A 13.6-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is 0.40 kg · m2, and its radius is 0.151 m. When the reel is turning, friction at the axle exerts a torque of magnitude 3.42 N · m on the reel. If the hose is pulled so that the tension in it remains a constant 23.1 N, how long does it take to completely unwind the hose from the reel? Neglect the mass of the hose, and assume that the hose unwinds without slipping.
Answer will be in seconds
To find the time it takes to completely unwind the hose from the reel, we need to calculate the angular acceleration of the reel using the given torque and moment of inertia. Then, we can use the equation of motion for rotational motion to find the time.
First, let's calculate the angular acceleration. The torque exerted by friction at the axle is equal to the moment of inertia multiplied by the angular acceleration:
τ = I * α
Where τ is the torque (3.42 N · m), I is the moment of inertia (0.40 kg · m^2), and α is the angular acceleration (unknown).
Rearranging the equation, we can solve for α:
α = τ / I = 3.42 N · m / 0.40 kg · m^2 = 8.55 rad/s^2
Now, let's find the angular velocity (ω) of the reel using the equation:
τ = I * α = I * (dω / dt)
Where dω / dt is the time derivative (rate of change) of angular velocity.
τ = I * (dω / dt)
Rearranging the equation, we get:
dω = τ / I * dt
Integrating both sides of the equation from 0 to ω (angular velocity) and from 0 to t (time), we have:
∫[0 to ω] dω = ∫[0 to t] τ / I * dt
ω - 0 = (τ / I) * (t - 0)
Simplifying the equation:
ω = (τ / I) * t
Now, let's find the angular displacement (∆θ) of the reel. The angular displacement can be obtained using the equation:
∆θ = ω * t
Substituting the value of ω from the previous equation, we have:
∆θ = (τ / I) * t * t
We know that the length of the hose is equal to the circumference of the reel, which is given by:
C = 2πr
Where C is the circumference, and r is the radius of the reel.
Given the radius (0.151 m), the length of the hose is:
L = 2πr = 2 * 3.14159 * 0.151 m = 0.953 m
As the hose unwinds without slipping, the angular displacement ∆θ is equal to the length L of the hose. Therefore:
∆θ = L = (τ / I) * t * t
Substituting the known values:
0.953 m = (3.42 N · m / 0.40 kg · m^2) * t * t
Simplifying the equation and solving for t:
t = sqrt((0.953 m * 0.40 kg · m^2) / (3.42 N · m))
t = sqrt(0.3842 s^2)
t ≈ 0.62 s
Therefore, it takes approximately 0.62 seconds to completely unwind the hose from the reel.