A pulley, with a rotational inertia of `1.0xx10^(-3)kg*m^(2)` about its axle and a radius of 10 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as `F=0.50t+0.30t^(2)`, with F in newtons and t in seconds. The pulley is initially at rest. At t = 3.0 s what are its (a) angular acceleration and (b) angular speed?

Explanation

To solve the problem step by step, we will find the angular acceleration and angular speed of the pulley at t = 3.0 seconds.

### Given Data:
- Moment of inertia of the pulley, \( I = 1.0 \times 10^{-3} \, \text{kg m}^2 \)
- Radius of the pulley, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Force acting on the pulley, \( F(t) = 0.50t + 0.30t^2 \)

### Step 1: Calculate the Torque
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