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Torque Causing Angular Acceleration on a Uniform Disk

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data
    /* Intro, personal: I am taking Studio Physics 1 at FSU this summer semester. We have an online homework assigned weekly via a program called ExpertTA. This question has been giving me the HARDEST time and I seriously just need some help on it. I have completed and submitted sections a-g and all were correct. Part h), however, I must be doing wrong or something because I believe that I have tried every possibility that I can come up with. Whoever attempts to give this a shot, I genuinely appreciate it. I'm not one to usually ask for help, but like I said... I am stumped. */

    PROBLEM:
    A uniform disk of mass m and radius R can rotate about an axle through its center. Four forces are acting on it as shown in the figure. These forces are labeled F1, F2, F3, and F4. F2 and F4 act a distance d from the center of mass while the other two act on the outer edge of the disk. These forces are all in the plane of the disk.

    naeahrpk.zc2.png

    RANDOMIZED VARIABLES:
    m= 3.6 kg
    R= 24 cm = 0.24 m
    F1= 2.5 N
    F2= 2.5 N
    F3= 6.5 N
    F4= 4.5 N
    d= 3.5 cm = 0.035 m

    a) Write an expression for the magnitude τ1 of the torque due to force F1
    ANSWER: τ1 = F1R

    b) Calculate the magnitude τ1 of the torque due to force F1, in N·m
    ANSWER: τ1 = 0.6

    [c) Write an expression for the magnitude τ2 of the torque due to force F2
    ANSWER: τ2 = F2d

    d) Calculate the magnitude τ2 of the torque due to force F2, in N·m
    ANSWER: τ2 = 0.0875

    [e) Write an expression for the magnitude τ3 of the torque due to force F3
    ANSWER: τ3 = 0

    [f) Write an expression for the magnitude τ4 of the torque due to force F4
    ANSWER: τ4 = dF4sin(53)

    g) Calculate the magnitude τ4 of the torque due to force F4, in N·m
    ANSWER: τ4 = 0.1258

    h) Calculate the angular acceleration α of the disk about its center of mass in rad/s2. Let the counter-clockwise direction be positive.
    PREVIOUS ANSWERS:
    Xα=3.95
    Xα=25.65
    Xα=25.6519
    Xα=25.66

    2. Relevant equations

    τ=rFsin(θ)
    τ=Iα
    α=τNET/I
    [itex]\downarrow[/itex] Moment of Inertia of a uniform disk about its center of mass
    I=[itex]\frac{1}{2}[/itex]mR2

    3. The attempt at a solution
    I don't recall how I got the first answer that I submitted, but I do for the last three.
    Since the directions of the torques varies, the net torque will be calculated by adding the two ccw torques and subtracting the cw torque
    τNET124
    Next I set up the angular acceleration equation using the net torque just written and the moment of inertia formula
    α=τNET/I = (τ124 )/ ([itex]\frac{1}{2}[/itex]mR2)
    This equation produces the answer in terms of m/s2, but the question clearly asks for radians so, using the equation where 1 rad = 0.24 m = R
    α=(τ124 )/ ([itex]\frac{1}{2}[/itex]mR2)·[itex]\frac{1 rad}{R}[/itex]
    This SHOULD give me the correct answer. And, plugging all of the numbers in, I get 25.6519 but that is allegedly incorrect.
    Please help me.
     
    Last edited: Jul 17, 2014
  2. jcsd
  3. Jul 17, 2014 #2

    jbriggs444

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    Science Advisor

    What leads you to think that the result of this equation is in meters per second squared? What are the units of torque? What are the units of moment of inertia? What are the units of the quotient?
     
  4. Jul 17, 2014 #3
    Error

    I made an error in translating my work, I'm going to go back and fix it now.
    Basically, τ3 is 0, so in my writing out the work that I did I accidentally wrote τ3 instead of τ4
    Fixing that now
     
  5. Jul 17, 2014 #4

    verty

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    Homework Helper

    With full precision I get 25.6512463... . Perhaps the rounding is the problem.
     
  6. Jul 17, 2014 #5
    Hm, okay. Lemme work this out.
    Units of torque?
    τ = (#) N·m = (#) (kg·m·m/s·s) · m = (#) kg·m3·s-2
    m = (#) kg
    R = (#) m
    R2 = (#) m2
    Units of moment of inertia?
    I = 0.5·m·R·R = (#) kg·m2
    α = τ/I = kg·m3·s-2/kg·m2 = kg·kg-1·m3·m-2·s-2 =(cancelling) [STRIKE]kg·kg-1[/STRIKE]·m[STRIKE]3·m-2[/STRIKE]·s-2 = m·s-2

    So that's my reasoning for why I believe the equation results in m/s2
     
  7. Jul 17, 2014 #6
    Honestly I thought so as well and that's why I tried a few variations of 25.65... they all came back as incorrect though. This homework program usually accepts answers within at least a tenth of a decimal point of the correct answer. For example is 25.65 WAS the correct answer it was looking for and I entered 25.5 or 25.7 it would accept it. I guess that's why I believe I'm not actually close to the correct answer.
     
  8. Jul 17, 2014 #7

    jbriggs444

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    Science Advisor

    Try that again. What are the units of the Newton?
     
  9. Jul 17, 2014 #8
    Oh my..
    N = kg·m/s2
    I confused Newtons with Joules.
    Wow.

    So if α = τNET/I
    = (τ123 )/ (0.5mR2)
    = (0.6 J - 0.0872 J + 0.1258 J) / (0.5 · 3.6 kg · 0.24 m · 0.24 m) = 0.6383 J / 0.10368 kg·m2
    = 6.16 (kg·kg-1·m2·m-2·s-2
    = 6.16 ([STRIKE]kg·kg-1[/STRIKE]·[STRIKE]m2·m-2[/STRIKE]·s-2
    = 6.16 /s2

    Which I'm assuming is in radians. Woooow. Kickin myself haha.

    And that is correct I just submitted it. Thank you SO much. So frustrating that it was literally such a little mistake but it confused me for HOURS.
     
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