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Homework Help: Braking torque (can someone check this please)

  1. Dec 30, 2009 #1
    1. The problem statement, all variables and given/known data
    a light shaft carries a steel disk 400m dia, 50mm thick,density 7800kg/m_3 calculate :

    (a) if the frictional resistance is equivalent to a torque of 1.5Nm, determine the braking torque to bring the disk to rest from 43.98 rad/s in 4 secs.

    (b) what would be the time for the system to stop, if the motor was turned off with no braking applied ?

    i also had work out the change in angular momentum which = 148 kgm_2/s

    and change in angular kinetic energy = 3.7 KJ

    2. Relevant equations

    acceleration = final velocity - initial velocity/ time

    Torque = moment of inertia x acceleration

    3. The attempt at a solution

    Question (a)
    acceleration = 0 - 43.98/4 = 10.99 rad/s

    Torque = moment of inertia x acceleration

    : moment of inertia = torque / acceleration = 1.5Nm / 10.99 rad/s = 136.48 x 10 ^-3

    Torque = moment of inertia x acceleration = 136.48 x 10 ^-3 x 10.99 rad/s = 1499.91 Nm

    Question (b)

    acceleration = final velocity - initial velocity/ time

    : time = final velocity - initial velocity/acceleration = 0 - 43.98/10.99 = 4 sec

    this does not look right to me but where have i went wrong any help would be most appreciated thanx again.....
    Last edited: Dec 30, 2009
  2. jcsd
  3. Dec 31, 2009 #2


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    The first line quoted is where you went wrong. The alpha = 10.99 rad/s2 is the TOTAL angular acceleration required in order to have a stopping time of 4 s. But 1.5 Nm is the torque provided by friction only, in the absence of any active braking. This torque (probably) does NOT provide an angular acceleration of 10.99 rad/s2. In short: you divided the wrong torque by the wrong angular acceleration. The torque that goes with the 10.99 is the total (applied + frictional) torque that will give you sufficient alpha to stop in 4 s. This is an unknown that you are supposed to solve for. To do so, you need to determine I (the moment of inertia of the disc) independently using some other method.
  4. Jan 1, 2010 #3
    thanks for your reply i will let you no how i get on :redface:
  5. Jan 1, 2010 #4

    I = 48.98 x 20 x 10^-3 = 979.6 x 10 ^-3 kgm^2/s

    Torque = moment of inertia x acceleration

    : T = 979.6 x 10 ^-3 kgm^2/s x 10.99 = 10.76 Nm

    Applied torque + frictional torque = 10.76 Nm + 1.5 Nm = 12.26 Nm Bracking Torque...

    Hope this is correct...
  6. Jan 1, 2010 #5


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    Hi series111,

    Sorry, I didn't see your latest post until now. I'll address your moment of inertia calculation in a little bit. But first, I think that there is an error here:

    The torque you just calculated before that line (10.76 Nm) is what provides 10.99 rad/s^2. That acceleration is all you need, which means that that torque is all you need, in total. Since friction is helping you, you don't have to apply as much yourself. In fact you only need to apply enough torque to make up the difference between what friction already provides and the total amount you need. Hence, you should be subtracting instead of adding:

    Total torque = Applied torque + frictional torque

    ==> Applied torque = total torque - frictional torque = 10.76 Nm - 1.5 Nm

    ...which is assuming that 10.76 Nm is correct, which it may not be (I'm gonna check your work more closely in a sec). I hope that this makes sense. If it doesn't, please let me know.
    Last edited: Jan 1, 2010
  7. Jan 1, 2010 #6


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    Okay, I'm back already:

    Can you please explain what 'K' is? It is possible that your formula is correct, but I can't say for sure, because I don't know what that factor means. This object is a disc. In fact, it is a thin disc, where the word 'thin' in this context is referring to the fact that its thickness is much much less than its width. As a result, we can approximate the mass distribution as being 2D. You can easily look up the moment of inertia of such an object (as well as many other basic geometric solids). I think almost any introductory physics or engineering mechanics textbook would have it (or even better yet, the internet).

    It's hard for me to troubleshoot this as well, because you haven't included any units with those numbers, making them essentially meaningless: 48.98 what? Please, please get into the habit of carrying units through every single step of your calculations. This practice is essential. Not only does it make sure that other people will know what you are talking about, but it also provides a crucial check against mistakes for you. If the quantity you are calculating doesn't end up with the units it is supposed to have, then you know something is amiss. All of your equations should always be dimensionally consistent (meaning that the thing on the left-hand side has the same dimensions as the thing on the right-hand side).

    For the time being, all I can say is that I agree that the mass of the disc will definitely be part of the moment of inertia calculation. It is not provided to you in this problem. However, you do know its density, and you can easily figure out its volume.
  8. Jan 2, 2010 #7
    k ^2= radius of gyration hence k^2 = d^2/8

    The equation = (I) moment of inertia = mass x radius of gyration

    I calculated the mass by :

    volume = pie dia^2/4 x length = pie 400 x 10 -3m ^2/ 4 x 50x 10-3m = 6.28 x 10-3m

    density= mass / volume :

    mass= density x volume = 7800 kg/m3 x 6.28 x 10-3m = 48.98 kg

    k ^2= radius of gyration = 400x10-3m^2/8 = 20x10-3m^2

    (I)moment of inertia=(M) mass x (K^2) radius of gyration :

    (I)moment of inertia =(M) 48.98 kg x (K^2)20x10-3m^2 = 979.6 x 10 ^-3 kgm^2/s

    angular acceleration = final velocity-initial velocity/time :

    angular acceleration = 0-43.98 rad/s / 4sec = 10.99rad/s^2

    Torque = moment of inertia x acceleration

    : T = 979.6 x 10 ^-3 kgm^2/s x 10.99rad/s^2 = 10.76 Nm

    Applied torque = total torque - frictional torque = 10.76 Nm - 1.5 Nm = 9.26 Nm

    i have also looked at Question (b) and found the equation:

    time = 2 pie square root mass / force

    so i have calculated 2 pie square root 48.98 kg/ 10.76 Nm = 13.4 secs

    is this the correct formula or am i totally wrong..

    i hope i have inturpreted your explination correctly thanks again for being patient and taking the time out to help me :smile:
    Last edited: Jan 2, 2010
  9. Jan 3, 2010 #8


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    In that case, I think that your equation is perfectly correct, because the moment of inertia of a thin disc is given by I = (mr2)/2 = [m(d/2)2]/2 = [(md2)/4]/2 = mk2

    Just FYI, the name of the Greek letter [itex] \pi [/itex] is spelled 'pi.' Of course, that's just a minor point. Your math looks fine. I get 49.01 kg, but that difference is probably attributable to rounding errors. One question: in your calculations, you seem to have used 400 mm as the disk diameter, but you typed 400 m in the original problem statement. Was the latter just a typo?

    This part in red seems to be an error. There is no 's' in the units. The units for moment of inertia are kgm2, which should be obvious because you are multiplying something in kg by something in m2. That having been said, the numerical value seems fine. I also get about 0.980 kgm2.

    All of this looks okay. I got 10.77 Nm for the total required torque, but that can be attributed to our slightly different numbers and is not a big deal.

    Again, dimensional analysis is your friend. Mass/force has dimensions of acceleration-1, right? Therefore, (mass/force)1/2 has dimensions of (acceleration)-1/2 = (time2/length)1/2 = (time/(length)1/2). Therefore, your equation cannot be correct, because it does not give a result having dimensions of time.

    It seems that you're still not getting a very important point. When I say include units in your calculations, I don't mean to just manually put in whatever units you think ought to be there (which is what you did on the right-hand side of the calculation above). I mean actually work out what a kg/(Nm) is, and put that in the result. If you had done so, you would have realized that the answer is not a second, and therefore something must be wrong. By the way, not only is this equation incorrect, but you have also applied it incorrectly, because you used the torque, and torque is not the same thing as force. I'm sorry if it sounds like I'm being harsh or nitpicky, but paying attention to what physical quantities are on either side of an equation and whether they are consistent is really going to help you in the future. You could have caught these errors by being a little bit more careful with units.

    So how to actually calculate the stopping time? Remember in part (a) how you were given a stopping time and a required change in angular velocity, and you used that to determine the angular acceleration, and hence the required torque? Well, did it ever occur to you that you could just do the same thing, only "backwards?" This time the torque is given, and hence the angular acceleration. Therefore, given that, and the required change in angular velocity, you can calculate the require stopping time. In other words, the exact same physics is being used. The only change between parts (a) and (b) is which quantities are given vs. which ones you have to solve for.
  10. Jan 4, 2010 #9
    So torque =(M) mass x (K^2) radius of gyration x angular acceleration

    :angular acceleration= (M) mass x (K^2) radius of gyration / torque

    : angular acceleration= (M) 48.98 kg x (K^2)20x10-3m^2 / (t) 1.5 Nm= 653.06 x 10 -3 rad/s^s

    final velocity = initial velocity + angular acceleration x time

    :time = final velocity-initial velocity x angular acceleration

    : time = 0-43.98 rad/s x 653.06 x 10 -3 rad/s^s= 28.7 seconds...

    hope this is correct and thanks again for being patient in helping me as you can gather i have just been introduced to the subject and by the way the kgm^2/s
    for the moment of inertia is in my handout from class so i assumed this to be correct ?????
    Last edited: Jan 4, 2010
  11. Jan 4, 2010 #10


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    You messed up on your algebra.

    [tex] \textrm{torque} = (\textrm{moment of inertia})\cdot (\textrm{angular acceleration}) [/tex]

    [tex] \Rightarrow \textrm{angular acceleration} = \frac{\textrm{torque}}{\textrm{moment of inertia}} [/tex]

    What you wrote is the reciprocal of that. Again, you would have caught this error if you had worked out your units properly, because you would have realized that your result didn't have units of angular acceleration. I am starting to feel like my message is not getting through.

    You also messed up on this equally simple manipulation:

    [tex] \textrm{ang. accel.} = \frac{\textrm{change in ang. vel.}}{\textrm{time}} [/tex]

    [tex] \Rightarrow \textrm{time} = \frac{\textrm{change in ang. vel.}}{\textrm{ang. accel.}} [/tex]

    The '^' operator is used in plain text environments to mean "raise it to the power of." Therefore, s^s means ss, NOT s*s. Your use of it here is wrong.

    It's no problem -- I am happy to help. However, I am wondering about why you are not taking my advice regarding units. Also, you seem to be having trouble with basic algebra, and basic kinematics. Furthermore, all of the errors you are making are ones that would have been revealed through basic dimensional analysis steps (i.e. the units don't work out, therefore it must be wrong). I think that it would help me to help you better if I understood what your background was. What is your background in mathematics and physics? Are you a high school student or a university student? If it is the latter, are you enrolled in an engineering program? What year?

    Clearly it is not correct. No matter which moment of inertia formula you look up, it involves multiplying a mass times a length2. What are the units of the result?
  12. Jan 4, 2010 #11
    Thanks for your help i will just battle on and wont bother you again cheers...
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