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Frictional Torque?

  1. Feb 22, 2016 #1
    I honestly don't know how to find the friction, much less frictional torque, when given things like this.

    1. The problem statement, all variables and given/known data

    "If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72rev/s, friction in the bearings causes the wheel to stop
    in just 12s.
    If the moment of inertia of the wheel about its axle is 0.30kg\m2 , what is the magnitude of the frictional torque?"

    2. Relevant equations

    I know:
    Στ=Iα
    α=Δω/Δt
    I=mr2

    3. The attempt at a solution
    Like I said, I have no idea how I am supposed to use the given variables to find friction.

    This is an attempt I tried:
    α=Δω/Δt
    =0.72/12
    =0.06m/s2

    So I can now find torque with the angular acceleration:
    ∑τ=Iα
    = (0.30kg/m2)(0.06m/s2)
    = 0.018Nm

    That's a small number that I'm pretty sure isn't right. Even if it is, I don't know where to go from here to find friction. If someone could just guide me/give me equations then I'd appreciate it

    thanks!
     
  2. jcsd
  3. Feb 22, 2016 #2

    berkeman

    User Avatar

    Staff: Mentor

    Couple things wrong there. You are given the angular speed in revolutions per second, not in radians per second (ω). And the units of α are 1/s^2, not m/s^2.

    You are correct to be using this equation to solve this problem: [tex]\tau = I \alpha[/tex]

    Once the wheel is set spinning, the only torque acting on it is from bearing friction.
     
  4. Feb 22, 2016 #3

    TSny

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    Homework Helper
    Gold Member

    If you put in the units that go with the 0.72 and the 12, do you get m/s2? Should you get m/s2 for α?

    Also, is Δω positive or negative for this problem?

    [Sorry, I posted just after berkeman. He's got you covered!]
     
  5. Feb 22, 2016 #4
    So .72rev/s = 1.44π(rad)/s or 4.52rad/s

    So then I put that into my equations:

    α=Δω/Δt
    =4.52rad/sec / 12s
    =0.38rad/sec2

    ∑τ=Iα
    = (0.30kg/m2)(0.38rad/s2)
    = 0.79Nm

    And it doesn't state which direction the wheel was spun, so I am assuming the negative/positive is irrelevant
     
  6. Feb 22, 2016 #5

    berkeman

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    Staff: Mentor

    Closer, but it looks like you copied the units of the moment of inertia I incorrectly in your original problem statement, and that carried through the calculations. The units of I are not kg/m^2...
     
  7. Feb 22, 2016 #6
    Sorry it's kg*m2. But how does that affect my equation? Torque is measured in N*m... Do I have to multiply 0.30 by 9.8m/s2 because of gravity to put it in Newtons?
     
    Last edited: Feb 22, 2016
  8. Feb 22, 2016 #7

    berkeman

    User Avatar

    Staff: Mentor

    It's just that the units of your answer were wrong, so that led me back to where the units were wrong in your OP.

    All else looks good. Always check your units for consistency at each step. It can help to find other mathematical errors.
     
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