A 25.0 kg propeller may be considered to be a 4.5 m long, thin rod, spun about its center. Assuming the propeller starts from rest, What angular velocity will it have if a torque of 1100 N·m is applied for 1.2 s?
A) 31 rad/s
B) 26 rad/s
C) 7.8 rad/s
D) 2.6 rad/s

Question

Quizplus · Accepted Answer

The moment of inertia (I) for a rod spun about its center is (1/12)ML^2. For the given propeller, I = (1/12)(25 kg)(4.5 m)^2 = 42.1875 kg·m^2. Using the formula τ = Iα, where τ is torque and α is angular acceleration, we find α = τ/I = 1100 N·m / 42.1875 kg·m^2 = 26.07 rad/s^2. The angular velocity (ω) is given by ω = αt = 26.07 rad/s^2 * 1.2 s = 31.28 rad/s, which rounds to 31 rad/s.

A 250 Kg Propeller May Be Considered to Be a 4