A thin circular disk density of 30 lbftm and thickness of 02

A thin circular disk (density of 30 lb/ft^m) and thickness of 0.2 feet, has a weight attached to it as shown below. If the cord is wrapped around the disk and starting from rest, the block is released, what is the angular velocity of the disk after 4 seconds?

Solution

We have a weight of 75 lb attached by a string over a pulley. Here, the weight attached will impart some torque on the pulley.

The pulley has a density of 30 lb/ft³, the volume being R³t

Hence, the weight of the pulley = 30 x 3.14 x 1 x 0.2 = 18.84 lb

That is the inertia of the pulley = MR²/2 = 18.84 / 2 = 9.42 lb-ft²

Now, we know that Torque = Inertia x angular acceleration

or, angular acceleration = Torque / Inertia = 75 x 32 / 9.42 = 254.777 rad/s²

Therefore the angular velocity of the disk in 4 seconds would be acceleration x time = 1019.108 rad/s