Question

A flywheel (I=50kg-m²) starting from rest acquires an angular velocity of 200.0 rad/s while subject to a constant torque from a motor for 5 s. What is the angular acceleration of the flywheel? What is the magnitude of the torque?

a) 40.0 rad/s², 1000 N·m
b) 50.0 rad/s², 800 N·m
c) 60.0 rad/s², 750 N·m
d) 70.0 rad/s², 700 N·m

Answer 1

The angular acceleration of the flywheel is 40.0 rad/s², calculated using α = ω / t. The torque, calculated using τ = I×α, is 2000 N·m. Note that this calculated torque does not match any of the multiple-choice options given.

Step-by-step explanation:

To find the angular acceleration (α) of the flywheel, we can use the formula α = ω / t, where ω is the final angular velocity and t is the time taken to reach this velocity.

Since the final angular velocity (ω) is 200.0 rad/s and the time (t) is 5 s, the angular acceleration is 200.0 rad/s ÷ 5 s = 40.0 rad/s².

Next, to calculate the magnitude of the torque (τ), we use the formula τ = I×α, where I is the moment of inertia of the flywheel.

Given I = 50 kg-m² and α = 40.0 rad/s², the torque is 50 kg-m² × 40.0 rad/s² = 2000 N·m.

Therefore, the correct answer is (a) 40.0 rad/s² for the angular acceleration and 2000 N·m for the torque, which is not listed among the options provided. It is possible there was a typo in the options given for torque, as none of the options correspond with the calculation based on the given data.