Help with Torque Calculation for High Speed Drill

[wickid]

Hello,

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.

 

[Delta]

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 $$T=\frac{33000HP}{2{\pi}N}$$, and will be in lb. ft.

 

[wickid]

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 $$T=\frac{33000HP}{2{\pi}N}$$, and will be in lb. ft.

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 * $${2{\pi}$$ = 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.

 

[FredGarvin]

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

$$HP = \frac{T \omega}{5252}$$


That will give you the resultant torque in ft-Lbf.

 

[Delta]

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.

$$T=\frac{P}{\omega}$$

$$T=\frac{HP*550}{N*2{\pi}/60}=\frac{60*550HP}{2{\pi}N}=\frac{33000HP}{2{\pi}N}$$

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.

 

[krab]

Yes, $\tau=P/\omega$ and $\omega=2\pi f$, 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:
$$\tau=\frac{275\,\mbox{ft.lbs/sec}}{2\pi\times 26.7\,\mbox{/sec}}$$
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.

 

[wickid]

Thank you for your explanation.

Using your conversions and formula, am I correct in saying that the answer is Torque = 1.64 ft-lbs?