AP Physics 1 Exam Question - Moment of Inertia

A thin rod of mass m and length L is pivoted at one end and is free to rotate about it. The rod is released from rest in a vertical position and allowed to rotate.

1. Derive an expression for the moment of inertia of the thin rod about the pivot point.
2. Calculate the moment of inertia for a thin rod with mass m = 2 kg and length L = 1.5 m.

Answer:

1. Deriving the expression for the moment of inertia:

We can consider the thin rod to be made up of small mass elements along its length. The moment of inertia of each small mass element dm about the pivot point is given by dI = dm * r^2, where r is the perpendicular distance of the mass element from the pivot point.

To find the moment of inertia of the entire rod, we need to integrate this expression over the length of the rod:

I = ∫dI = ∫(dm * r^2)

For a thin rod, the mass element dm at a distance x from the pivot is dm = (m/L) * dx, where dx is an infinitesimally small length element.

Substituting this into the integral, we have:

I = ∫((m/L) * dx * r^2)

To evaluate this integral, we need to find an expression for r. For a thin rod pivoted at one end, the distance r is equal to x.

I = ∫((m/L) * x^2 * dx)

Evaluating the integral from 0 to L, we get:

I = (m/L) * ∫(x^2) dx  [Integrating from 0 to L]
  = (m/L) * [(1/3)x^3] [Applying the power rule]
  = (m/L) * (1/3)(L^3)   [Substituting limits]
  = (1/3)mL^2 

Therefore, the moment of inertia of a thin rod pivoted at one end is given by I = (1/3)mL^2.

2. Calculating the moment of inertia:

Given:

• m = 2 kg
• L = 1.5 m

Using the derived expression for the moment of inertia I = (1/3)mL^2, we can substitute the given values to calculate the moment of inertia:

I = (1/3)(2 kg)(1.5 m)^2
  = (1/3)(2 kg)(2.25 m^2)
  = 1.5 kg·m^2

Therefore, the moment of inertia for a thin rod with m = 2 kg and L = 1.5 m is 1.5 kg·m^2.