# Torque and rotational dynamics

1. Nov 29, 2006

### inner08

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Torque = alpha * I
time = w/alpha
f = ucF
Torque = -fR

3. 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 realised 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!

2. Nov 29, 2006

You could think of it this way: $$M=\alpha_{1} I$$, and $$M+300=\alpha_{2} I$$. 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.)

3. Nov 29, 2006

### Staff: Mentor

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!)​

4. Nov 29, 2006

### AlephZero

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.

5. Nov 29, 2006

### inner08

Thanks a bunch! I got it :)