Torque and rotational dynamics

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inner08
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Homework Statement


A wheel is turning initially at 1200 turns/min and stops in 4 min because of friction. If we add a supplementary torque of 300 N*m, the wheel stops in 1 min.

a) What is the moment of inertia of the wheel?
b) What is the torque of friction?


Homework Equations


Torque = alpha * I
time = w/alpha
f = ucF
Torque = -fR

The Attempt at a Solution



I figured i'd start by converting the 1200 turns/min into rad/sec

(1200 *2pi)/60 = 126 rad/sec (this gives me angular speed - w)

Then I figured i'd find the angular acceleration using the angular speed that I just found with the time it takes to stop (in this case its 4min or 240seconds).

alpha = 126/240
= 0.525

Now at first I thought I could just substitute these in the equation Torque = alpha * I but it doesn't work out.
I then realized that they are saying that there is a SUPPLEMENTARY torque and it then stops in 1min. I thought it would then be something like "initial torque + 300 = something...". Does that make any sense?

Any help to clear this up would be appreciated!
 
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inner08 said:
Now at first I thought I could just substitute these in the equation Torque = alpha * I but it doesn't work out.
I then realized that they are saying that there is a SUPPLEMENTARY torque and it then stops in 1min. I thought it would then be something like "initial torque + 300 = something...". Does that make any sense?

Any help to clear this up would be appreciated!

You could think of it this way: [tex]M=\alpha_{1} I[/tex], and [tex]M+300=\alpha_{2} I[/tex]. The additional torque creates a new net torque, which creates a new angular deceleration. Try to work something out from here on. (P.S. Note that the moment of inertia is, of course, the same.)
 
You are given two (related) situations with two different accelerations. Find both accelerations, then write two "Torque = alpha * I" equations. Solve both equations together and you'll get your answers.

(Oops... radou beat me to it!)​
 
You are nearly there. Yes, Torque = alpha * I is a good equation to use.

You have got two cases. In the first one you have the unknown friction torque (call it T) and the wheel stops in 4 min. You worked out the acceleration alpha correctly.

In the second case you have an additional torque so (as you said) the torque is "initial torque + 300 = something" ... well, the "something" is just "T+300". This time the wheel stops in 1 min, so the value of alpha is different.

Using Torque = alpha * I for the two cases will give you 2 equations in the 2 unknowns (T and I) which you can solve.
 
Thanks a bunch! I got it :)