Two Disks Collide (Torque)

  • Thread starter Arman777
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  • #1
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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 differance 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
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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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.
 
  • #3
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You stopped looking after considering A. What happens to B ?
Ok for B I found ##1504.98J## so the differance 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 cant use symbols after this ?
 
  • #5
BvU
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Homework Helper
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I found ##W=\tau\delta\theta## I cant 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.
 
  • #6
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179
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 wouldnt do this probably...

Thanks
 

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