A flywheel with a moment of inertia of 1.0 times 10 to the two kilograms per meter squared, that has an initial angular velocity of 52.36 radians per second, is brought to rest by frictional torque in a time of 120 seconds. What is the average frictional torque on the flywheel?
We were told that the moment of inertia of the flywheel is 1.7 times 10 to the two kilograms meter squared; we’ll call that capital 𝐼. The flywheel has an initial angular velocity of 52.36 radians per second; we’ll call that 𝜔 sub 𝑖. We’re told that the flywheel comes to rest in a time of 120 seconds; we’ll call that 𝑡. We want to know the average frictional torque that acts on the flywheel; we’ll call 𝜏 sub 𝑎𝑣𝑔. If we walked in an aerial view of the flywheel spinning, we’re told that at first it has an initial angular velocity that is positive. However, due to a frictional torque that acts on the flywheel to slow it down, over time the flywheel comes to a rest in a time of 120 seconds.
In this exercise, we’ll define clockwise motion as motion in the positive direction. To solve for the average torque exerted on the flywheel, let’s recall the relationship between torque and moment of inertia 𝐼. The torque exerted on a rotational object is equal to its moment of inertia multiplied by its angular acceleration 𝛼. So in our case, torque is equal to 𝐼, which we know in our given, times 𝛼, which we do not yet know. But we can recall the definition for angular acceleration 𝛼. 𝛼 is equal to the change in angular speed 𝜔 divided by the change in time, Δ𝑡. So we can replace 𝛼 in our equation with 𝜔 sub 𝑓 minus 𝜔 sub 𝑖 divided by 𝑡. 𝜔 sub 𝑓, the final angular speed of the flywheel, must be zero because we’re told it’s brought to rest. So the frictional torque on the flywheel, so the average frictional torque on the flywheel, 𝜏 sub 𝑎𝑣𝑔, equals the moment of inertia of the flywheel 𝐼 multiplied by negative 𝜔 sub 𝑖 divided by 𝑡.
When we plug in for 𝐼, 𝜔 sub 𝑖, and 𝑡, and calculated this value, we find that the average torque is negative 44 newton meters. This result is negative because it acts in the counterclockwise direction, and it has two significant figures because that’s the fewest number of significant figures of our given information.