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Expressing moment of inertia in terms of m,a,r and g

  1. Nov 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find an expression for the moment of inertia (I) of the platform in terms of acceleration of the mass (a), the value of the hanging mass (m), the radius of the spindle (r) and the constant (g).

    Diagram: http://i.imgur.com/rv3zFYG.jpg

    2. Relevant equations
    F=ma t=Iα a=rα


    3. The attempt at a solution
    Net force: mg-T=ma (T is tension of spring)
    Net torque: T-tf=Iα (T is tension of spring, tf is torque)
    T=1/2ma

    Apparently the answer is: I= (g/a-1)mr^2 but I'm having some trouble getting there.
     
  2. jcsd
  3. Nov 5, 2013 #2

    Simon Bridge

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    tension in the what? "string" perhaps?

    T is not a torque so you cannot subtract a torque from it - dimensions don't match. What is the relationship between torque and force?

    How did you get that?

    You have the right approach - start with free-body diagrams for each element.
    But you need to be careful about your math.

    i.e. for the mass: +ve = "down": ##mg-T=ma##

    i.e. how many torques are acting on the flywheel?
     
  4. Nov 5, 2013 #3
    How did you calculate T=1/2ma?

    Also note that tf=T*r


    WELCOME TO PF!!
     
  5. Nov 5, 2013 #4
    I believe I solved it thanks to all of your help. The forces acting on the mass are gravity (mg) and the tension of the string (T). There is one torque acting on the spindle ( tf). The net torque on the mass is mg - T= ma which can be rewritten as T= mg - ma. The net torque on the spindle is tf =Iα. tf can be rewritten as Txr and I can substitute mg - ma for T, so now I have (mg - ma)xr =Iα. a =αr and I can rearrange this so α =a/r, which I can then substitute into (mg - ma)xr =Iα. Now I have (mg - ma)xr =I(a/r), using some algebra I can rearrange the equation into (mg - ma)xr x(r/a). Now I have (mg - ma)r^2/a, with more algebra I have (mg/a-ma/a)r^2 =I which simplifies to [(mg/a)-m]r^2= I. I pull out m and have (g/a-1)mr^2 =I, please tell me this is right...?
     
  6. Nov 5, 2013 #5

    Simon Bridge

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    Well done - though it's a little hard to read.
    I'd have just left it as $$I=\frac{g-a}{a}mr^2$$ but it is nice if it looks like the model answer.
    (you want the mr^2 together because that is the moment of inertia of a point mass...)

    Just a tip - it is easier to read your algebra if you lay it out like you would when you write.
    i.e give each statement it's own line like this:

    The forces acting on the mass are gravity (mg) and the tension of the string (T).
    There is one torque acting on the spindle ( tf).
    The net torque on the mass is mg - T= ma which can be rewritten as

    T= mg - ma

    The net torque on the spindle is tf =Iα.
    tf can be rewritten as Txr and I can substitute mg - ma for T, so now I have

    (mg - ma)xr =Iα.

    a =αr and I can rearrange this so α =a/r, which I can then substitute into (mg - ma)xr =Iα.
    Now I have

    (mg - ma)xr =I(a/r)

    ... using some algebra I can rearrange the equation into (mg - ma)xr x(r/a).
    Now I have

    (mg - ma)r^2/a

    ... with more algebra I have

    (mg/a-ma/a)r^2 =I

    ... which simplifies to

    [(mg/a)-m]r^2= I

    ... I pull out m and have

    (g/a-1)mr^2 =I
     
  7. Nov 5, 2013 #6
    Huzzah! Thank you very much for both helping me and organizing my mess!
     
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