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Frictional losses: Why does frictional torque depend on ω?

  1. Dec 6, 2006 #1
    I'm doing a lab in my physics class on rotational motion. Part of the lab was understanding that the device we are working with is not frictionless, and that we will have frictional losses. We got the thing spinning at 500 deg/s, and had a device give us measurements of the angular velocity every second until it reached 150 deg/s (it took about 15 minutes).

    Then, using the relation ω=ω0*e^(-βt)
    where ω0 was the initial angular velocity and β is some constant.

    We then graphed the values of ln(ω) vs time, and found a linear relationship:
    ln(ω)=ln(ω0) -βt

    Thus, from the slope, we determined a value for β, and therefore the frictional torque depends on angular velocity. However, I still don't understand why.

    Can you guys help explain this?
     
  2. jcsd
  3. Dec 6, 2006 #2
    I don't really get your question. Are you asking is to why the torque depens on omega ? That is quite straightforeward so i guess you must be meaning something else. Could you explain any further ?

    marlon
     
  4. Dec 6, 2006 #3
    For the same reason air resistance depends on velocity, think of the two situations as analogous. The faster you go, the greater the force of air resistance. Are you taking a calc based course?
     
  5. Dec 6, 2006 #4
    What does kinetic friction normally depend on? The coefficient as well as the force between the object and the surface. Think about it with relation to your case. What would change the faster it spins?
     
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