The Physics Philes, lesson 59: Work and Power, Torque Style
I’ve been rambling on and on for the last several weeks about rotational and translational motion. This week we’re…not changing the subject, exactly. Instead, it’s time to bring the concepts of work and power into the mix.
In real life, we can do work on a rotating object. Like when you pedal a bike or push a merry-go-round. We can express the work being done in terms of torque and angular displacement.
Let’s say that you’re exerting a force on the rim of a pivoted disk, like that hypothetical kid pushing a merry-go-round. The disk rotates an infinitesimal angle dθ about a fixed axis during a time interval dt. The work done by that initial tangential force while the a point on the rim moves an distance ds is equal to the force multiplied by the distance. If the angle dθ is measured in radians, then the distance is equal to the product of the radius and the angle in radians, and the work done is equal to the tangential force times the radius times the angle in radians.
Remember that torque is equal to the tangential force times the radius. We we can re-write he above equation to include torque:
This basically means that the word done is equal to the torque multiplied by dθ.
But that’s just general. There is a special equation when we’re working with a constant torque. Since the angle changes by a finite amount, the work done is equal to the torque multiplied by the difference between angle one and angle two:
These equations are correct for any force, no matter its components. If the tangential force had component parallel to the axis of rotation or a component directed toward or away from the axis, the component would do no work. The displacement of the point of application has only a tangential component, thus it would make no contribution to the torque about the axis of rotation.
This is fine, I guess, but I promised power at the top of this post. POWAH!!! To find the power, we need to take the first equation and divide each side by the time interval during which the angular displacement took place:
Notice that the left side of the equation is the rate of doing work, otherwise known as POWAH!!! Also notice that the fraction on the right is angular velocity. That means we get a pretty simple-looking equation:
But enough talk! Let’s do a simple problem.
You’ve got a car. The power output of the engine is 200 hp at 6,000 rpm. What is the torque?
Calm down. You can do this. We’ll solve the problem together.
Our target variable is the torque. We’re given the power and the angular velocity, so we just need to use the power equation above to solve this problem.
First, let’s convert horsepower to watts. (Horsepower, incidentally, is 746 watts.)
(If anyone has any tips on how to make letters not italicized in LaTeX, please leave them in the comments.)
Next, we need to convert revolutions per minute into radians:
In case you can’t tell, that is 628 rad/s. Now we just have to rearrange the power equation to find the torque:
There you have it! The corresponding torque is 237 newton meters. See? I told you we’d get through this together.
Tune in next week when I do another physics thing! What will it be? TIME WILL TELL.
Featured image credit: Charles Knowles