note: Make sure to check the latest formula sheet and equations provided by the College board for the AP Physics 1 Exam, as they can change over time.