Inside Collection (Book): Accessible Physics Concepts for Blind Students
Summary: This module explains the mathematics of torque in a format that is accessible to blind students.
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 the mathematics of torque 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 .
I will begin this discussion with a brief review of what we have learned about the rotation of rigid bodies and go from there into a discussion of torque.
One of the objectives of this review is to summarize the angular kinematic variables that are used to describe rotational motion and relate them to the translational kinematic variables that we already know about.
Given a system of particles, we can describe motion as having two components:
Terminology
When an object rotates, it experiences an angular displacement, which I will refer to as theta in this review.
(Textbooks typically use the Greek letter theta for this purpose. However, your Braille display probably won't display the Greek letter theta. Therefore, I will spell it out when I use it in text, and will replace it by the character "Q" when I use it in an equation.)
The time rate of change of the angular displacement is the angular velocity, which I will refer to as omega in this review.
The time rate of change of the angular velocity is the angular acceleration, which I will refer to as alpha in this review.
Similarity to translational motion
These definitions are very similar to definitions from earlier modules having to do with translational motion. For example, when an object moves, it experiences a displacement. The time rate of change of the displacement is the velocity, and the time rate of change of the velocity is the acceleration.
This similarity derives from the fact that rotational motion can be described by a single angular displacement, theta, just as linear motion can be described by a single spatial displacement, x.
Constant angular acceleration
For constant angular acceleration, we can derive a set of equations that are analogous to corresponding equations for translational motion:
Q = Q0 + w0*t + (1/2)*a*t^2
w = w0 + a*t
w^2 = w0^2 + 2*a*(Q  Q0)
where
Corresponding translational equations
For reference, here are the three translational equations that correspond to the equations given above . These equations were explained in an earlier module that involved the constant translational acceleration of gravity.
h = h0 v0*t + 0.5*g*t^2
v = v0 + g*t
v^2 = v0^2 + 2*g*(hh0)
where
Motion of a point that is a fixed distance from the rotational axis
It is customary to define counter clockwise rotation as the positive direction of rotation. That will be the case in this review.
Consider a rotating disk. Since all points on the disk are rotating together, we can determine the linear displacement, speed and acceleration of any point on the disk in terms of the corresponding angular parameters.
Parameters for a point on a rotating disk
Consider a point on the disk that is a distance R from the axis of rotation. Assume that the disk has rotated through an angle theta. The point will have moved through a distance, s, along the circumference of a circle of radius R. The distance s is equal to the product of the radius and the angular displacement measured in radians:
s = R * Q
where
Relationship between speed and angular velocity
With a little calculus, we can find a relationship between the tangential speed of the point along its path around the circle and the angular velocity of the disk:
v = R*w
where
Relationship between tangential acceleration and angular acceleration
By using a little more calculus, we can find a relationship between the tangential acceleration of the point along its circular path and the angular acceleration of the disk:
y = R*a
where
Kinetic energy in rotation
Consider the hypothetical case of a rigid object made up of a set of point particles connected by rods with zero mass. In other words, ignore the mass of the mechanism that holds the point particles in a rigid geometry.
Assume that this object rotates about a fixed axis with a constant angular velocity, omega. Also assume that you know the mass of each particle and that you know the distance of each particle from the axis of rotation.
Kinetic energy of each individual particle
We know how to compute the speed of each particle along its circular path.
The kinetic energy of each particle would be
Ki = (1/2)*mi*vi^2
where
Kinetic energy for translational motion
Hopefully you recognize the above equation as being the same as the following equation that we saw in an earlier module involving the kinetic energy of an object in translational motion.
KE = 0.5*m*v^2
where
Kinetic energy for the object
The object described above is made up of a system of particles.
The total kinetic energy of the system is the sum of the kinetic energy values of each of its particles.
The kinetic energy for the system is
Ks = (1/2)*(sum from i=0 to i=N(mi*ri^2))*w^2
where
Rotational inertia or moment of inertia
Define the rotational inertia, or the moment of inertia, whichever term you prefer as
I = sum from i=0 to i=N(mi*ri^2)
where
Rotational kinetic energy based on rotational inertia
From this, we can determine that
Ks = (1/2)*I*w^2
where
Similar to translational kinetic energy
Note the similarity between the kinetic energy of a rotating system and the kinetic energy an object undergoing translational motion:
Ke = (1/2)*I*w^2
Kt = (1/2)*m*v^2
where
Similarities between rotational inertia and mass
The mass of an object tells us how its kinetic energy is related to the square of its velocity.
The rotational inertia of a rotating object tells us how its kinetic energy is related to the square of its angular velocity.
The rotational inertia plays the same role in rotational motion that mass plays in translational motion.
Differences between mass and rotational inertia
While mass and rotational inertia play similar roles, there are also major differences between the two including:
Consider a point mass that is constrained to move in a circle. Let the mass be acted upon by an arbitrary force F. We learned earlier that in order for the mass to be moving in a circle, there must be a component of force, (the centripetal force), that is directed toward the center of the circle.
Tangential force, mass, radius, and angular acceleration
If we assume that the speed of the mass is changing, there must also be a component of the force that is tangential to the circle at the location of the point mass acting on the mass. This force is required to produce acceleration. Therefore, we can write:
Ft = m*y
From above ,
y = r*a
Substitution yields
Ft = m*r*a
Tangential force, mass, radius, and angular acceleration.  


Tangential force, radius, angular acceleration, and moment of inertia
We confirmed earlier that the rotational analog of mass is the rotational inertia . In the case of a single mass,
I = m*r^2
Substitution yields
Ft = m*r*a, or
Ft = (I/r^2)*r*a, or
Ft = (I/r)*a
Multiplying both side by r yields
r*Ft = I*a
Tangential force, radius, angular acceleration, and moment of inertia .  


Similar to Newton's second law
The equation in Figure 2 looks a lot like Newton's second law for translational motion, which is often expressed as
Force = mass * acceleration
In this case, the rotational inertia, I, is analogous to mass and the rotational acceleration, a, is analogous to translational acceleration.
If this similarity holds, that must mean that the term on the left side is analogous to force in the tangential motion scenario.
Torque, rotational inertia, and angular acceleration
The term on the left , consisting of the product of the force and the distance from the point of application of the force to the axis of rotation is commonly known as the torque. The common symbol for torque is the Greek letter tau, which I will replace with the character T in equations in this module.
T = r*Ft = I*a , or
T = I*a
Torque, rotational inertia, and angular acceleration.  


Torque produces angular acceleration
Thus, we have determined that a torque, which is the product of a tangential force and the distance from the application point of this force to the axis of rotation, produces an angular acceleration.
This is beginning to look a lot like Newton's second law for rotation. We will refine it some more later to improve the analogy. To do that, we will need to define torque and angular acceleration as vector quantities. We will accomplish that using the cross product of two vectors that you learned about in an earlier module.
The convention for positive rotation
If you imagine a rotating object from a viewpoint in which the rotation axis is perpendicular to the page, it is conventional to define a counter clockwise rotation as the positive direction of rotation. That is the convention that I will use in this module.
You learned about the righthand rule involving vectors in an earlier module. There is a similar righthand rule that we use to describe rotating objects.
The righthand rule for angular velocity
If you curl the fingers of your right hand in the direction of rotation of the object, your thumb will point in the direction of the angular velocity vector of the object.
In other words, if some object is spinning in the counterclockwise direction in the xy plane, curling the fingers of your right hand in this direction results in your thumb pointing in the +z direction which we define to be the direction of the angular velocity vector.
The rule for angular acceleration
If the magnitude of the angular velocity increases in time, then the angular acceleration vector has the same direction as that of the angular velocity. If the magnitude of the angular velocity decreases in time, then the angular acceleration vector has the opposite direction as that of the angular velocity.
Defining torque as a vector quantity
The magnitude of a torque is the product of two terms:
Define theta as the angle between the line and the force vector F (not the tangential component of the force vector but the force vector itself). Then the magnitude of the tangential force vector is given by
Ft = Fmag*sin(theta)
where
Define a vector R
Let R represent a vector that lies along the line from the axis of rotation to the point where the force acts with its tail at the axis of rotation. I will refer to the magnitude of this vector as Rmag
Doing a little algebra, we can write
T = r*Ft, or
Tvec = Rmag*Fmag*sin(theta)
where
The cross product
Why do I refer to Tvec as a vector in the above equation?
You learned in an earlier module that the cross product of two vectors A and B is given by
AxB = Amag*Bmag*sin(angle between A and B)
where
The torque vector
Comparing the torque vector with the cross product , we determine that
Tvec = RxF
The torque vector .  


The direction of the torque vector
Recall from the earlier module that the direction of Tvec is perpendicular to both R and F and obeys the righthand rule in terms of its absolute direction.
The relationship of torque and rotational inertia
Combining this with what we learned earlier about the relationship among torque, rotational inertia, and rotational acceleration, we can write
Tvec = I*A
A general equation for net torque.  


A general equation for net torque
The equation shown in Figure 5 is a general equation for net torque. The net torque about an axis of rotation is equal to the product of the rotational inertia about that axis and the angular acceleration.
Similar to Newton's second law
This equation is Newton's second law applied to a system of particles in rotation about a given axis. It makes no assumptions about constant rotational velocity.
One of the objectives of this module is to develop concepts involving rotational motion that are analogous to concepts from earlier modules that involve translational motion.
A Lazy Susan
On my dining room table, there is a device that is commonly called a Lazy Susan. In case you are unfamiliar with such devices, it is essentially a turntable. By a turntable, I mean a rather large disk mounted on bearings so that it is free to turn in a plane that is slightly above but parallel with the top of the dining room table.
The purpose of the Lazy Susan
The purpose of a Lazy Susan is to make it easier to serve food at the dining room table. Various dishes are placed on it . When someone wants a helping of carrots, for example, instead of saying "Please pass the carrots," they simply turn the Lazy Susan until they can reach the bowl of carrots and help themselves.
Not a module about carrots
However, we won't be discussing how to serve carrots in this module. In this module, I will use the Lazy Susan, in its empty state between meals, to discuss various aspects of rotating rigid objects.
To make it easier to type the material under discussion, I will refer to the Lazy Susan as a turntable. (For some reason, I can type turntable much more quickly than I can type Lazy Susan.)
Low angular acceleration when coasting
My turntable has pretty good bearings. It is also rather heavy for its size and therefore has a relatively large rotational inertia or moment of inertia, whichever term you prefer.
If you give it a good spin, it will spin for quite a while before all of its rotational energy is dissipated through friction in the bearings and air resistance. By default, therefore, its angular acceleration is low. In other words, the rate of change of its angular velocity is small.
A perpetual motion machine
If we could find a way to eliminate all of the frictional forces acting on the turntable, including air resistance, then it would spin forever. In that case, we would have invented what has been called a perpetual motion machine. The rate of change of angular velocity would be zero, meaning that its angular acceleration would also be zero.
Similar to Newton's first law
This reminds us of a moving body that satisfies Newton's first law, which can be paraphrased something like the following:
In other words, that law tells us that absent a force to the contrary, a moving body will continue to move with no change in velocity.
If we could eliminate all of the frictional forces acting on my turntable (which is a rotating body), we might like to say that
However, that would not be a true statement.
Houston, we have a problem
Assume that you do the following to our hypothetical frictionless turntable while it is spinning. Using one finger from each hand, press on opposite sides of the turntable, applying an equal force directed towards the center of the turntable on each side of the turntable.
No net force
In this case, the turntable would not experience any net forces. Assuming that you are facing north when you do this, and that you press on the east and west sides of the turntable, the frictional forces generated by your fingers would be directed in opposite directions.
If the turntable were spinning counter clockwise (when viewed from the top), the frictional force created by your finger on the east side would be directed toward the south. The frictional force created by your finger on the west side of the turntable would be directed toward the north.
From a translational viewpoint, at least, there would be no net force applied to the turntable. The force that points to the north would be cancelled by the force that points to the south. The force that points to the east would be cancelled by the force that points to the west. Therefore, the turntable would be in translational equilibrium.
Acceleration would no longer be zero
Despite that, the velocity would be decreased meaning that the acceleration would no longer be zero. From this we might conclude that just because a rotating object is in translational equilibrium, it is not necessarily in rotational equilibrium.
The world of torque
We have just entered the world of torque with this hypothetical experiment. Torque is a quantity that plays a role in rotation that is analogous to the role that force plays in translation.
However, torque is not separate from force. It is not possible to exert a torque without exerting a force.
Torque is a measure...
In anecdotal terms, torque is a measure of how effective a force is at changing the angular velocity of an object. Stated differently, torque is a measure of how effective a force is at causing an object to have a nonzero angular acceleration.
Angular acceleration can be either positive or negative
For an object that is rotating about a fixed axis (or is capable of rotating about a fixed axis), such as my turntable, torque can either increase or decrease the angular velocity of the object.
Net translational force was zero
When you pressed your fingers on my turntable as described earlier, the net translational force applied to the turntable was zero. Therefore, the turntable remained in translational equilibrium, meaning that it didn't go sliding towards the edge of the table.
Net torque was not zero
However, the net torque was not zero, so it was not in angular equilibrium. The kinetic frictional forces generated on each side of the turntable resulted in the same algebraic sign of nonzero angular acceleration.
Each force caused the angular velocity to decrease. The two torques created by the kinetic friction forces were not only equal, they had the same algebraic sign. Therefore, the turntable experienced a nonzero net torque.
How are force and torque related?
To begin with, torque is proportional to the magnitude of the applied force, but that is not the end of the story. The rest of the story involves exactly where and in what direction the force is applied.
Create a torque to close a door
For example, consider applying a force for the purpose of creating a torque to close an open door. Initially, the door has zero angular velocity.
You could apply a force in any direction at any point on either side or the edge of the door that you are tall enough to reach. However, with regard to the objective of closing the door, it would matter very much where and in what direction you applied the force.
A very intuitive topic
The interesting thing about this topic is that you already know all about it from a practical and intuitive viewpoint. You would already know intuitively where and in what direction to push on the door to cause it to close with a minimum expenditure of energy.
You probably wouldn't push on the edge of the door
If you pushed on the edge of door, directing your force directly at the hinges (the axis of rotation), the door wouldn't move. While this might prove to be a good form of isometric exercise, it would not be an effective way to close the door.
A force in any other direction
A force that is applied to the door, (even on the edge of the door) acting in any direction other than directly toward the hinges could be decomposed into two components:
The radial component is wasted effort
The radial component would make no contribution to the development of the torque required to change the angular velocity of the door and cause the door to close. Only the perpendicular component would contribute to the development of such a torque.
Could develop torque at any point on the door
So now you know that you could apply a force at any point on the door, and so long as that force has a component that is perpendicular to the surface of the door, the perpendicular component would contribute to the development of torque.
Location, location, and location
However, it is also important where you push on the door to apply the force. If you push on what we normally consider to be the inside surface of the door, it might create a torque, but that torque may have the wrong sign or direction to cause the door to close. In fact, that would cause the door to open even further.
Once you realize that you must push on what we would call the outside surface of the door, it would still be important where you push and apply the force. Suppose for example that you were to push at a point that is only one inch away from the hinge. You know intuitively that even for a lightweight door, you might have to apply a very strong force to cause the door to close by applying the force at that location.
Where would you push?
You would probably push on the door at a point somewhere between the center of the door and the outer edge of the door.
If the door happened to be a really heavy door, you would probably push on the door at a point as close to the outer edge as possible. This would make it possible for you to cause the door to close with the minimum effort on your part.
The magnitude of the torque
In an attempt to codify your intuition, your instinctive knowledge, or perhaps your acquired knowledge into something more mathematical, we will define the magnitude of the torque as
torque = r*F
The conventional symbol for torque is the Greek letter tau. However, your Braille display probably won't display that Greek letter, so in this module, I will represent torque with the letter T as shown in Figure 6 .
The sign convention for torque
In this module, I will use a sign convention such that a force whose perpendicular component, when acting alone, would cause the object to rotate in a counter clockwise direction as a positive torque.
If that torque is the only torque acting, it would cause a positive angular acceleration.
A definition of torque
You saw a mathematical definition of torque earlier . Figure 6 shows a somewhat less mathematical definition of torque.
The definition of torque.  


The units can be confusing
The units for torque can be confusing because the SI unit for work or energy in joules is also N*m. However, even though torque and energy have the same units, they have entirely different meanings. Torque is not a form of energy.
Imagine a puck sliding in a circular groove that has been cut in the ice at an ice rink. A cross section of the grove is rectangular so that the puck just fits from side to side and sets level on the bottom of the groove. When a puck slides inside the groove, it will move in a large circle.
Apply a force to the puck
If you apply a force to the puck in (almost) any direction, a component of that force will directed toward or away from the center of the circle. For any case where the direction of the force doesn't lie on a line from the puck to the center of the circle, there will also be a component of the force that is perpendicular to that line, which will make it tangential to the circle.
Construct a graph board simulation
Use your graph board and create a Cartesian coordinate system with the origin near the lowerleft corner of the graph board. Use pushpins and pipe cleaners to draw a quarter of a circle, with the center of the circle at the origin. Make the radius approximately onehalf of the smallest dimension of the graph board. This circle should include the entire upperright quadrant of your Cartesian coordinate system.
Identify the location of the puck
Now insert a pushpin at a point somewhere on the circle about mid way between the intersection of the circle and the x and y axes of the coordinate system.
Imagine that this is the puck mentioned above that is constrained to move in a circle. Label this point P.
A radial line from the center
Use a pipe cleaner or a rubber band to draw a line from the puck to the center of the circle. Label this line r.
Create a force vector
Make a little loop at one end of a pipe cleaner that is about half the radius of your circle and place the loop around the pushpin that represents the puck at P.
Leave it loose enough that it can be rotated around the pin. Imagine that this is a vector that describes a force being applied to the puck with the tail of the vector at the puck.
Point the force vector at the center
Begin by pointing the force vector directly at the center of the circle. You will probably be able to imagine that since the puck is not free to move directly to the center, a force in this direction will not cause the puck to move.
The technical reason that it won't cause the puck to move is because the force doesn't have a component that is tangent to the circle at the location of the puck.
Rotate the force vector
Now rotate the force vector clockwise by about 30 or 40 degrees and pin it down so that it won't move. Label the tip of the force vector F.
Draw the tangential component of the force vector
This may be the most difficult part of this exercise for a blind student. Use your protractor (or some other method that you know about) to find a point on the line labeled r such that a line through that point and perpendicular to r goes through the tip of the force vector. Mark that point with a pushpin and label it Q.
Draw a line from Q to F
Use a pipe cleaner to draw a line from Q to F. That line represents the component of the force vector that is tangent to the circle at the location of the puck. (Actually it is parallel to the tangential component of the force vector, but that is OK. It is still the correct length and points in the correct direction.) The direction of that tangential force component is from Q to F.
This is the component of the force vector that causes the puck to move. Label this vector Ft for tangential force.
The radial component of force
Use a pipe cleaner to draw a line from P to Q. This is the component of the force vector that points directly from the puck to the center of the circle. This component won't cause the puck to move.
A right triangle
If you examine your vector diagram at this point, you can determine that the points labeled P, Q, and F represent the vertices of a right triangle, with the right angle at the point Q.
The length of the tangential force vector
Label the interior angle at P with an A. Now you should be able to determine that the length of the tangential vector named Ft is equal to the product of the force F and the sine of the angle A.
Ft = F*sin(A)
where
The torque
Referring back to Figure 6 , we find that the torque produced by this force is equal to the product of the distance from the center to P and the tangential or perpendicular component of the force vector.
Therefore,
T = Ft*r, or
T = F*sin(A)*r
where
This section contains several example scenarios involving torque.
The turntable discussed earlier, which has a radius of 24 cm, is spinning clockwise. You press your fingers on the east and west sides of the turntable with equal forces of 6.67 N. The coefficient of friction between the turntable and your fingers is 0.75. What is the net torque on the wheel?
Solution:
The 13.3 N force on each side of the table creates a tangential kinetic friction force on each side of the table equal to
Ft = 6.67 N * 0.75
Each force is in the opposite direction of the direction of rotation.
The net torque is equal to the sum of the torques.
Each torque is equal to the product of the force and the distance from the center of the turntable to the point at which the force is applied.
T = 2 * Ft * r, or
T = 2 * 6.67 newtons * 0.75 * 24 cm
Entering this expression into the Google calculator gives us
T = 2.4 joules
However, this is one case where the Google calculator gives us a misleading answer. We know that torque is not measured in joules. Instead, torque is measured in N*m. Therefore, the net torque on the turntable is
T = 2.4 N*m
When viewed from above, the scenario is a door that is open. From above, the wall to which the door is attached can be represented by a horizontal line that runs from west to east. The door can be represented by a line segment at an angle of about 45 degrees south of east. The line segment (door) is attached to the wall at the upperleft end of the line segment. That is the point where the door is hinged, and that point is the axis of rotation for the door.
Assume that the axis of rotation extends out of the page towards you.
The door will need to rotate about 45 degrees counter clockwise to become flush with the wall and be closed.
A person is standing on the north side of the wall pulling on a rope that is attached to the door. The rope is attached 11.5 cm from the hinge and makes a 45 degree angle with the surface of the door. That person pulls on the rope with a force of 51 N.
Using the door hinges as the axis of rotation, find the magnitude of the torque that is exerted on the door. What is the sign of the torque.
Solution:
This solution is based on the cross product from Figure 4 .
The magnitude of the torque is given by
T = r * F * sin(angle), or
T = 11.5 cm * 51 newtons * sin(45 degrees)
Entering this expression into the Google calculator gives us
T = 4.15 N*m
The torque will cause the door to rotate in a counter clockwise direction. Therefore, the torque has a positive sign.
Three objects are rotating about their centers. All three objects have a mass of 10 kg. The three objects have the following shapes:
A. A solid disk with a moment of inertia given by
I = (1/2)*m*r^2
where
B. A disk with a round hole in the center with a moment of inertia given by
I = (1/2)*m*(r1^2 + r2^2)
where
C. A square plate with a moment of inertia given by
I = (1/12)*m*(h^2 + w^2)
where
Find the net torque required to cause each object to accelerate at a rate of 10 radians/sec^2.
Solution:
All three solutions are based on the general equation for torque given in Figure 5 .
A. T = I * A, or
T = (1/2)*m*r^2 * A, or
T = (1/2)*10kg*(2m)^2 * 10 radians/second^2
Entering this expression into the Google calculator gives us
T = 200 N*m
B. T = I * A, or
T = (1/2)*m*(r1^2 + r2^2) * A, or
T = (1/2)*10kg*((1m)^2 + (2m)^2) * 10 radians/second^2, or
T = 250 N*m
Note that because more of the mass is located close to the outer edge of the disk, the moment of inertia is higher and more torque is required to achieve the same acceleration for the same mass.
C. T = I * A, or
T = (1/12)*m*(h^2 + w^2) * A, or
T = (1/12)*10kg*((3.54m)^2 + (3.54m)^2) * 10 radians/second^2, or
T = 2.09 N*m
Note that the square in part C was designed to have the same surface area as the disk in part A. The mass in both cases was uniformly distributed throughout the entire surface. Under those conditions, a square has a slightly higher moment of inertia than a disk and thus requires a slightly greater torque to achieve the same acceleration.
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.
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.
This section contains a variety of miscellaneous information.
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 preprinted 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
"Blind students should not be excluded from physics courses because of inaccessible textbooks. The modules in this collection present physics concepts in a format that blind students can read […]"