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Torque Question

  1. Jun 20, 2005 #1

    I have been stuck on what is probably a fairly easy physics question involving Torque.

    A high speed drill develops 0.500 HP at 1600 RPM. Wjat torque is applied to the drill bit?

    I have the formula as T=P/Omega

    P=.5 HP which I am unsure if I need to convert to another measure of unit
    Omega = 1600 RPM which I am pretty sure needs to be converted into Degree

    If someone could please help me set the problem up with the proper units, I would really appreciate it.
  2. jcsd
  3. Jun 20, 2005 #2
    It depends what units you want the Torque in.

    You would have to be aware that 1HP is equal to 550 ft.lb/s, you also need to convert RPM to rad/s (multiply by 2pi/60), which is indicated by the word omega (angular velocity).

    This produces the formula [tex]T=\frac{33000HP}{2{\pi}N}[/tex], and will be in lb. ft.
  4. Jun 20, 2005 #3

    Thank you for your response,

    I am a bit confused as to how you got 33000HP.
    Also to get Rad/S, I thought we would take 1600 RPM/60 s = 53.33 RPS * [tex]{2{\pi}[/tex] = 167.467 rad/s

    Does the N refference Newtons in the divisor, or is it a number or another unit?

    Also when using HP, we are converting to ft lbs, if dividing by omega in Rad/S aren't we using both british and metric units? Shouldn't we use degrees/s instead of Rad/s as they are both british units?

    Please forgive me is I am missing something really basic, and I appreciate your assistance.
  5. Jun 20, 2005 #4


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    If you want to keep the HP and rpm units a direct calculation is:

    [tex]HP = \frac{T \omega}{5252}[/tex]

    That will give you the resultant torque in ft-Lbf.
  6. Jun 20, 2005 #5
    The formula you have (T=P/omega) is is based on a metric system (i.e. standard SI units). HP is more of a imperial measurement.

    The conversion of HP to ft.lb/s is by a factor of 550, the conversion of RPM to rad/s is by a factor of 2pi/60. And I should of mentioned that N represents the RPM. If these factors are included in the formula you gave above you get.



    On the note about degrees. Radians are always used in angular velocities and torques etc whatever the bias of other force units.

    It may be best/easier (or more logical) to forget about the formula above and simply convert the HP to its lb.ft/s, the RPM to rad/s and plug them into your T=P/omega equation.
  7. Jun 20, 2005 #6


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    Yes, [itex]\tau=P/\omega[/itex] and [itex]\omega=2\pi f[/itex], where f is the revolution frequency; it is 1600 min^-1 or 26.7 s^-1. P=0.5 hp=275 ft-lbs/s. Now: Don't worry about systems of units! Just see whether or not the units cancel:
    [tex]\tau=\frac{275\,\mbox{ft.lbs/sec}}{2\pi\times 26.7\,\mbox{/sec}}[/tex]
    Notice that the inverse seconds in the numerator and denominator do cancel. That's what you want. Do the math and you are left with ft.lbs.
  8. Jun 20, 2005 #7
    Thank you for your explanation.

    Using your conversions and formula, am I correct in saying that the answer is Torque = 1.64 ft-lbs?
    Last edited: Jun 20, 2005
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