AP Physics 1 Exam Question

A small object of mass 0.2 kg is attached to a string and moves in a horizontal circle with a radius of 0.5 meters. The object completes one full revolution in 2 seconds. Assuming no external torque acts on the system, calculate the angular momentum of the object.

Solution:

The angular momentum of an object is given by the formula:

L = Iω,

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Since the object is moving in a circular path, we can calculate the moment of inertia using the formula:

I = mr²,

where m is the mass of the object, and r is the radius of the circular path.

Given that the mass of the object is 0.2 kg and the radius of the circular path is 0.5 meters, we can substitute these values into the equation:

I = (0.2 kg)(0.5 m)² = 0.05 kg·m².

We can calculate the angular velocity using the formula:

ω = 2π/T,

where T is the period of revolution.

Given that the object completes one full revolution in 2 seconds, we can substitute these values into the equation:

ω = 2π/2 = π rad/s.

Now, we can calculate the angular momentum using the formula:

L = Iω = (0.05 kg·m²)(π rad/s) ≈ 0.157 kg·m²/s.

Therefore, the angular momentum of the object is approximately 0.157 kg·m²/s.