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Rotating Cylinder of Variable Density

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Solid cylinder: H=0.14m, R=0.05m. Mass density ∂=900-(900r/0.05), where r is distance from axis of the cylinder.
    A string of negligible mass and length 0.85m is wound around the cylinder, which is set spinning by a horizontal pull on the string with F=2.5N. The cylinder starts from rest and the string is pulled off. Assume no friction between the cylinder and surface on which it rests.
    a) Compute the mass, and moment of inertia around its axis, of the cylinder.
    b) What is the final velocity and angular velocity of and about the CoM of the top?
    c) Compute the final KE of the cylinder. Show this corresponds to work done by force F.


    2. Relevant equations
    M=∫dm
    τ=Fl=αI
    ω=ω0+αt

    3. The attempt at a solution
    a) Not sure at all about either of these, they look wrong to me but not sure what to do...
    dm = 0.5*R2*h dθ *∫∂dr =0.1575
    M = [2π,0]∫0.1575 dθ = 0.99kg

    I = 0.5*M*R2 = 1.24*10-3

    b)
    τ = Fl=2.5N*0.05m = 0.125Nm
    α = τ/I = 101rad.s-2
    ω = αt
    Don't know where to go from here...
     
  2. jcsd
  3. Nov 26, 2011 #2
    Let's start with computing the mass correctly.

    M = integral(dm)

    What is dm? Write an expression for it.
     
  4. Nov 26, 2011 #3

    cepheid

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    Welcome to PF NotACrook

    For the mass, you have [tex] M = \int_V\,dm [/tex]where the V indicates that the integral is actually an integral over the volume of the cylinder. You can write dm as [itex]\rho dV [/itex] so that [tex] M = \int_V \rho\,dV [/tex]Now it's a matter of picking a suitable coordinate system, in this case cylindrical coordinates [itex] (r, \phi, z) [/itex]. In general, the density is a function of position in 3D space within the cylinder [itex] \rho = \rho(r, \phi, z) [/itex]. However, in this case, rho only depends on r and not on the other two coordinates: [itex] \rho = \rho(r) [/itex]. In other words, there is a cylindrical symmetry to the distribution of mass. The density is constant on rings that are centered on, and perpendicular to the cylinder's axis of symmetry. Since the integral is over the entire volume, it's a 3D (triple) integral, as you'll see once you express it as integrals over each of your three coordinates.
     
  5. Nov 26, 2011 #4

    gneill

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    You should specify the units for your density function. What are they? Grams per cubic centimeter? Kilograms per cubic meter? Tons per cubic millimeter? :smile:
     
  6. Nov 26, 2011 #5
    kg/m^3 for density, sorry.

    So,

    V=πr2h -> dV/dr=2πrh -> dV=2πrh dr
    2πh∫∫∫Vρ(r,Φ,z)*r dr

    Well... Setting around the center of the cylinder, boundaries for Φ would be 0 and h, r and z both R to -R. Would that be correct?

    40πh∫R-RR-Rh0 80r-9r2 dΦ dr dz

    ?
     
  7. Nov 26, 2011 #6

    gneill

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    If you take full advantage of the cylindrical symmetry the volume can be broken down into nested thin cylindrical shells.
     
  8. Nov 26, 2011 #7
    I'm.. honestly not sure how to do that. Could you give me the first step, possibly?
     
  9. Nov 26, 2011 #8

    gneill

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    What's the material volume of a thin cylindrical shell of radius r and thickness dr?

    attachment.php?attachmentid=41282&stc=1&d=1322333461.jpg
     

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  10. Nov 26, 2011 #9
    Yeah, went afk for dinner and worked on it a bit before checking back here, sorry. So, would I be correct in having that...

    dV = hπ(r+dr)^2-hπr^2=hπ(2rdr+dr^2)
    I seem to remember things like dr^2 can be discounted as being pretty much non-existant?
    M = ∫ρ dV = 2πh∫ρr dr = 2πh∫1600r-18000r2 (Hey, I was using 900/5 instead of 900/0.05 above. Well that was stupid and explains a lot.)
    M = 400πh∫R08r-90r2 = 1.1
    ?

    And was I=MR2 the right equation for MoI?
     
  11. Nov 26, 2011 #10

    gneill

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    It's even simpler than that. If the shell is very thin (as in "dr" thin), then a differential approximation for the volume of the shell is the surface area at radius r multiplied by dr. It's as though you've taken a rectangular sheet of paper of thickness dr and formed a cylinder out of it. The surface area is just [itex] 2 \pi r h[/itex], so the mass element is [itex] 2 \pi h r \rho(r) dr[/itex].
    Yes, the moment of inertia of a thin-walled cylinder is MR2.
     
    Last edited: Nov 26, 2011
  12. Nov 26, 2011 #11
    Ok, taking M=1.1, I=0.5MR2 as correct, would I be correct in stating these equations for the later parts of the question (plugging in numbers of course):

    VCM:
    a=F/M
    V2=U2+2aS=0+2SF/M
    VCM=SQRT(2SF/M)

    Angular Velocity about CM:
    ω=VCM/R

    Kinetic Energy and Work Done:
    K=0.5MVCM2 + 0.5ICMω2
    W=FS
     
  13. Nov 26, 2011 #12

    gneill

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    Remember that you have to integrate the dm's with the density function to sum up the contributions of all the nested cylinders from radius 0 through R. The final expression for the moment of inertia will be a bit more complicated than that of a single cylindrical shell!
    I think you'll have to work out how long it takes the string to unwind from the cylinder, which tells you how long the force is applied, before you can determine the other quantities. It will require calculating the angular distance that the cylinder turns w.r.t. time (so [itex] \alpha,~ \omega,~ \text{and}~\theta [/itex]).
     
  14. Nov 26, 2011 #13
    From what I can find, 1/2MR2 (noting the 1/2) is the equation for moment of inertia of a solid cylinder - since I already have the mass and its symmetrical with its density, won't that work?

    Wouldn't that end up with the same answer for VCM except using a different equation (V=U+at or similar)? Its a constant force, and if I'm remembering right the linear acceleration ends up the same regardless of where the force is applied, its just angular acceleration that changes, and for the rest since we have distance and acceleration time is implicit.

    You're probably right, I'd just like to know why rather than accepting it and finding out nothing.
     
    Last edited: Nov 26, 2011
  15. Nov 26, 2011 #14

    gneill

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    You have to keep in mind that the moment of inertia depends upon the distribution of the mass as well as its total. If the cylinder were of uniform density it would be okay.

    I'm thinking that the time that the force is applied depends upon the moment of inertia of the cylinder, since that will determine how long it takes for the string to unwind. The cylinder may move (linearly) many times the string's length before it detaches.

    Also, since the surface is frictionless, the rate of rotation of the cylinder won't be related to the linear velocity (except via the mass versus moment of inertia). In fact, the problem doesn't specify if the cylinder is lying down or standing upright on its end!
     
  16. Nov 26, 2011 #15
    I'd have thought that, so long as you calculate MoI around the edge and not halfway through, the distribution of mass would be fairly irrelevant so long as it is symmetrical? Hm, probably just a conceptual difficulty on my end.

    From the context, standing upright can be assumed.

    I'll work on it tomorrow (midnight does not do my maths any good) - thanks for all your help. Not entirely sure how to work out the time but I guess I'll have to find something...
     
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