This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.

This module explains torque, work and energy in a format that is accessible to blind students.

In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:

The minimum prerequisites for understanding the material in these modules include:

I recommend that you open another copy of this document in a separate browser window and use the following links to easily find and view the figures while you are reading about them.

I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .

One of the textbooks that I have read uses a very familiar example to illustrate that torque can do work. The example is that of a person pulling on the rope on a power mower or outboard engine to try to get it started.

If you are unfamiliar with that scenario, many small internal combustion engines use a rope wrapped around a pulley to start the engine. When the user pulls the rope, a torque is created on the pulley by the rope. The torque causes an angular displacement of the pulley, which in turn causes certain parts inside the engine to move. If you are lucky and everything is working properly, the engine starts.

The machine fights back

However, the machine fights back and the compression in the cylinders creates a resistive torque. If the user pulls hard enough, the torque created by the user overcomes the resistive torque, the pulley turns, the parts inside the engine move appropriately, and hopefully the engine starts running.

When the engine refuses to start...

Clearly when the engine refuses to start, it becomes apparent very quickly that torque can do work on a human. A few dozen pulls on the rope will cause even the most physically fit user to become exhausted.

The force does the work

The textbook point out that it is actually the force and not the torque that does the work. However, torque and force are related in a very definitive way, and the textbook points out that it is often easer to calculate the amount of work done on the basis of torque rather than making the calculation on the basis of force.

Review -- what is work?

You learned in an earlier module on translational motion that the work done by a constant force is the product of the force and the displacement caused by that force. In other words,

Wt = Ft * d

where

A rotational analogy

Similarly, work done by a constant torque can be calculated as the product of the constant torque and the displacement caused by that torque.

A constant torque

It is important to note that the entire remaining discussion in this section applies only to the application of a constant torque. I will have a few words about a variable torque in the next section .

Power

The power generated or consumed by the application of a constant torque can be calculated as the product of the constant torque and the angular velocity.

A wheel scenario

Imagine a force being applied to a point on the outer edge of a wheel to cause an angular displacement of the wheel. As you will recall from an earlier module, the torque produced by the force is equal to the product of

The point moves through a circular arc

When the force causes an angular displacement of the wheel, the point at which the point is applied moves through a circular arc. The length of that circular, often referred to by s, can be measured. The work done is equal to the product of

The work resulting from the application of the perpendicular force is given by the equation shown in Figure 1 .

Work as a function of torque

Now that we have the work as a function of the perpendicular force and the length of the arc, let's rewrite it in terms of torque and displacement.

Torque

We know that torque is equal to

T = r * Fp

where

Arc length

We also know that the arc length is given by

s = r * A

where

Through substitution

W = Fp * s, or

W = (T/r)*r*A, or

The work done by a constant torque is given by the equation shown in Figure 2.

Power

As in the translational case, power is a measure of the work done per unit of time. If we divide both sides of the above equation by time, we get

(W/t) = T*(A/t)

where

Thus, the power generated or consumed by applying a constant torque is given by the equation shown in Figure 3 .

A torque doesn't have to be constant to do work. In fact, the torque generated by the user with the starter rope on the power mower discussed in the previous section probably isn't constant.

However, if the torque is not constant, you cannot use the equations developed in the previous section to compute the work done by the torque.

Maybe you can use calculus

If the torque as a function of time can be described by a function that you can integrate using integral calculus, you can use calculus to compute the work done by the torque. However, in the real word, this is probably rarely the case.

Maybe you can use a computer

If you are in the business of computing work done by a variable torque, the most likely case is that you will have equipment that allows you to sample the torque and displacement values at uniform intervals of time and to save the values of the samples for digital processing. Then you can use any one of several digital methods to approximately integrate the product of the torque function and the displacement function.

Find the amount of work that must be done to bring the wheel from rest to an angular velocity of 8.38 radians/sec

Solution:

Recall from a previous module that the rotational kinetic energy for a rotating object is given by

Ks = (1/2)*I*w^2

We could rewrite this equation as

deltaKs = (1/2)*I*(w0 - wf)^2

where

However, since the initial kinetic energy value is zero, that would simply complicate the algebra. Therefore, we will stick with the original equation .

We either have, or can calculate values for all of the terms in this equation. Substituting the values given above gives us

Ks = (1/2)*I*w^2 , or

Ks = (1/2)*((1/2)*M*R^2)*w^2 , or

Ks = (1/2)*((1/2)*80kg*(0.5m)^2)*(8.38 radians/sec)^2

Entering this expression into the Google calculator gives us

Ks = 351 joules

This is the amount of work that must be done to bring the wheel from rest to an angular velocity of 8.38 radians/sec

If the motor that drives the wheel delivers a constant torque of 10 N*m during this time, how many revolutions does the wheel turn in coming up to speed.

Solution:

We know how to relate the displacement angle and the work for a constant torque using the equation in Figure 2 .

W = T*A

where

In this case, we know the amount of work and the value of the torque and need to find the angle. Therefore,

A = W*joules/T*n*m

However, this gives us the angular displacement in radians. We need to scale to convert it to revolutions.

A = (W*joules/T*n*m)/2*pi, or

A = (351joules/10newton meters)/(2*pi), or

A = 5.59 revolutions

This is the number of revolutions that the wheel turns in coming up to speed.