Friction Torque and Lost Energy in Colliding Disks: Where Did it Go?

[Arman777]

Homework Statement

Disks A and B each have a rotational inerita $0.300 kg.m^2$ about the central axis and a radius of of $20.0cm$and are free to rotate on a central rod through both of them.To set them spinning around the rod in the same direction,each wrapped with a string that is pulled for $10.0s$ (the string detaches at the end).The magnitudes of the forces pulling the strings are $30.0N$ for a disk A and $20.0N$ for a disk B.After the strings detach,the disks happen to collide and frictional force between them to the same final angular speed in $6.00s$
What are (a) magnitude of the average frictional torque that brings them to the final angular speed and (b) the loss in kinetic energy as the torque acts on them? (c) Where did the "lost energy" go ?

Homework Equations

$τ=rFsinθ$
$τ=I∝$

The Attempt at a Solution

First I found the angular accelerations for both objects
For A-
$τ=rF=I∝$

$∝=\frac {rF} {I}$

$∝=\frac {0.2m.30N} {0.3 kgm^2}=20\frac {rad} {s^2}$
For B-
$∝=\frac {0.2m.20N} {0.3 kgm^2}=13,3\frac {rad} {s^2}$

For angular speed (w) which they pulled $10 sec$, so $w-w_0=∝t$
$w_A=20.10=200\frac {rad} {s}$
$w_B=13,3.10=133\frac {rad} {s}$

They are coming equal in 6 sec so the change in $w_A$ is $200-(\frac{200+133} {2})=33.5\frac {rad} {s}$

so angular acceleration is $∝=\frac {33.5} {6}=5.58\frac {rad} {s^2}$
so
$τ=I∝=0.3.5.58=1.67N.m$

For (b)

Initial rotational kinetic energy for A is from $\frac 1 2Iw^2=(0.5).(0.3) (200^2)$
Last kinetic energy for A is $\frac 1 2Iw^2=(0.5).(0.3) (166.5^2)$
The difference is $1841,66J$ which its not the correct answer..

Or I tried $Δ(E_r)=W=τΔθ=1.67Nm.33.5\frac {rad} {s}.6s=335.67J$
but answer says its $333J$
Where I did wrong ?

And for answer c is of course heat

Thanks

 

[BvU]

Where I did wrong

You stopped looking after considering A. What happens to B ?
The 333 or 336 comes about because you calculate with numbers instead of symbols. The advice is to work with symbols as long as feasible; then -- in the final expression -- a lot of things cancel and you get a more accurate answer.
Additional benefit: less typing on the calculator, less chance for typing errors and easier approximate checks on the numerical outcome.

 

[Arman777]

You stopped looking after considering A. What happens to B ?

Ok for B I found $1504.98J$ so the difference between them should give us the result whch its approximatly $336J$

The 333 or 336 comes about because you calculate with numbers instead of symbols

I found $W=τΔθ$ I can't use symbols after this ?

 

[Arman777]

If logic is correct then I don't think 333J or 336J matter so much..?

 

[BvU]

I found $W=\tau\delta\theta$ I can't use symbols after this

You can express both factors in terms of given variables. If you do that, the correct answer is exactly 1000/3 J. You should do that exercise.

In such a calculation in an exercise, a 1% deviation is considered an error. Loss of score for no purpose.

 

[Arman777]

Ok I found by $W=I(w^2)$

$0.3.({200-(\frac {1000} {6}))^2}=W$

from there I found,but in exam or somewhere else I wouldn't do this probably...

Thanks