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Torque/Moment of inertia. Find the mass of the sphere.

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    A sphere (radius = 0.60cm) can be rotated about an axis through its center by a torque of 25 N*m, which accelerates it uniformly from rest through a total angular displacement of 180 rev in a time of 15 s. What is the mass of the sphere?

    2. Relevant equations

    My text book says that this is the formula for the [solid] sphere: I=(2/5)MR^2

    3. The attempt at a solution

    Well, I don't really know what to do here, but I tried this: I=(2/5)MR^2
    I = M* (2/5)(.60)^2
    I = .144 m.
    However, that doesn't make a lot of sense; I'm pretty sure there has to be another equation that needs to be used. I need someone to explain this to me and guide me through this process. Thank you.
     
  2. jcsd
  3. Dec 13, 2011 #2

    Simon Bridge

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    Welcome to PF.

    Think how you would do the equivalent problem in normal linear motion:

    A force of 25N accelerates a block uniformly from rest to a distance of 180m in time 15s. What is the mass of the block?

    Got it?

    It's exactly the same, but using rotational equivalents - so instead of force you have torque, instead of mass you have moment of inertia, instead of distance you have angle.
     
  4. Dec 13, 2011 #3
    Could you give me the formulas that I have to use? In order that they would be used. I'm confused, and the fact that it's well past midnight probably plays a big role as well [right now]. :(
     
  5. Dec 13, 2011 #4

    Simon Bridge

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    Concentrate: what equations are associated with a constant acceleration?
     
  6. Dec 13, 2011 #5
    a = change in velocity/ change in time?
     
  7. Dec 13, 2011 #6

    Simon Bridge

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    Whenever you see the words "constant acceleration" or "uniform acceleration" you should think kinematics and you use kinematic equations... or you draw a velocity-time diagram and use grade-school geometry. (Google "constant acceleration"!)

    For instance - in the linear problem in post #2:
    The v-t graph is a line starting at (v,t)=(0,0) and ending at (u,15) (assign variables for everything you don't know). The acceleration is the slope of this line and the displacement is the area under it. The problem wants you to find the mass - so you find the acceleration, then use F=ma (since you are given F).

    slope: [itex]a = \frac{(u-0)}{(15-0)} = \frac{u}{15}[/itex] (rise over run)
    area: [itex]d = 180 = \frac{15u}{2}[/itex] (half base times height)

    two equations, two unknowns.
    you should be able to complete it from there.


    The rotational form of this will be an angular velocity (ω) vs time graph, the area under the graph will be the angular displacement (θ) and the slope will be the angular acceleration (α).

    Your strategy should be the same as above, but you have an extra step on the end when you convert moment of inertia (I) to mass.

    It some stage you will have to stop thinking of physics problems in terms of finding the right equation and plugging in the numbers. There are way too many equations to memorize for this to work well. Instead, try to remember the physical relationships and use those to construct equations - then you only have to memorize the ones you use lots.
     
  8. Dec 15, 2011 #7
    So, do I use a = change in w / change in t. Which would be 180 rev/ 15 sec = 12 rev/sec. Then, Torque = I * acceleration. ---> 25 N*m/12 rev/sec = 2.1 (what units would go here?)
    And then, m = I /2/5 R^2 = 2.1/(2/5)(.60)^2 = 14.6.

    does this make any sense and is there anything right here?
     
  9. Dec 15, 2011 #8

    Simon Bridge

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    It will help you understand if you just draw the graph.

    [itex]\alpha = \frac{\Delta \omega}{\Delta t}[/itex] fine, but [itex]\Delta\omega \neq 180\text{revs}[/itex]

    I can see your confusion: you are used to thinking of the "rev" as shorthand for "revolution per second" but here you are told that this is an angular displacement of 180 revolutions.

    You need to convert angles into radiens to be useful. 1.revolution = 2π.radiens
    In rotating systems, distance is angle is radiens, speed is radiens per second and acceleration is radiens per second-squared.

    List what you know:
    [itex]\theta = 360\pi\; rad[/itex]
    [itex]\omega_0 = 0\; rad.s^{-1}[/itex]
    [itex]\tau = 25.Nm[/itex]
    [itex]T=\Delta t = 15 s[/itex]
    [itex]r=0.6 m[/itex]

    [itex]\omega_f = \text{?}[/itex]
    [itex]\alpha = \text{?}[/itex]
    [itex]m = \text{?}[/itex]

    Some useful relations:
    (1) [itex]\tau = I\alpha[/itex]
    (2) [itex] d=v_0t+\frac{1}{2}at^2[/itex]
    (3) [itex] a = r\alpha[/itex]
    (4) [itex] I_{sphere}=\text{look it up}[/itex]

    I don't think I can give you any more hints without doing it for you!

    What kind of school makes you work through December?
     
    Last edited: Dec 15, 2011
  10. Dec 15, 2011 #9

    Simon Bridge

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    OK: Merry Xmas...

    from (2): [itex]360\pi=0+\frac{1}{2}\alpha (15)^2 \Rightarrow \alpha = (3.2)\pi \; rad.s^{-2}[/itex] (why?)

    from (4): [itex]I = \frac{1}{2}mr^2[/itex] (check this - look it up)

    put into (1): [itex]\tau = \frac{1}{2}mr^2 \alpha[/itex][tex]\Rightarrow m = \frac{2\tau}{r^2\alpha}=\frac{2(25)}{(0.6)^2(3.2) \pi }=13.816 \; kg[/tex]
     
  11. Dec 16, 2011 #10
    Hey, thanks!... I'm just more confused than I should be at this point, and especially with this problem.
     
  12. Dec 16, 2011 #11

    Simon Bridge

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    Cavendum doctori!
    I have been known to maliciously post wrong answers to catch people out.
    Check what I have written. No responsibility for advice not taken.
     
  13. Dec 16, 2011 #12
    If your help was real and correct - Thank you;
    if it wasn't - then screw you and your life.
    I'm not in a mood for any kind of crap right now.
    Thank you very much.
     
  14. Dec 16, 2011 #13
    Btw, I did check what you wrote. Those two formulas match to the ones my professor gave me.
     
  15. Dec 16, 2011 #14

    Simon Bridge

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    Seriously? No wonder you get confused!
    My advise is always real and always good ;) but you won't learn if you don't do some work to find things out for yourself.

    What did your professor tell you the moment of inertia of a sphere is?
    Remember it is against the rules for me to do your homework for you?
    If you don't understand what I did, you'll get it wrong.
    I think that's the best I can do. Good luck.
     
  16. Dec 16, 2011 #15
    I thought u said that u wrote the info incorrect on purpose. I know the sphere moment of inertia is 2/5mr^2. I appreciate ur help, and as I said above if it's correct than I do thank u, only if u were messing with it than (whatever I wrote). I just dont get that stuff cuz I haven't studied enough :(
     
  17. Dec 16, 2011 #16

    Simon Bridge

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    Does that sphere moment of inertia match the moment of inertia I gave you?
    Should it?
     
  18. Dec 16, 2011 #17
    It doesn't, but only the fraction doesn't. But I still think that I didn't do the problem completely correct. I already handed it in.
     
  19. Dec 16, 2011 #18

    Simon Bridge

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    That fraction is important! I had deliberately used the I for a cylinder to make sure I wasn't doing all the work for you. Had you blindly copied what I did, you'd have deserved the lost marks: you can see that right? (But it wouldn't have been enough to fail the question.)

    You still have to go to school in December? You poor thing!
    In NZ, college is out in October and exams finished by November.
    We don't go back till March.

    Apart from that - doing exactly what I did, but with the correct fraction, and no arithmetic errors, you'd get the right answer.

    Well - good luck and happy xmas.
     
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