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Torque required to rotate something

  1. Dec 25, 2009 #1
    Hi guys, was doing some thinking on torque, we all know that we can calculate torque by using T= Force x radius. But let's say if i have a cylinder of mass x kg which has a radius of r, sitting vertically on the floor (flat surface on the floor). How do i determine how much torque is required to start rotating this cylinder?
    I am thinking that it will probably be similar to pushing / pulling a box on a surface, the box will only start moving once the static friction is exceeded, it is similar in this case that cylinder only rotates once static friction is exceeded, which means F=mu x N should come into this, i am thinking maybe Torque required to turn = mu x N x radius??

    Thanks all~
     
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  3. Dec 25, 2009 #2

    rcgldr

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    Re: Torque required to ratate something

    The required torque is related to the rolling resistance. When at rest, there's a tiny flattened spot on the bottom of the cylinder, and a tiny curved derformation in the surfrace the cylinder rests on. This results in in both a (linear) force and torque a that oppose movment. The smaller the deformation and the more elastic the surfaces, the lower the rolling resistance. If the rolling resistance is very small then almost all of the applied torque is being used to increase linear and angular kinetic energy.
     
  4. Dec 25, 2009 #3

    Redbelly98

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    Re: Torque required to ratate something

    Jeff, the OP is talking about a cylinder with a flat face resting on the floor.
     
  5. Dec 25, 2009 #4

    russ_watters

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    Re: Torque required to ratate something

    So then static friction...
     
  6. Dec 25, 2009 #5
    Re: Torque required to ratate something

    If i am trying to slide the whole cylinder, then i understand it will start sliding when the force exerted exceed the static friction.

    But i wanted to know the torque that will start rotating the cylinder and not sliding it, any idea how i can calculate that?

    Thanks!
     
  7. Dec 25, 2009 #6
    About what axis will the cylinder start to rotate initially? Where are you applying the force?
     
  8. Dec 25, 2009 #7

    Doc Al

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    Re: Torque required to ratate something

    Try this: Assume that the cylinder's weight is uniformly distributed over its area. Set up an expression for the torque required to overcome static friction for an element of that area and integrate to find the total torque required.
     
  9. Dec 25, 2009 #8

    Redbelly98

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    Well, it would be an easy calculation for a hollow cylinder. Since all the mass is concentrated at the radius, torque would simply be μmgr.

    You can think of a solid cylinder as being made up of many hollow cylinders, all nested together. Then add up the contributions of all the cylinders, which should mean doing an integral to get the total torque.
     
  10. Dec 25, 2009 #9

    tiny-tim

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    Merry Christmas!

    Hi Ask1122! :smile:

    (have a mu: µ :wink:)
    If it was a cylindrical shell, then the friction torque would be µsmgR.

    So divide the cylinder into shells of radius r and width dr, find the mass of each shell, and then integrate the friction torque for each shell, from r = 0 to R. :wink:
     
  11. Dec 25, 2009 #10

    Redbelly98

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    Since the CM is not sliding, the rotation axis would be the central axis of the cylinder. Also, the question is about torque, not force, to start the rotation.
     
  12. Dec 25, 2009 #11
    Thx for the answers all! Please see if i got the idea correctly:
    Assuming if it is a solid cylinder, then i will integrate the following function (mu*m*g)
    mu * pi * r^2 * height of cylinder * rho (density of cylinder) * gravity
    = mu * pi * h * rho * g * r^2
    after integration (from 0 to r)= mu * pi * h * rho * g * r^3 * 1/3 = Torque required to rotate the cyclinder

    Is it right?

    Thanks all!
     
    Last edited: Dec 25, 2009
  13. Dec 25, 2009 #12

    Doc Al

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    Again, write an expression for the torque for an element of area. (Circular rings are fine.) Once you get that expression, then you can integrate.

    You don't need the density of the cylinder (just assume its mass is M), but you do need its weight per unit area.
     
  14. Dec 25, 2009 #13

    tiny-tim

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    Hi Ask1122! :smile:

    (what happened to that µ i gave you? :confused:)
    No, your volume is wrong, you need the volume of a cylindrical shell.

    Hint: if you're going to integrate, then the volume will have to have a dr in it, won't it? :wink:
     
  15. Dec 25, 2009 #14
    hmm i dont understand, if i assume mass is M, then wouldn't i take away the r in the equation, but i need the r term because i am integrating using the area of the ring?

    Let me write the equation out in full this time:
    F (static friction) = µmg (thx for the µ tiny ;) )
    F dr = µmg dr
    therefore, F dr = T = integral sign µ*pi*r^2*h*rho*g dr
    T = µ*pi*h*rho*g*r^3*1/3 | 0 to r
    T = µ*pi*h*rho*g*r^3*1/3 - 0
    therefore, T = µ*pi*h*rho*g*r^3*1/3

    I am not sure where did the volume go wrong??

    Thanks!
     
  16. Dec 25, 2009 #15

    Doc Al

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    :confused: Calling the mass of the cylinder M has nothing to do with the r term in the expression for torque.

    What's 'm' the mass of? Hint: 'mg' has to be the weight allocated to the element of area. That must have an area term and thus a 'dr'.
    Not sure what you are doing here.

    Do this, step by step:

    Define the element of area.
    Define the amount of the cylinder's weight allocated to that element of area.
    Define the static friction allocated to that element of area.
    Write an expression for the torque required to overcome the static friction for that element of area.

    Then you'll have a correct expression for the torque that you can then integrate.
     
  17. Dec 25, 2009 #16

    tiny-tim

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    Hi Ask1122! :smile:

    (have an integral: ∫ and a pi: π and try using the X2 tag just above the Reply box :wink:)
    πr2hdr isn't the volume, is it? :wink:

    (and you can write density = M/πR2h)
     
  18. Dec 27, 2009 #17
    Thx for the advices, but i am having some difficulties in finding the right expression for the element. Please see if what i am doing makes sense.
    I will use the circumference as my elements which will be 2πr, there will be infinite amount of circules from the centre point to the outer diameter and summing them up will give me area of the cyclinder base, therefore area = 2πr dr.
    All circles will have the same height h, thus, area * h * ρ will give me my mass.
    And m*g*µ will give the static friction force for that particular element, times that by r will give the torque required.

    So my integration formula will be as follow:
    T = ∫ µ*r*g*ρ*h*2πr dr (limits are from centre to radius r = 0 to r)
    therefore: T = µ*g*ρ*h*πr³*2/3 - 0

    is this correct?

    Thanks!
     
  19. Dec 27, 2009 #18

    tiny-tim

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    Hi Ask1122! :smile:

    (just got up :zzz: …)
    ah, now I see it all written out, yes that's fine (with R instead of r, of course). :smile:

    (I was confused when you left out the dr, and I thought you'd got an r2 into the area :confused:)

    ok, now rewrite ρ in terms of M and R. :wink:
     
  20. Dec 27, 2009 #19
    Cool! so it will become T = 2/3 *µ*N*R for a solid cylinder :D
    Ok, so one down and a million other questions to go!
    Thx for the help all! :D
     
  21. Dec 27, 2009 #20

    Redbelly98

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    That's a correct answer, though it seems the answer should be in terms of mass M rather than the normal force N.

    I.e., substitute N = ___?
     
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