A 2.1-kg 12-cm-radius cylinder, initially at rest, is free to rotate about the axis of the cylinder. A rope of negligible mass is wrapped around it and pulled with a force of 18 N.
(a) Find the magnitude of the torque exerted by the rope.
N · m
(b) Find the angular acceleration of the cylinder.
rad/s2
(c) Find the angular velocity of the cylinder at t = 0.30 s.
rad/s
To find the answers to these questions, we need to use some principles of rotational motion. The torque exerted by a force on a rotating object can be found using the formula:
Torque (τ) = Force (F) x Lever Arm (r)
(a) Let's start by finding the magnitude of the torque exerted by the rope. In this case, the force applied by the rope is 18 N and the lever arm is equal to the radius of the cylinder, which is 12 cm (or 0.12 m).
τ = F x r
τ = 18 N x 0.12 m
τ = 2.16 N · m
So, the magnitude of the torque exerted by the rope is 2.16 N · m.
(b) Next, let's find the angular acceleration of the cylinder. The torque can also be related to the moment of inertia (I) and the angular acceleration (α) using the formula:
τ = I x α
For a cylindrical object, the moment of inertia can be calculated as:
I = (1/2) x m x r^2
Where m is the mass of the cylinder and r is the radius of the cylinder.
I = (1/2) x 2.1 kg x (0.12 m)^2
I = 0.01512 kg · m^2
Now, we can rearrange the torque equation to solve for α:
α = τ / I
α = 2.16 N · m / 0.01512 kg · m^2
α ≈ 143.04 rad/s^2
So, the angular acceleration of the cylinder is approximately 143.04 rad/s^2.
(c) Finally, let's find the angular velocity of the cylinder at t = 0.30 s. The relationship between angular velocity (ω), initial angular velocity (ω0), angular acceleration (α), and time (t) is given by the equation:
ω = ω0 + αt
Since the cylinder is initially at rest, ω0 = 0. Plugging in the values, we get:
ω = 0 + (143.04 rad/s^2) x (0.30 s)
ω ≈ 42.91 rad/s
So, the angular velocity of the cylinder at t = 0.30 s is approximately 42.91 rad/s.