A pulley with a rotational interia of 4.4*10^-3 kg m^2 about its axle and a radius of 6 cm is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F=0.32t + 0.27t^2 where F is in newtons if t is given in seconds. The pulley is initially at rest. At t=3.9s what is the angular velocity?

To find the angular velocity of the pulley at t = 3.9s, we need to follow these steps:

Step 1: Find the torque acting on the pulley.
The torque (τ) is given by the formula τ = Iα, where I is the moment of inertia and α is the angular acceleration.
In this case, the moment of inertia (I) is given as 4.4 × 10^-3 kg m^2.

Step 2: Find the angular acceleration (α).
The angular acceleration can be calculated using the torque and moment of inertia: α = τ / I.

Step 3: Find the net force at t = 3.9s.
The net force (F_net) acting on the pulley can be obtained by substituting t = 3.9s into the force equation: F_net = 0.32t + 0.27t^2.

Step 4: Find the torque at t = 3.9s.
The torque acting on the pulley can be calculated using the equation τ = r × F_net, where r is the radius of the pulley. In this case, the radius (r) is given as 6 cm, which is equivalent to 0.06 m.

Step 5: Find the angular acceleration at t = 3.9s.
Using the torque and moment of inertia from steps 1 and 4, we can substitute them into the equation α = τ / I.

Step 6: Find the angular velocity at t = 3.9s.
The angular velocity (ω) can be obtained by integrating the angular acceleration with respect to time:
ω = ω_initial + ∫α dt, where ω_initial is the initial angular velocity (which is zero since the pulley is initially at rest).

By following these steps, you should be able to find the angular velocity of the pulley at t = 3.9s.