About this video
This video models a physical pendulum (a nailed bowling pin) using torque rather than forces. Gravity provides a restoring torque -d m g sin(theta); under the small-angle approximation sin(theta) is about theta, so the equation matches the spring system and gives simple harmonic motion with period T = 2 pi sqrt(I / (d m g)). It then recovers the simple pendulum result T = 2 pi sqrt(l/g) and extends the method to a torsion pendulum with T = 2 pi sqrt(I/k).
You'll be able to
- analyze a swinging rigid body using net torque instead of forces.
- explain why gravity gives a restoring torque and why the normal force at the pivot gives none.
- apply the small-angle approximation to get simple harmonic motion and a period.
- specialise the physical-pendulum result to a simple pendulum and a torsion pendulum.
Key words
- physical pendulum
- A rigid body swinging about a pivot, with its mass distributed rather than concentrated at a point.
- restoring torque
- A torque that points back toward the equilibrium orientation, causing oscillation.
- small-angle approximation
- For small angles in radians, sin(theta) is approximately theta, which turns the equation into simple harmonic motion.
- torsion pendulum
- An object hung from a string that twists and untwists, with a restoring torque proportional to the twist angle.
Transcript
- [Instructor] Take a bowling pin, put a nail through the top, and let it oscillate. You get a system called a physical pendulum. In fact, the simple pendulum which you might be familiar with is just a special case of a physical pendulum. In the simple pendulum, we model all the masses being concentrated at a single point.
But of course, for real objects like this bowling pin, the mass is distributed throughout the object, so we can't make that approximation. So in this video, we'll see how to predict the motion of a physical pendulum and derive an expression for the period. All right, so this vertical position is the equilibrium position.
So let's call that pollution as theta equals zero. And now suppose we pull it aside by a small angle and let it go. It'll start oscillating about this equilibrium position, right? So our goal is to predict exactly what that motion looks like and give out an expression for the period.
So where do we start? Well, as always, the best place to start is Newton's second law. So what are the different forces acting on our pin? Well, gravity acts downwards through the center of mass. Here, m is the mass of the bowling pin and g is the gravitational field.
And the nail will exert in normal force on the pin. What is the direction of that force? Well, both the direction and the magnitude of that normal force will depend on the orientation and even the speed of this pin. So it's going to be a little complicated. So I'm just going to draw a random direction over here.
And now to calculate the acceleration, we'll have to decompose these forces. And because the pin is moving along a circle, there will be a tangential acceleration. There will also be a centripetal acceleration, or a radial component of the acceleration. Okay, things are getting pretty complicated.
What do we do? Well, instead of thinking in terms of translational motion and forces, we can think in terms of rotational motion and torques. Since the object rotates about this point, let's calculate the net torque about that point where it's nailed. So how do we calculate the torque?
Well, suppose the center of mass is at a distance D from this point of rotation. Now the torque produced by the gravity is simply the force of gravity multiplied by the perpendicular distance from the pivot, this particular distance. Now, of course, we could also calculate torque by taking the component of the force perpendicular to D and multiplying it by D.
That also works. Both approaches are equal and they yield the same result. But over here, I find this a little bit more intuitive mathematically. The geometry is a little bit more intuitive over here. So let's use this. Let's stick to this. Okay, so for the torque due to gravity, the force is mg, but what is d perpendicular?
Well, if this angle is theta, then sine theta is just d perpendicular divide by d. So that means d perpendicular is d times sine theta. So the torque due to gravity is going to be d times sine theta times the force of gravity. Okay, what about the torque due to the normal force?
Well, it's acting at the axis of rotation. And so the perpendicular (indistinct) is zero, so it doesn't produce any torque. It's kind of like pushing at the hinge of a door. It does not make the door rotate. So if we ignore air resistance and everything, then look, gravity is the only force producing torque about this point, which means this itself is the net torque.
But this is just the magnitude of the torque. What about its direction? Well, notice if we displace our pin this way, the torque tries to rotate it in this, in the opposite direction. Similarly, if we had displaced it to the left, the torque would try to rotate it towards the right in the opposite direction.
So the torque acts in the opposite direction to the angular displacement, and therefore we need to put a negative sine over here. And this makes me really happy because it immediately tells me that the torque is pointing towards the equilibrium position always, which means it is a restoring torque, which means even if we didn't know already, we can predict that this system must oscillate.
But of course, now we need to figure out exactly what kind of oscillation this is. Well, for that, we use Newton's second law in the rotational form. Instead of saying net force equals mass or inertia times acceleration, we say net torque equals rotational inertia times angular acceleration.
So that equals this much. If we rearrange for alpha, we get this. And just like how acceleration is secondary to of position, angular acceleration is the secondary way of angular position. Now, if we solve this differential equation, we will get theta as a function of time, and that'll tell us exactly what kind of oscillation this is.
But I don't have to solve it because the moment I look at this, I say, "Hey, this looks very, very analogous to the differential equation that we sold for the spring system." Of course, over here, the secondary ratio of x was proportional to x. So if the secondary ratio of theta was proportional to theta itself, that would have been awesome because then everything is analogous.
But instead I have a sine theta. Is there some way I can make this theta? Because then everything that we did earlier in our previous video for the spring, same thing would apply over here. And the thing is, we can small angle approximations. If the angle theta is small enough, then sine theta can be approximated to be theta.
Remember, the theta plugged into sine here must be in radians. But how small should this theta be? Well, that depends entirely on the precision that we need. The higher the precision, the smaller the angle should be. But as long as the angle is sufficiently small, sine theta approximates to theta, which means our equation now becomes analogous to the spring system.
And so I don't really have to solve it. I know over here, X changes sinusoidally as a function of time. So under small angle approximations, theta will also change sinusoidally as a function of time. In other words, the physical pendulum undergoes a simple harmonic motion.
And now how do we derive the expression for the period? Well, in the case of the springs, we found that we could represent this coefficient as just the angular frequency squared. And from that, we calculated the period. We can do the same thing over here. So great idea to pause the radio and see if you can find the expression for angular frequency, and from that, calculate the period.
All right, so just like the springs over here, this coefficient can be represented as omega squared. We're just reaping the fruits of all the hard work we've done before. So from this, omega square equals dmg divide by I, and omega equals square root of dmg divided by I.
The period is two pi divided by omega. Substituting, we get the expression for the period. Does this make intuitive sense? Well, let's see. Well, notice that increasing the rotational inertia increases the period. Does that make sense? Yeah, it does. Because if rotational inertia is high, it means it has a higher resistance to being accelerated.
So it will accelerate very slowly, angularly accelerate very slowly, and it'll take a longer time to finish its oscillation and therefore the period would be higher. In contrast, if these increase, then the period decreases. Does that make sense? Yes, because if any of these increase, then the net torque increases, that tends to make the system oscillate faster by putting a higher angular acceleration.
So that tends to make the period smaller. So all of this makes intuitive sense. But of course, remember, rotational inertia is also related to d and m, and that depends on the shape of the pendulum. So the exact expression for the period depends a lot on the shape and how the mass is distributed.
Okay, but remember how a simple pendulum is a special case of this physical pendulum? This means that I should be able to derive an expression for the period of a simple pendulum using this expression, right? Well, yeah, let's see if we can do that. So let's say we have a simple pendulum of length l, and the mass of the bob is m.
Again, great idea to pause the video. Find what d and i will be for this particular case, substitute, and get the expression for the period. Pause and try. All right. Well, we are assuming that all the mass is concentrated over here. So this itself is the center of mass, which means the distance d, which is the distance from here to here, is just l in this particular case.
Okay, and what is rotational inertia in this particular case? Since the point mass is moving in a circle, we can just use the expression for the rotational inertia of a point mass, which in this case would be just ml squared. So if we plug in, the expression for the period becomes, well, this is the rotational inertia divided by, for d, I substitute l, so l times mg.
Well, the mass cancels, one l cancels, and so the expression we end up with is two pi squared of l divided by g, and there we have it, we have derived the expression for the period of a simple pendulum. Now we could stop over here, but just to appreciate how general this method of deriving the expression for the period is, let's consider one more case.
Suppose instead of hanging our bowling ball, oh, sorry, bowling pin from a nail, we suspended using a string like this. Now let's twist it by some angle. As a string twist, it naturally tries to untwist itself and that restoring effect causes the object to oscillate back and forth, right?
This is called a torsion pendulum. But the name isn't really important. What's interesting is, can we use the same logic to derive an expression for its period? The answer is yes. Once again, let's begin by identifying the forces. We have gravity acting downwards and the string pulls upwards, the tension force.
But this time, these two forces balance each other because we know the object isn't accelerating up or down, it's only rotating in the horizontal plane. So these forces are balanced. So where does the torque even come from? Well, it actually comes from the string itself.
If we twist the string by some angle, say delta theta, then the string tries to untwist, and in doing so, it exerts a torque. So that itself is the net torque. It's coming from the string itself. And a key property over here is that this restoring torque is directly proportional to the angle of twist.
That kind of makes sense, right? If you don't twist it itself, then there is no restoring torque. The more you twist, the more the restoring torque. And the negative sine is basically saying the torque is in the opposite direction of the twist. K over here is a constant that depends on the string.
If the k value is higher, that means for a given twist, the restoring torque is pretty high. So in other words, it's harder to twist it. So it's analogous to the stiffness constant of a spring, okay? So once again, let's apply Newton's second law in the rotational form.
But of course, before we do that, let's choose the equilibrium angle as theta equal to zero. Then we can call this angle of twist as just some theta. And so then this will become just negative times k theta. Okay, now if you apply the newton second law, this net torque should equals I, rotational inertia, times the angular acceleration, which is just the secondary ratio of theta.
We can rearrange this and we can say, "Hey, this secondary ratio should equal negative omega square times theta, which means omega squared equals k or I, or omega is equal to the squared of k or I. And since the period is just two pi over omega, substituting, we get the period of our torsional pendulum.
There you have it. Beautiful, right? So the big lesson over here is once we recognize that the system has a restoring torque, we'll see if we can somehow, under some approximation, make it proportional to the displacement or the angular displacement itself. If that is the case, we are done.
We will get simple harmonic motion and we can analyze it very analogous to the way we analyze the springs or the spring system.
Video © Khan Academy, used under CC BY-NC-SA. Watch on Khan Academy. Puzzicle is not affiliated with Khan Academy.