A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 2.86 m/s2.
(a) How much work has been done on the spool when it reaches an angular speed of 7.10 rad/s?
(b) How long does it take the spool to reach this angular speed?
(c) How much cord is left on the spool when it reaches this angular speed?
To solve this problem, we can use the principles of rotational motion and work-energy theorem. Let's break it down step by step:
(a) To find the work done on the spool, we need to calculate the change in its rotational kinetic energy. The formula for rotational kinetic energy is:
K = (1/2) * I * ω^2
where K is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid cylinder is given by:
I = (1/2) * m * r^2
where m is the mass of the spool and r is the radius.
Plugging in the values, we have:
I = (1/2) * (1.00 kg) * (0.500 m)^2
I = 0.125 kg⋅m^2
Now, we need to find the initial and final angular velocities (ω) of the spool. The initial angular velocity is 0, and the final angular velocity is given as 7.10 rad/s.
Using the work-energy theorem, we have:
Work = ΔK = K_final - K_initial
Work = (1/2) * I * ω_final^2 - (1/2) * I * ω_initial^2
Work = (1/2) * I * (ω_final^2 - ω_initial^2)
Plugging in the values, we get:
Work = (1/2) * (0.125 kg⋅m^2) * [(7.10 rad/s)^2 - (0 rad/s)^2]
Work = (1/2) * (0.125 kg⋅m^2) * (7.10^2 rad^2/s^2)
Work ≈ 0.313 J
Therefore, the work done on the spool when it reaches an angular speed of 7.10 rad/s is approximately 0.313 Joules.
(b) To find the time it takes for the spool to reach this angular speed, we can use the formula for angular acceleration:
α = (ω_final - ω_initial) / t
Rearranging the formula, we have:
t = (ω_final - ω_initial) / α
Plugging in the values, we get:
t = (7.10 rad/s - 0 rad/s) / 2.86 m/s^2
t ≈ 2.482 s
Therefore, it takes approximately 2.482 seconds for the spool to reach an angular speed of 7.10 rad/s.
(c) To find out how much cord is left on the spool when it reaches this angular speed, we need to calculate the distance traveled by the cord. Since the cord is being pulled with a constant acceleration, we can use the kinematic equation:
Δx = v_initial * t + (1/2) * a * t^2
In this case, the initial velocity (v_initial) is 0, the acceleration (a) is 2.86 m/s^2, and the time (t) is 2.482 s (as calculated in part b).
Plugging in the values, we have:
Δx = 0 * 2.482 s + (1/2) * (2.86 m/s^2) * (2.482 s)^2
Δx ≈ 8.82 m
Therefore, when the spool reaches an angular speed of 7.10 rad/s, approximately 8.82 meters of cord is left on the spool.