Question

A 20.0 N force is applied to a cord wrapped around a pulley of mass 1.00 kg and radius

33.0 cm. The pulley is observed to accelerate uniformly from rest to reach an angular speed of 30.0/ radians/s in 3.00 s. If there is a frictional torque at the axel, of 1.10 mN, determine the moment of inertia of the pulley.

Answer 1

To find the moment of inertia of the pulley, use the equation: Torque = Moment of Inertia x Angular Acceleration. Substituting the given values into the equation and solving for Moment of Inertia, we find it to be 7.26 kg · m2.

Step-by-step explanation:

To determine the moment of inertia of the pulley, we need to use the equation:

Torque = Moment of Inertia x Angular Acceleration

The torque applied to the pulley is equal to the force applied (20.0 N) multiplied by the radius of the pulley (33.0 cm = 0.33 m). The torque due to friction is given as 1.10 mN (1.10 × 10-3 N·m). The total torque is the sum of the applied torque and the torque due to friction.

Substituting the values into the equation, we have:

Force x Radius + Torque due to friction = Moment of Inertia x Angular Acceleration

(20.0 N)(0.33 m) + (1.10 × 10-3 N·m) = Moment of Inertia × (30.0 rad/s - 0 rad/s) / 3.00 s

Solving for the Moment of Inertia, we find:

Moment of Inertia = (20.0 N · 0.33 m + 1.10 × 10-3 N·m) / ((30.0 rad/s - 0 rad/s) / 3.00 s)

Moment of Inertia = 7.26 kg · m2