Final answer:
The torque generated is 48.0 Nm, the linear velocity at the rim after 4 seconds is 54.56 m/s, and the cylinder makes about 17.3 revolutions during this period.
Step-by-step explanation:
To solve the problem described, we need to apply concepts from rotational motion. (a) The torque (τ) generated by a force applied tangentially to the rim of a cylinder is given by τ = r * F, where r is the radius and F is the force. The radius is half the diameter, so we have r = 80.0 cm / 2 = 40.0 cm = 0.40 m. Plugging in the values, τ = 0.40 m * 120.0 N = 48.0 Nm.
(b) To find the linear velocity (v) after 4.00 s, we first need the angular acceleration (α) which is α = τ / I, where I is the moment of inertia of the cylinder. For a solid cylinder, I = 1/2 * m * r^2. Substituting the given values, I = 1/2 * 44.0 kg * (0.40 m)^2 = 3.52 kgm². Thus, α = 48.0 Nm / 3.52 kgm² = 13.64 rad/s². Using the kinematic equation v = α * t, we get v = 13.64 rad/s² * 4.00 s = 54.56 m/s at the rim of the cylinder.
(c) The number of revolutions is found by using the relationship between angular distance (θ in radians), angular velocity (ω), time (t), and angular acceleration (α): θ = 1/2 * α * t^2. One revolution is 2π radians, so the number of revolutions n is θ / 2π. Substituting the values in, we get θ = 1/2 * 13.64 rad/s² * (4.00 s)^2 = 108.5 rad. Therefore, n = 108.5 rad / 2π rad/rev ≈ 17.3 revolutions.