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# Angular Momentum -- Torque, Work and Energy for Blind Students

Module by: Richard Baldwin. E-mail the author

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

Note: You are viewing an old version of this document. The latest version is available here.

## Preface

### General

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.

### Prerequisites

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:

### Viewing tip

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.

#### Figures

• Figure 1 . Work done by perpendicular component of force.
• Figure 2 . Work done by constant torque.
• Figure 3 . Power generated or consumed by a constant torque.

### Supplemental material

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 .

## Discussion

This section will begin by developing the equations from which you can compute the work done by a constant torque that causes a known displacement. Then it will provide a brief discussion of the situation where the torque is not constant.

### Constant torque

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

• Wt represents translational work in joules or newton-meters
• Ft represents translational force in newtons
• d represents the displacement in meters

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 distance from the center of the wheel to the point where the force is applied and
• the component of the force that is perpendicular to a line drawn from that point to the center of the wheel.

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 length of the circular arc and
• the perpendicular component of applied force.

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

Figure 1: Work done by perpendicular component of force.
Work done by perpendicular component of force.
 W = Fp * s where W represents the work done by the perpendicular force Fp is the perpendicular component of force described above s is the length of the circular arc through which the point moves

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

• T represents torque
• r represents the distance from the center of the wheel to the point where the perpendicular force is applied
• Fp represents the perpendicular force

Arc length

We also know that the arc length is given by

s = r * A

where

• s represents the arc length
• r represents the distance from the center of the wheel to the point where the perpendicular force is applied as before
• A represents the angle of displacement measured in radians

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.

Figure 2: Work done by constant torque.
Work done by constant torque.
 W = T*A where W represents the work done by a constant torque T represents the constant torque A represents the angle of displacement measured in radians resulting from the application of the constant torque Work can be either positive or negative. If the torque and the angular displacement have the same sign, the work is positive. Otherwise, the work is negative.

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

• W/t = work per second or power
• A is the angular displacement in radians
• t is time in seconds
• A/t is the displacement in radians per second, which we recognize as angular velocity

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

Figure 3: Power generated or consumed by a constant torque.
Power generated or consumed by a constant torque.
 P = T*w where P represents power in watts (joules per second or newton-meters per second) T represents torque in newton meters w represents angular velocity in radians per second

### Variable torque

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.

## Example scenario

I once visited a factory where mirrors were made. At one of the stations on the manufacturing line, a person used a large horizontal grinding wheel to grind a bevel on the edge of the mirror.

Assume that the grinding wheel is a uniform disk with:

• A moment of inertia, I, equal to (1/2)*M*R^2
• M = mass = 80 kg
• R = radius = 0.0.5 meters

### Part 1

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

• where Ks represents the kinetic energy for the system
• I represents the rotational inertia for the system
• w represents the angular velocity of the system

We could rewrite this equation as

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

where

• deltaKs represents the change in kinetic energy
• w0 represents the initial kinetic energy
• wf represents the final kinetic energy

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

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

### Part 2

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

• W represents the work done by a constant torque
• T represents the constant torque
• A represents the angle of displacement measured in radians resulting from the application of the constant torque

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.

## Do the computations

I encourage you to repeat the computations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

## Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.

## Miscellaneous

This section contains a variety of miscellaneous information.

### Note:

Housekeeping material
• Module name: Angular Momentum -- Torque, Work and Energy for Blind Students
• File: Phy1330.htm
• Keywords:
• physics
• accessible
• accessibility
• blind
• graph board
• protractor
• refreshable Braille display
• JavaScript
• trigonometry
• force
• torque
• work
• energy

### Note:

Disclaimers:

Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.

I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.

Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.

-end-

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