Question

A propeller consists of two blades, each 3.0 m in length and mass 120 kg each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30 seconds at a constant rate. What is the angular momentum of the propeller at t=10s?

a. 3600 kg m/s
b. 1800 kg m/s
c. 1200 kg m/s
d. 600 kg m/s

Answer 1

The angular momentum of the propeller at t=10s is 90595.52 kg m^2/s.

Step-by-step explanation:

The angular momentum of the propeller at t=10s can be calculated by considering the conservation of angular momentum. Initially, the propeller is at rest, so its initial angular momentum is zero. Over the course of 30 seconds, it rotates up to 1200 rpm, which can be converted to 125.66 rad/s. Using the equation for angular momentum L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity, we can calculate the moment of inertia of the propeller as follows:

I = 2 * (1/3) * m * r^2, where m is the mass of each blade and r is the length of each blade. Plugging in the given values, we get: I = 2 * (1/3) * 120 kg * (3.0 m)^2 = 720 kg m^2.

Since the angular momentum is conserved, we can use the formula L = Iω to find the angular momentum at t=10s:

L = I * ω = 720 kg m^2 * (125.66 rad/s) = 90595.52 kg m^2/s.