Balance: it’s something you probably strive for in your day-to-day life, like finding a healthy school-life balance with your busy schedule. Just like your daily life filled with homework, friends, hobbies, and other responsibilities, all sorts of different forces around you are constantly in action. Your schedule probably feels balanced when you have just the right amount of time for each task, and forces are in a state of equilibrium when balanced to a net force of zero. On the other hand, your schedule probably feels hectic and unbalanced if you don’t have enough hours in a day to get everything done, causing you to fall behind on assignments and your schedule to fall out of place. Unbalanced torque is the same—if a force isn’t balanced to equilibrium, we’ll witness changes in motion as a result.
Torque is the twisting and rotating force of levers, spinning fans, and orbiting planets. When torque is unbalanced, we observe changes in rotational motion. In this article, we’ll review what torque is, review the equations we use for unbalanced torque calculations, discuss common problem types, and walk through some examples of unbalanced torque.
What Is Torque?
Let’s start off our discussion of unbalanced torque by reviewing the definition of torque. Recall that torque is a force that causes some type of rotational motion, such as a twisting or balancing motion.
Torque is a force applied at some distance away from a pivot point, resulting in a change in the rotational motion of an object.
Torque is a vector quantity measured in units of newton meters, represented by the symbol \(\mathrm{N\,m}\), or \(\mathrm{\frac{kg\,m^2}{s^2}}\). You regularly witness, experience, and apply torque in daily life. Each time you use a faucet handle, doorknob, or steering wheel, you’re applying force at a distance that causes a change in motion!
Pushing on the door from the location of the doorknob applies torque, rotating the door about its hinges, Wikimedia Commons CC BY-SA 3.0
Let’s look at how torque causes the familiar motion of pushing open a door. A door pivots about its hinges, the small bit of hardware that attaches the door to a frame and confines its range of motion. When you twist on a doorknob, you unlatch the door from its frame, which then allows you to apply a pushing force. The distance between the hinges and the point you apply that force is called the lever arm. The result of your push is an applied torque about the hinges perpendicular to the force applied with your push, and the door swings open.
A lever arm is the perpendicular distance (right angle) from the axis of rotation or fulcrum to the location of an applied force.
This distance is usually expressed using the variables \(r\) or \(d\). The placement of doorknobs on the side opposite the pivot points is no mistake—increasing the distance between the pivot point and the location we apply force reduces the total amount of force required to turn the door. That’s torque in action! Try pushing a door open at different distances from the hinges to see this for yourself.
Balanced and Unbalanced Torque Definitions
Now that we’ve brushed up on our understanding of torque, let’s define two more related concepts: balanced and unbalanced torque.
Balanced torque occurs when the torques and forces on either side of a pivot point in a system are equal, resulting in the static equilibrium of the system, either at rest or with zero rotational acceleration.
Mathematically, we can write balanced torque as:
\begin{align*} \Sigma \tau=\tau_{\mathrm{net}}=0\end{align*}
where \(\tau_{\mathrm{net}}\) is the net torque, expressed with the lowercase Greek letter tau.
In everyday life, balanced torque can look like an unmoving shelf mounted to a wall, supported by a central bracket. Placing similarly weighted objects on either side of the central support will keep the shelf in static equilibrium with balanced torques.
Let’s contrast this with unbalanced torque, which we define as follows:
Unbalanced torque occurs when any torques and forces in a system do not cancel out, resulting in rotational acceleration in the direction of the net torque.
Mathematically, we can write unbalanced torque as:
\begin{align*} \Sigma \tau &\neq 0 \\ \tau_{\mathrm{net}}&\neq 0 \end{align*}
Building off our shelf example, we can create unbalanced torques by stacking a bunch of objects on the left end of the shelf while leaving the right end empty. If the length of the shelf is \(L\), then the weight of the stacked objects due to gravity applies a net torque at a distance of \(\frac{L}{2}\) away from the center. The unbalanced torque causes a rotational acceleration and the shelf will pivot—assuming the shelf could pivot about its support and not simply break off the wall, of course.
Equations for Unbalanced Torque
You’ve already seen most equations for torque—the torque is unbalanced when the net torque \(\tau_{\mathrm{net}}\) is nonzero in a calculation. Let’s start off by recalling the equation for torque written as a product of the applied force \(F\), the radial distance between the pivot and the point of applied force \(r\), and the angle \(\theta\) between the two:
\begin{align*} \tau=r_{\perp} F=rF\mathrm{sin\theta} \end{align*}
We also have the equation to solve for torque with the moment of inertia \(I\) and the rotational acceleration \(\alpha\):
\begin{align*} \tau=I \alpha \end{align*}
where the units for the moment of inertia are in \(\mathrm{kg\,m^2}\) and the units for rotational acceleration are in \(\mathrm{\frac{rad}{s^2}}\). Because the torque depends on the moment of inertia, we cannot ignore the shape of an object. We can use both of these equations to calculate the value of the applied torque in an unbalanced system. But remember, this is true for the net torque, so make sure to pay attention to what you are calculating throughout a problem! We have one final set of equations we can use for torque calculations:
\begin{align*} \tau_1&=\tau_2 \\ F_1r_2&=F_2r_2 \end{align*}
and, if the force in question is the gravitational force:
\begin{align*} m_1gr_1&=m_2gr_2 \\ m_1r_1&=m_2r_2 \end{align*}
These equations are useful for finding out what mass or radial distance is required to balance a system with unbalanced torques.
Rotation Caused by Unbalanced Torque
Now that we’ve looked at the equations behind unbalanced torque problems, let’s go into more detail about what this looks like conceptually. Remember the equation we previously introduced, \(\tau=I \alpha\)? The moment of inertia, \(I\), is a measure of the resistance to changes in the angular velocity of an object. This is an important property determined by the mass and the distribution of mass from the object’s axis of rotation.
If we hold the moment of inertia constant, meaning the quantity will remain unchanged, then we have the following relationship between torque and angular acceleration:
\begin{align*} \tau \propto \alpha \end{align*}
or in words, torque is proportional to angular acceleration. This means if the torque increases, the angular acceleration will as well. This relationship also makes clear that if the net torque is zero, there cannot be any angular acceleration!
The right-hand rule shows the direction of the torque vector. Pointing your index and middle fingers in the direction of the radius and the applied force while holding your thumb out will show the direction of the torque, Wikimedia Commons CC BY-SA 3.0
So, when we have a nonzero torque in a system, this unbalanced twisting force will result in a change of rotational acceleration, causing the system to spin about its axis of rotation in the direction of the net torque vector. The direction of the torque vector is perpendicular to the force and the radius \(r\) from the axis. If you’re uncertain about the direction of the torque, use the right-hand rule, a handy trick to verify the direction of a perpendicular vector:
Point your index finger in the direction of the radial distance.
Point your middle finger in the direction of the applied force.
Hold your thumb straight out. Your thumb will point in the direction of the torque vector.
If the applied force is rotational about a central axis, instead curl your fingers from the direction of the radial distance towards the applied force. This will indicate a clockwise or counterclockwise motion. Holding your thumb outwards will point in the direction of the torque.
You’ll be encountering the right-hand rule often throughout your physics studies, so make a note of this trick now, even if you don’t memorize it yet! This is an important tool for cross products, which you’ll learn more about in calculus and more advanced physics courses.
Unbalanced Torque Calculations and Problem Types
You’ll likely be encountering some unbalanced torque problems during your physics studies. Let’s go through an overview of some problem types you might see and the steps you’ll need to take to tackle these torque calculations.
For any problem, you’ll want to make sure to always define what your system is and consider all the horizontal and vertical components of any forces involved. Torque can be positive or negative, too, so watch your signs!
Balancing a Seesaw or Lever in Non-equilibrium
A classic physics problem with unbalanced torque involves balancing a seesaw, or similar setup with a plank on top of a fulcrum. You might be asked to balance the seesaw to rest in rotational equilibrium parallel to the ground, based on a set of initial conditions given.
Let’s think through the steps we might need to take to approach this problem type.
What information do we have? You’ll likely be given three values, with one mass or radial distance unknown.
What quantity are we trying to solve for? Determine what the missing quantity is, and remember that the goal is to balance the torque on both sides of the seesaw. Draw a diagram if you need to.
Which equation should we use? This depends on the information initially given. In many cases, this will be \(m_1r_1=m_2r_2\), but pay attention to what variables you’re starting with!
Isolate the unknown variable and solve.
Balancing Torque within a Mobile
Balancing a mobile is a problem type similar to balancing a seesaw, but often with multiple tiers of dangling masses, each tier being its own lever. In a mobile problem, the number of objects hanging from each rod might not be the same. These types of torque calculations are essentially an extension of the simple seesaw scenario.
Creating a mobile involves balancing hanging masses on different lever arms, Daniel X. O’Neil via Flickr CC BY 2.0
How might we solve this type of problem? Again, let’s consider the different components and what we need to do to balance the torques.
Draw or examine a diagram of the mobile. In these problems, you’ll always be given the lengths of the rods as well as the lever arms (the distance from the string to each hanging object). The lever arms will not always be equal to one-half the rod length! You will also be given the weight on one side of the lowest tier to being.
Ignore the weight of each rod and piece of string and consider only the weight of the objects hanging from the rods. Remember that the force in question is gravity, so if we’re given weight to begin with, we’ll want to use the \(F_1r_1=F_2r_2\) form of the torque equation.
Beginning from the lowest hanging tier, apply the torque balancing equation and solve for the missing weight.
Work your way up each tier and calculate the next weight that will keep the mobile in equilibrium. As you work your way up, make sure to use the total weight from the previous tier to find the next level’s value.
Fill in your diagram with each value until you’ve calculated all of the weights. You’ve now balanced the hanging mobile!
Second Law Problems with Torque and Rotational Inertia
One final type of problem you’ll likely encounter is analogous to Newton’s second law. Recall that Newton’s second law of motion states that the sum of the forces is proportional to acceleration and inversely proportional to the mass of an object:
\begin{align*} a=\frac{F_{\mathrm{net}}}{m} \end{align*}
where \(a\) is the linear acceleration. Newton’s second law can easily be extended to form the second law of rotational motion, knowing the sum of the torques is proportional to the angular acceleration and inversely proportional to the moment of inertia:
\begin{align*} a=\frac{\tau_{\mathrm{net}}}{I} \end{align*}
In these types of problems, you’ll likely be working with different information to start with. We can still balance torques, even if we don’t begin knowing masses and radial distances—remember that as long as the torques on either side are set equal to one another:
\begin{align*} \tau_1=\tau_2 \end{align*}
then we can solve for whichever unknown variable remains.
You won’t be balancing a system to static equilibrium in many torque problems. Instead, you’ll be tasked with finding the net torque knowing the object shape and information about the rotational acceleration or some other variation of this torque calculation.
Unbalanced Torque Examples
Let’s finish off our discussion of unbalanced torque by walking through an example, starting with the problem of an unbalanced seesaw.
A child weighing \(\mathrm{30\,kg}\) sits on one end of a \(\mathrm{2.5\,m}\) long seesaw. The seesaw’s fulcrum is placed exactly at the center. At what distance from the pivot should a second child weighing \(\mathrm{42\,kg}\) sit to balance the seesaw parallel to the ground?
We want to solve to make the torques equal on either side of the equation:
\begin{align*} F_1r_1=F_2r_2 \end{align*}
and since we’re given weight, we can plug these numbers in for the gravitational force causing the torques. Remember that, in everyday language, we usually express weight in kilograms or pounds. However, weight due to gravity is actually newtons or pound-force. Another component to think about is the angle; in this scenario, the angle for the seesaw to be parallel to the ground is \(\theta=90^\circ\), and because \(\mathrm{sin(90^\circ)=1}\), the angle doesn’t change our calculation.
The child on one end sits at a distance of \(\mathrm{\frac{2.5\,m}{2}=1.25\,m}\). Putting this all together, we find:
\[\begin{align} r_2 &= \frac{F_1r_1}{F_2} \\ \mathrm{0.89\,m} &= \mathrm{\frac{30\,N\cdot1.25\,m}{42\,N}} \end{align}\]
So, the second child will have to sit much closer to the center of the seesaw in order to reduce the torque caused by their weight and make the system stable.
Let’s go through another example, this time calculating the net torque for a rotating object with nonzero acceleration.
A solid sphere with mass \(m=3.0\,\mathrm{kg}\) and radius \(r=0.50\,\mathrm{m}\) has a moment of inertia of \(I=\frac{2}{5}mr^2\). If the net torque on the sphere is \(50\,\mathrm{N\,m}\), what is the angular acceleration?
For this problem, we’ll want to use the following torque equation:
\begin{align*} \tau_{\mathrm{net}} &=I\alpha \\ \mathrm{\frac{\tau_{net}}{I}}&=\alpha \end{align*}
Plugging in our known values, we find:
\begin{align} \alpha &= \mathrm{\frac{50\,\frac{kg\,m^2}{s^2}}{\frac{2}{5}(3\,kg\cdot(0.5\,m)^2)}} \\ &= \mathrm{83\,\frac{rad}{s^2}} \end{align}
Remember that radians are dimensionless units, so our units cancel to \(\mathrm{\frac{1}{s^2}}\).)
Let’s do one more quick example by looking at how the shape of an object impacts rotational acceleration and torque.
We release three objects from rest at the top of a hill: a hollow sphere (thin spherical shell), a thin cylindrical shell with negligible thickness, and a solid cylinder. The moments of inertia for these objects are \(I_{\mathrm{sph.\, shell}}=\frac{2}{3}mr^2\), \(I_{\mathrm{cyl.\, shell}}=mr^2\), and \(I_{\mathrm{solid\, cyl.}}=\frac{1}{2}mr^2\). Which object will reach the bottom first? Which object will be last?
For this problem, we want to compare the relative moments of inertia. Remember the relationships between angular acceleration, the moment of inertia, and torque:
\begin{align*} \tau &\propto \alpha \\ \alpha &\propto \frac{1}{I} \end{align*}
Knowing this, we expect the object with the smallest moment of inertia to reach the bottom of the hill the fastest. Putting this together, we find:
\begin{align*} I_{\mathrm{solid\, cyl.}} &\geq I_{\mathrm{sph.\, shell}} \geq I_{\mathrm{cyl.\, shell}} \\ \alpha_{\mathrm{solid\, cyl.}} &\geq \alpha_{\mathrm{sph.\, shell}} \geq \alpha_{\mathrm{cyl.\, shell}} \end{align*}
Since the solid cylinder has the most mass closest to its axis of rotation, the solid cylinder will accelerate the fastest and reach the bottom first. The cylindrical shell has its entire mass at a distance \(r\) away from its axis of rotation and has the greatest moment of inertia.
When calculating torque and solving static equilibrium problems, keep in mind that the torque is calculated at a specific pivot point. In static equilibrium conditions, the net torque will always be zero about all pivot points, which means we have the freedom to choose a pivot point where the calculations are simplest, like where some torques are already canceled out!
Unbalanced Torque - Key takeaways
- Torque is a vector quantity measuring a twisting applied force to a body, resulting in a change in rotational motion.
- Balanced torque occurs when any acting forces on either side of a fulcrum or point of rotation are equal, resulting in no change in motion.
- Unbalanced torque occurs when any applied forces don’t balance out to equilibrium and have a net force in one direction, resulting in changes in rotational acceleration.
- Some common types of problems involving unbalanced torque include balancing a seesaw, balancing a multi-tiered hanging mobile, and other applications of Newton’s second law for rotational motion.
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