A 15 M LENGTH OF HOSE IS WOUND AROUND A REEL WHICH IS INITIALLY AT REST. THE MOMENT OF INERTIA OF THE REEL OF 0.44KGM^2 AND ITS RADIUS IS 0.160M. WHEN THE REEL IS TURNING FRICTION AT THE AXLE EXERTS A TORQUE OF MAGNITUDE 3.40NM ON THE REEL. IF THE HOSE IS PULLED SO THAT THE TENSION IN IT REMAINS A CONSTANT 25N, HOW LONG DOES IT TAKE TO COMPLETELY UNWIND THE HOSE FROM THE REEL? NEGLECT THE MASS AND THICKNESS OF THE HOSE ON THE REEL AND ASSUME THAT THE HOSE UNWINDS WITHOUT SLIPPING.

To find out how long it takes to completely unwind the hose from the reel, we need to consider the torque exerted by the friction at the axle, as well as the tension in the hose.

The torque exerted by the friction at the axle is given as 3.40 Nm. Since torque is the product of force and distance from the axis of rotation, we can calculate the lever arm distance (r_friction) between the axle and the point where the friction force acts.

Using the formula torque = force * r_friction, we can rearrange the equation to solve for r_friction:

r_friction = torque / force
r_friction = 3.40 Nm / 25 N
r_friction = 0.136 m

Now, we need to calculate the torque exerted by the tension in the hose. The tension in the hose is given as 25 N. The torque exerted by the tension force can be calculated using the formula torque = force * distance from the axis of rotation.

In this case, the distance from the axis of rotation is the radius of the reel, given as 0.160 m.

torque_tension = force * radius
torque_tension = 25 N * 0.160 m
torque_tension = 4 Nm

Since the reel is initially at rest, the net torque acting on the reel is the difference between the torque exerted by the tension and the torque exerted by the friction:

net torque = torque_tension - torque_friction
net torque = 4 Nm - 3.40 Nm
net torque = 0.60 Nm

To find the angular acceleration of the reel, we can use the formula torque = moment of inertia * angular acceleration:

angular acceleration = net torque / moment of inertia
angular acceleration = 0.60 Nm / 0.44 kgm^2
angular acceleration = 1.3636 rad/s^2

Now, we can use the equation of motion for rotational motion:

θ = ω_i * t + (1/2) * alpha * t^2

where:
θ = angle of rotation (360 degrees or 2π radians, since the hose needs to completely unwind)
ω_i = initial angular velocity (0 radians/s, since the reel is initially at rest)
alpha = angular acceleration (1.3636 rad/s^2, calculated above)
t = time it takes to unwind the hose (what we want to find)

Substituting the values into the equation:

2π = 0 * t + (1/2) * 1.3636 * t^2

Simplifying the equation by canceling out the zero term:

2π = (1/2) * 1.3636 * t^2

Rearranging the equation to solve for t:

t^2 = (2π) / (0.5 * 1.3636)
t^2 = 9.1764
t = sqrt(9.1764)
t ≈ 3.03 seconds

Therefore, it takes approximately 3.03 seconds to completely unwind the hose from the reel.