A cylinder of mass M=58 kg, radius R=1.3 m and moment of inertia I=(1/2)MR2 is initially rotating around its central axis with an angular velocity ωi= 1.8 rad/s. A tangential force F of constant magnitude applied to the edge of the cylinder during 5.9 s brings it to rest. What is the magnitude of the force F?

To find the magnitude of the force F, we can use the concept of torque and the equation of rotational motion.

1. First, let's calculate the initial rotational kinetic energy (KEi) of the cylinder. The rotational kinetic energy is given by the formula KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.

KEi = (1/2)(1/2)MR^2 * ωi^2 = (1/4) * 58 kg * (1.3 m)^2 * (1.8 rad/s)^2

2. Next, let's find the change in kinetic energy (ΔKE) during the deceleration process. Since the cylinder comes to rest, the final rotational kinetic energy (KEf) will be zero. Therefore, ΔKE = KEf - KEi = -KEi.

3. The work done (W) by the applied tangential force F is equal to the change in kinetic energy. The work-energy theorem states that W = ΔKE.

W = ΔKE = -KEi

4. The work done by the force F is given by the formula W = Fd, where d is the displacement along which the force is applied. In this case, the displacement is given by the formula d = ωi * t, where t is the time taken to bring the cylinder to rest. Therefore, we can rewrite the equation as W = F * ωi * t.

F * ωi * t = -KEi

5. Finally, we can rearrange the equation to solve for the magnitude of the force F.

F = -KEi / (ωi * t)

Now, substitute the given values: M = 58 kg, R = 1.3 m, I = (1/2)MR^2, ωi = 1.8 rad/s, and t = 5.9 s into the equation to find the magnitude of the force F.