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Torsional Pendulum

If you've ever been on a tire swing, you'll know that you don't only go back and forth on it, but it can also rotate for the full motion sickness experience. But there is something peculiar about the rotational motion. Sometimes the tire seems to stop turning and start rotating in the other direction: it feels like you are essentially also going back and forth in the rotational sense of the tire's motion! This feeling is absolutely correct, as this is an example of a torsional pendulum. Read on to find out more about torsional pendulums.

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Torsional Pendulum

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If you've ever been on a tire swing, you'll know that you don't only go back and forth on it, but it can also rotate for the full motion sickness experience. But there is something peculiar about the rotational motion. Sometimes the tire seems to stop turning and start rotating in the other direction: it feels like you are essentially also going back and forth in the rotational sense of the tire's motion! This feeling is absolutely correct, as this is an example of a torsional pendulum. Read on to find out more about torsional pendulums.

Torsional Pendulum Tire swing StudySmarterFig. 1 - A tire swing is a classic example of a torsional pendulum.

Torsional Pendulum Definition

A torsional pendulum is much like a regular pendulum except that it exhibits rotational motion instead of translational motion.

A torsional pendulum is a bob hanging off a string such that it can rotate freely.

See also the illustration below. To set a torsional pendulum in motion, you need to rotate the bob somewhat and let it loose. The bob will start to rotate back to its equilibrium position, after which it will 'overshoot' and rotate out of its equilibrium position in the other direction than you had rotated it originally. Its rotation will be slowed down again, and the bob will keep rotating in this oscillatory way.

Torsional Pendulum Theory

The theory behind the workings of a torsional pendulum all boils down to the fact that our string has some stiffness. That is, if we introduce a twist in our string, then it will have the tendency to un-twist if we let everything go and let the string do its thing. In fact, the string has the tendency to un-twist harder if we twist it harder, and this relationship is exactly linear.

The string in a torsional pendulum exerts a torque that is proportional to the angle through which the lower end of the string is turned while the upper end of the string remained fixed.

Now we can also see the resemblance with a regular pendulum: the rotational equivalent of force (torque) is proportional to the rotational equivalent of displacement (angle). Thus, we could question whether or not it would exhibit simple harmonic motion. Let's find out if that's the case by setting up some equations and solving them.

Torsional Pendulum Equations

To speak of a torsional pendulum, we assume that the torque exerted by the string on the bob is proportional to the angular displacement of the bob. This means that the string has a certain torsion constant, \(c,\) which is equal to the torque exerted by the string divided by the angular displacement of the bob. As you might have been able to derive from its definition, the torsion constant \(c\) is measured in \(\mathrm{\tfrac{N\,m}{rad}}\). The stiffer the string is, the higher the torsion constant is. We can now set up the following equation:

\[\vec{\tau}=-c\vec{\theta},\]

where \(\vec{\tau}\) is the torque exerted by the string and \(\vec{\theta}\) is the angular displacement of the bob. The minus sign is there to indicate that the torque exerted by the string is in the direction opposite to the angular displacement of the bob: after all, the string wants to un-twist itself. Therefore, we can say that this is a restoring torque in the same sense, that the force in a classical pendulum is a restoring force.

You may remember Newton's second law in angular form, which implies that the torque exerted on our bob is equal to the bob's moment of inertia \(I\) times its angular acceleration, \( \alpha \), so we have the following differential equation:

\begin{aligned}I\textcolor{#00b695}{\alpha} & = \textcolor{#56369f}{\tau}\\[8pt]I\textcolor{#00b695}{\frac{\mathrm{d}^2\vec{\theta}}{\mathrm{d}t^2}} &=\textcolor{#56369f}{-c\vec{\theta}},\end{aligned}

where \(t\) is time.

As our bob spins back and forth, the sense of the bob's angular displacement vector and of torque changes. Using the right-hand thumb rule, we can see that angular displacement's direction is vertically up as it twists to the right, and vertically down as it untwists (turns to the left). This tells us the direction of torque because we know it acts in the opposite direction as \( \vec{tau} = -c\vec{\theta}. \) Nevertheless, in both cases torque it is oriented vertically. Therefore, we are guaranteed that the motion occurs in the same horizontal plane. Since this is the case, we can leave aside the vector notation which is complicated and unnecessary for this case. Instead, we can continue by using a scalar differential equation where the algebraic sign keeps all the information needed to describe the pendulum motion.

\[\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}=-\frac{c}{I}\theta.\]

This looks suspiciously much like the differential equation governing the simple harmonic motion of the regular pendulum, and indeed, the solution to this equation describes simple harmonic motion:

\[\theta(t)=\theta_\text{max}\sin\left(\sqrt{\frac{c}{I}}t\right),\]

where \(\theta_\text{max}\) is the angular amplitude or the maximum angular displacement of the bob.

From this equation, we see that the angular frequency \(\omega\) is given by

\[\omega=\sqrt{\frac{c}{I}}.\]

The angular frequency of the simple harmonic motion of the torsional pendulum should not be confused with the angular speed of the bob itself! The latter is equal to \(\tfrac{\mathrm{d}\theta}{\mathrm{d}t}\) and changes with time, for example, it is zero when the bob twist to the maximum angular displacement before changing direction to start untwisting. On the other hand, angular frequency is constant and tells how fast the argument of the sine function changes. It is directly related to the frequency and the period of the oscillations.

The period \(T\) of the oscillatory motion is given by

\begin{aligned}&T=\frac{2\pi}{\omega}\\[6pt]&\boxed{T=2\pi\sqrt{\frac{I}{c}}}.\end{aligned}

We can always do a sanity check and see if the units work out in complicated expressions. The moment of inertia is measured in \(\mathrm{kg\,m^2}\) and the torsion constant is measured in \(\mathrm{\tfrac{N\,m}{rad}}\), so the right-hand side of the equation for the period is measured in

\[\sqrt{\frac{\mathrm{kg\,m^2\,rad}}{\mathrm{N\,m}}}=\sqrt{\mathrm{\frac{kg\,m}{N}}}=\sqrt{\mathrm{\frac{kg\,m}{\tfrac{kg\,m}{s^2}}}}=\mathrm{s}.\]

Energy in a Torsional Pendulum

We can find an expression for the potential energy stored in a torsional pendulum. We can define the potential energy as minus the work done by the string on the bob from its equilibrium position to its current position. The work \(W\) done by a torque \(\vec{\tau}\) is equal to \(\int\vec{\tau}\cdot\mathrm{d}\vec{\theta}\) so the potential energy \(E_\text{pot}\) in the torsional pendulum can be expressed as follows:

\begin{align*}&E_\text{pot}=-W\\&E_\text{pot}=-\int\vec{\tau}\cdot\mathrm{d}\vec{\theta}\\&E_\text{pot}=-\int(-c\vec{\theta})\cdot\mathrm{d}\vec{\theta}\\&\boxed{E_\text{pot}=\frac{1}{2}c\theta^2}.\end{align*}

Note that the potential energy of the torsional pendulum does not depend on the moment of inertia of the bob at all! It only depends on the torsion constant, \(c\) (the stiffness of the string), and the angular displacement of the bob, \(\theta\) (how twisted the string is). In short, the potential energy only depends on the string and not on the bob.

Torsional Pendulum Experiment

You can do an experiment with a torsional pendulum yourself if you have two things at your disposal: a tire swing and a windless day.

To actually have the torque exerted by the rope be proportional to the angular displacement of the tire, we should use an initial angular amplitude not larger than \(270^\circ\). That means that you should not rotate the tire more than 3/4 of a full turn from its equilibrium position. Now let go of the tire and measure the oscillation period with a stopwatch. You can best do this by measuring the time it takes for the tire to go from a stationary point to the next stationary point: this will of course be half a period.

Did you manage to measure the period of this simple harmonic motion? Good! Now, if we assume a typical value of a car tire's moment of inertia of \(I_\text{tire}=0.4\,\mathrm{kg\,m^2}\), you can calculate the torsion constant of the rope using the formula for the period of motion! If you manipulate this equation correctly, you should get

\[c=I\left(\frac{2\pi}{T}\right)^2.\]

Torsional Pendulum Clock

Torsional Pendulum Anniversary clock StudySmarterFig. 3 - An anniversary clock is another typical example of a torsional pendulum.

The torsional pendulum clock is an application of the torsional pendulum and was invented around 50 years after the invention of the torsional pendulum. Such a clock is also called an "anniversary clock" or "400-day clock" because it can last a year on a single winding. This is possible because they are much more efficient than traditional regular pendulum clocks: there is less mechanical friction at the pivot but also less air resistance at the bob. The flip side of these clocks is that the string's torsion constant is very dependent on environmental factors such as temperature, so their timekeeping is less accurate than traditional regular pendulum clocks.

Torsional Pendulum Example Calculation

Let's finish by looking at some examples of what calculations involving torsional pendulums can look like.

Suppose we have a torsional pendulum clock that has a period of \(T=10.0\,\mathrm{s}\) and a string with a torsion constant of \(c=4\times 10^{-4}\,\mathrm{\frac{N\,m}{rad}}\). If the bob consists of three equal masses that are at a distance of \(d=4\,\mathrm{cm}\) from the string, what is the mass \(m\) of one of those masses?

Solution

We know the period and the torsion constant, and we know that if we plug those into our equation for the period, we can find out the moment of inertia, which is related to the mass of the system. So, the first step is to solve for \( I \):

\begin{align*}T&=2\pi\sqrt{\frac{I}{c}}\\[6pt]\implies I&=\left(\frac{T}{2\pi}\right)^2c\\[6pt]&=\left(\frac{10.0\,\mathrm{s}}{2\pi}\right)^2\times 4\times 10^{-4}\,\mathrm{\frac{N\,m}{rad}}\\[6pt]&=0.001\,\mathrm{kg\,m^2}.\end{align*}

Recall that we can express the moment of inertia in terms of the masses

\begin{aligned} I& = \sum_i m_i (d_i)^2\\[6pt]I&=3md^2,\end{aligned}

so the mass value of each of these masses is given by

\[m=\frac{I}{3d^2}=\frac{0.001\,\mathrm{kg\,m^2}}{3\times (0.4\,\mathrm{m})^2}=0.2\,\mathrm{kg}.\]

Therefore, each of them has a mass value of \(m=0.2\,\mathrm{kg}\).

Torsional Pendulum - Key takeaways

  • A torsional pendulum is a bob hanging off a string such that it can rotate freely.
  • The string in a torsional pendulum exerts a torque that is proportional to the angle through which the lower end of the string is turned while the upper end of the string remained fixed. The proportionality constant is the torsion constant \(c\):\[\vec{\tau}=-c\vec{\theta}.\]
  • Becuase the torsion is proportional to the angular displacement and in opposite direction, it acts as a resoting torsion which causes the system to exhibit harmonic motion.
  • A torsional pendulum exhibits simple harmonic motion, and we can describe it with the equation \[\theta(t)=\theta_\text{max}\sin\left(\sqrt{\frac{c}{I}}t\right).\]
  • The associated angular frequency is \(\omega=\sqrt{\frac{c}{I}}\) and the period is \(T=2\pi\sqrt{\frac{I}{c}}\).
  • The potential energy of a torsional pendulum is stored in the twist of the string and is given by \(E_\text{pot}=\frac{1}{2}c\theta^2\).
  • Examples of torsional pendulums are a tire swing and a torsional pendulum clock.

References

  1. Fig. 1 - Tire Swing, Mulberry Street, Over-the-Rhine, Cincinnati, OH (47588664081) (https://commons.wikimedia.org/wiki/File:Tire_Swing,_Mulberry_Street,_Over-the-Rhine,_Cincinnati,_OH_(47588664081).jpg) by Warren LeMay (https://www.flickr.com/people/59081381@N03) licensed by Public Domain.
  2. Fig. 2 - Illustration of a torsional pendulum, StudySmarter Originals.
  3. Fig. 3 - Right hand rule applied to torsion pendulum, StudySmarter Originals.
  4. Fig. 4 - Haller torsion pendulum anniversary clock (https://commons.wikimedia.org/wiki/File:Haller_torsion_pendulum_anniversary_clock.jpg) by Graham Evans, licensed by CC BY-SA 2.0 (https://creativecommons.org/licenses/by-sa/2.0/deed.en).

Frequently Asked Questions about Torsional Pendulum

A torsional pendulum exhibits simple harmonic motion with an angular frequency, ω = (c/)^(1/2), where c is the torsion constant and I is the moment of inertia of the bob.

Two uses of a torsional pendulum are timekeeping for example in an anniversary clock and generating oscillatory motion for a mechanical ride, for example in a tire swing.

An example of a torsional pendulum is an anniversary clock.

The equation of motion for a torsional pendulum is τ = -, where τ is the torque exerted by the string, c is the torsion constant, and θ is the angular displacement of the bob.

A torsional pendulum is a bob hanging off a string such that it can rotate freely. It exhibits simple harmonic motion with a period that depends on the string's torsion constant and the bob's moment of inertia.

Test your knowledge with multiple choice flashcards

A tire swing can be used as ___.

Which clock keeps time more accurately?

The torque that the string exerts on the bob ___ the angular displacement of the bob.

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