A 15.5-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is 0.42 kg · m2, and its radius is 0.170 m. When the reel is turning, friction at the axle exerts a torque of magnitude 3.63 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.

Question

Amanda · Accepted Answer

Hi, for this one you will be doing
Applied torque minus frictional torque
(23.1)(.17)-3.63=.28 N-m
(summarizing where I got the numbers above from the problem;)
(remains constant at)(radius)-(torque w/ magnitude)

then you want to find the angular acceleration. Ac=T/ I (Torque over moment of intertia)
We know the torque is now .28 N-m and the moment of inertia from the problem is .42 so
(.28)/(.42)=.66 or .67 radians/s ^2 rounded for the angular acceleration.

Next we will use the hose length/ radius so 15.5/.17=91.1 radians

Finally to find the answer to our problem we will do;

91.1 radians=(1/2) (.67)t^2

We are solving for t. so
91.1=.355t^2

divide by .355 on each side
256.62=t^2
then square root to isolate t to get
16.02 s

I hope you can apply this strategy to the numbers on your problem that you get . Have fun with Physics :)

drwls · Answer

The applied torque minus the frictional torque is (23.1)(0.170)-3.63 = 0.28 N-m

Torque divided by moment of inertia is the angular acceleration, alpha. Its units are radians/s^2. Calculate it

To unreel all of the hose, you need to turn the reel through 15.5/0.17 = 91.1 radians.

Solve this equation for the required time, t:

91.1 radians = (1/2)*(alpha)*t^2