Calculate Mass & Energy Lost of Flywheel B for Homework

[DR1]

Homework Statement

a flywheel (A) with mass 100kg and radius of gyration 1200mm rotates at 150 rev min-1(clockwise). The kenetic energy of this flywheel it to be reduced by 20% by impacting with second flywheel (B) rotating at 80 revs min-1 (anti-clockwise), such that they have the same (clockwise) angular velocity after impact.

Q1A) calculate the required mass of flywheel (B) if its radius of gyration is 800mm

B) calculate the energy lost to the surroundings.

C) Is the impact elastic? give reasons

I have no idea were to start with this i found the moment of inertia for flywheel A but don't know if that will help 169kg m and thus found the kinetic energy of 20849.53 J and angular speed of 5pie rad s-1 I am also sure the angular speed of flywheel B is 2.7pie rad s-1

can anyone point me in the right direction please

 

[DR1]

does anybody have any ideas

 

[DR1]

Im guessing from the lack of response that this is not just me being stupid

does anyone have any idea I am sure someone is clever enough to help

 

[jrlaguna]

My problem is that I can't see how the collision is going to take place... CaN you be more specific?

I guess your teacher's idea is to use conservation laws, for example, conservation of angular momentum.

 

[DR1]

im sorry i can not be any more specific all the info i have is in the question

I have been studying a book as part of this assesment which covers conservation of energy so I am sure that is the correct route to take.

 

[jrlaguna]

If energy conservation was all the story, they'd not be asking you whether the collision is elastic. In general terms, momentum and angular momentum conservation are far more important than energy conservation in applications. Even in non-elastic collisions they're conserved, while energy is not. Try the conservation of angular momentum, you will get an answer.

My problem still is... how is the impact taking place? I mean, I don't know which geometry to imagine.

 

[DR1]

are you therefore implying that if i somehow use the formule

Ia(W1)a + Ib (W1)b=Ia(W2)a + Ib(W2)b

i will be able to start working this out my issue now is how do i use this effectively

as far as impact taking place would it help to imagine it as a pair of clutch plates joining
unfortunately the tutor is away at the moment hence the reason i am turning to you all for help

 

[jrlaguna]

Yes, this "clutch-type" collision sounds the most likely... And your formula sounds also nice, so why don't you just apply it? You know the two final angular speeds are equal...

 

[DR1]

Yes, this "clutch-type" collision sounds the most likely... And your formula sounds also nice, so why don't you just apply it? You know the two final angular speeds are equal...

the reason i am having an issue useing the formule is because I am am not sure if the question implies the final angular speeds are equal at 150 rev min-1 or have changed due to the impact

or am i over analyzing the question

 

[jrlaguna]

They want to reduce its kinetic energy, right? So you can find out the final angular velocity that we require.

 

[DR1]

ok so by reaaranging the formule as we know that the new KE after impact will be 20% less than before impact

ke=.5IW^(2) FOR W

i get the common angular volocity as being 14.04rads-1 does that sound about right or look correct

 

[jrlaguna]

Doesn't sound right to me. Too low. You reduce the kinetic energy by 20%, so do this:
(a) Find the initial KE: (1/2) I w^2
(b) Reduce that amount by 20%
(c) Find the new w, corresponding to the reduced KE.

Should be only a little bit lower than the initial 150 rpm

 

[DR1]

Doesn't sound right to me. Too low. You reduce the kinetic energy by 20%, so do this:
(a) Find the initial KE: (1/2) I w^2
(b) Reduce that amount by 20%
(c) Find the new w, corresponding to the reduced KE.

Should be only a little bit lower than the initial 150 rpm

ok i think i mucked up my initial ke calculation

redoing my figures takes the result down to 12.96rads-1 does that sound more like it

 

[jrlaguna]

sounds better to me too :) ok, you're in the right track!

 

[DR1]

ok fantastic so how would i get from that information to getting the mass of flywheel b

 

[jrlaguna]

Now, conservation of angular momentum.

 

[DR1]

Now, conservation of angular momentum.

without the moment of inertia for b the formule won't work or am i missing something

 

[jrlaguna]

In the equation for the conservation of angular momentum you have all but the moment of inertia for b, so you can get it...

 

[DR1]

In the equation for the conservation of angular momentum you have all but the moment of inertia for b, so you can get it...

just noticed that trying to work it out now

 

[DR1]

ok if i did it correctly i got an answer of Ib=86.27 but i not convinced my math is right

 

[jrlaguna]

I don't know either, but if you tell me your method I can tell you if it's right.

 

[DR1]

I don't know either, but if you tell me your method I can tell you if it's right.

putting in all info into Ia(W1)a + Ib (W1)b=Ia(W2)a + Ib(W2)b


144a(15.71)b + Ib (8.37)b=144a(12.96)a +Ib (12.96)b

2262.24+Ib(8.37)b=1866.24+Ib(12.96)b

2262.24-1866.24 + Ib(8.37)b=Ib(12.96)b

396=Ib (12.96)-Ib(8.37)

396=Ib(4.49)

Ib= 396/4.59

Ib= 86.27kg m^(2)

 

[jrlaguna]

seems to be ok... :) congrats!

 

[DR1]

seems to be ok... :) congrats!

so for the mass i reaarange I=mk^(2)

so m=I/k^(2)

=86.27/0.8^(2)

m=134.8kg

that sound ok