Introduction to Torque
If you’re reading about torque, you’ve probably already covered work, which measures a force applied that displaces an object in any direction on a Cartesian coordinate system (x/y coordinates). Now, what if we take that same object and rotate it around a fixed axis? When we do that, we are applying a torque on that object.
Let’s clarify the difference with examples. If there is a box and I am pushing it horizontally, I am doing work on it. If I push it onto an inclined plane to load it onto the back of a truck, it’s still work because the box is not fixed on any particular axis. However, if I take that box and put it on one end of a seesaw, with myself standing at the other, both the box and I are applying torques on the seesaw. The seesaw’s fulcrum fixes it on an axis that causes rotational motion.
Understanding Torque
Like work, torque is a force multiplied by a distance. However, the difference lies in how force and distance are multiplied, and what kind of distance is used in this scenario. For work, the force (F) and the distance (d) were multiplied by a dot product. Because it is the dot product of two vectors, it is a scalar value.
W=F\cdot d=Fdcos { \theta }
Torques, however, involve rotation along a fixed axis. Therefore, instead of a dot product, we need to use a cross product. We also can’t use the distance d that we applied in the previous work equation because we are not displacing the object a certain distance. We are rotating it along a fixed axis. To calculate torque we must introduce a new vector, r. Physically, it is the distance from the axis of rotation to the point where the force is applied.
\tau =r\times F=rFsin { \theta }
Torque is the cross product of two vectors, which means that torque is also a vector. This can make torque problems quite tricky and difficult to understand. We’ll go over an example later on in this article to show you how you can go ahead and solve one of these.
First, let’s look at the physical meaning of this equation for one second. Because of the cross product, we end up with an equation with a sine term, rather than the dot product for work, which netted us a cosine term.
That means that if the force occurs perpendicular to the direction of motion, the work done would be zero because that would mean our cosine term would be cos (90°) = 0. For torque, on the other hand, because of the sine term, there would be no torque if the force were applied parallel or anti parallel to the direction of motion, because sin (0°) = sin (180°) = 0.
Therefore, torque only occurs if the force applied to the object has a component that is perpendicular to the r vector. This means that, although it is a force multiplied by a distance, its units aren’t in joules (J) like work or energy. Instead, the units are just newton-meters (N•m). In this case, newton-meters and joules are not interchangeable due to their implications.
It should also be noted that torque can be calculated on different points on an object. The r vector in the torque equation can technically start at any point so long as it ends at the point on the object where the force is applied. As we will see in our free response example from an AP® Physics test, skillfully choosing the starting point for r can simplify torque problems.
Other Implications of Torque
As a result of torque’s rotational nature, there are certain terms that go along with it that have Cartesian analogs. This is necessary because we’re now operating on a different axis and plane.
First up, we have distance. The rotational equivalent would be the angular displacement (θ – lower-case Greek letter theta), and it is in radians (rad). Next is angular velocity (ω – lower-case Greek letter omega). In Cartesian velocity, we had units of m/s.
Because the axis is rotational for angular velocity, the units are in circular terms. In SI The SI unites would be radians per second (rad/s). Going off of that, we have angular acceleration (α – lower-case Greek letter alpha) in units of radians per second squared (rad/s2). It also causes angular momentum, but that’s a more complex topic covered in a different topic.
To reiterate, torque is a vector. So we can calculate the magnitude, but what is the direction? For the purpose of the AP® Physics 1 & 2 exam, you only need to know clockwise or counterclockwise. Now is the cue for you to turn to your nearest clock and figure out which direction is which way. Don’t pretend like you didn’t need to do that. It’s ok, though, only your computer is judging you.
Depending on which is more convenient for you and makes more sense for that particular problem, you can assign either direction as positive or negative. Traditionally, however, a positive direction is indicative of counterclockwise direction, while you assign clockwise motion with a negative direction. Again, it’s up to you to choose depending on which is more convenient for you.
Torque-Based Free-Body Diagrams
So you thought you never had to worry about free-body diagrams again after learning about forces, now did you? Well, sorry to inform you, but from that point on, free-body diagrams will govern everything you do. They are incredibly useful in determining physical quantities and properties. You will continue to struggle throughout this course if you cannot fully master free-body diagrams.
There is a torque analog of a free-body diagram. When you were using free-body diagrams to determine forces, you represented the object as a single point. For torques, however, you use a straight line to represent the object. This is because the object rotates about a fixed axis, such as a fulcrum. The distance from the axis of rotation is what determines torque, so on the diagram, you need also to indicate r, which is known as the torque arm. Below is an example.
The different arrows represent the forces that are being applied to the torque arm with the respective directions in which they are being applied. Notice that the length of the torque arm is also included, L. 0.5 L obviously represents half of the full length of the arm.
Similar to forces, if there’s no movement and it’s at equilibrium, there is no net torque. You can use this information to set up an equation that allows you to calculate certain pieces of information.
Torque and AP® Physics 1 & 2 Exam
Torque is such an integral part of physics. You can expect several conceptual questions on the multiple choice related to torque. Regarding the free-response question (FRQ), you should prepare for a full question related to rotational motion that incorporates almost all of everything that you’ve learned in the course so far.
Torque is only a small part of rotational motion, so when you get more into rotational motion topics such as angular momentum and rotational inertia, you’ll get a better idea and understanding of what these problems will look like.
We will work on an example from the AP® Physics C Mechanics exam from 2008. Although it is from the AP® Physics C exam, we will only cover the first two parts of the question, which are fair game for the AP® Physics 1 & 2 exam.
This is question 2 from the AP® Physics C mechanics exam from 2008.
The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is attached to a vertical support by afriction-lesshinge that allows the rod to swing up or down. The right end of the rod is supported by a cord that makes an angle of 30° with the rod. A spring scale of negligible mass measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.
(a) On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force vector originating at its point of application.
(b)Calculate the reading on the spring scale.
Well, that looks like quite a contraption. However, it’s important not to get too caught up in the appearance of a problem that you forget how to solve it methodically. It’s quite simple. CollegeBoard already gave us the first step in solving this problem. Let’s draw the forces on the lever.
Let’s start at the left. We are told that the rod is attached to a hinge. The hinge provides support up and to the right, so we’ll draw the arrow in that direction. We aren’t given an angle on the hinge, but we’ll figure out later it’s unimportant. It is important to draw these arrows such that they originate at the lever arm. Next, we have the cord, which is pointed up and to the left from the rod at 30°.
Next, we need to take into account the mass of the block, which we’ll denote as m. We will have that pointing downward from the right end of the rod. Finally, we take into consideration the mass of the rod itself, denoted by large M. As mentioned above, it’s a uniform rod, so we can assume its center of mass is at its actual physical center. Here is what the diagram should look like.
So, part (a) is taken care of. Let’s move onto part (b). Let’s first ask ourselves what the reading on the spring scale represents. It is attached to the cord that is attached to the rod. The cord is represented by the tension force, FT, as indicated by the diagram. So we are essentially solving for FT.
Now let’s remember that the rod is at rest. Therefore, the sum of all torques is zero.
\sum{\tau} = 0
Remember how we said we don’t need to know anything about FH? Here’s why: we can set the hinge as the starting point for our r vector. Therefore, r and r Fsin (θ) zero point on the arm. It’s quite convenient as well, seeing that it is actually at the far left of the rod.
Let’s set up the expression for the torques acting on the y-axis. The length of the rod is denoted by L.
\sum { { \tau }_{ y }=} -\dfrac { L }{ 2 } Mgsin({ 90 }^{ o })+{ F }_{ \tau }Lsin({ 30 }^{ o }) -Lmgsin({ 90 }^{ o }) = 0
Let’s do some algebra to eliminate negatives. We can also cancel every sin(90°) by recognizing that sin(90°) = 1.
{ F }_{ \tau }Lsin({ 30 }^{ o })=\dfrac { L }{ 2 } Mg + Lmg
Next, we can eliminate the length of the rod as it is present in each term in the expression. That allows us to solve for FT.
{ F }_{ \tau } = \dfrac { g(0.5M + m) }{ sin({ 30 }^{ o })}
Plug in the numbers and get our final answer for part (b), the reading on the spring scale.
{ F }_{ \tau } = \dfrac { (9.8 m/{ s }^{ 2 })(0.5(2.0 kg) + 0.50 kg) }{ sin({ 30 }^{ o }) } = 29N
On the AP® exam, this is only going to be a part of a problem of a full FRQ on rotational motion that is covered in other articles regarding angular kinematics such as angular momentum. Setting up the diagram is important, however, as that will dictate how well the rest of the problem goes. On the multiple choice part, you will find conceptual questions related to torque. Make sure you understand the concept of it fully.
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