Physical principles -- torque 2

Kinetics and Kinematics

Newton's three laws

When it comes to kinetics -- relating torque to motion -- Newton's same three laws exist. Torques substitute for forces, and we will see other rotational units substitute for the linear units we have seen earlier. By the way, "rotational" and "angular" usually mean the same thing when we talk about dynamics.

Newton's first law of angular motion: A body in rotation stays in rotation at the same speed and direction (same angular velocity) unless acted upon by some torque outside the object itself.

Exactly the same as for linear motion, except now we are talking about rotation around an axis.

Newton's second law of angular motion: If a torque is applied to a body, the body accelerates according to the relationship Τ=Iα

Similar to linear motion, except:

Obviously, we substitute torque for force.
Instead of mass, we have I, moment of inertia, the measure of rotational resistance to motion. Clubmakers to talk about "MOI", but physicists and engineers have called it I for centuries. We will discuss more below about what moment of inertia is.
Instead of simple acceleration, we have α, angular acceleration.

Newton's third law of angular motion: For every torque applied by one body to another, there is an equal and opposite reaction torque.

When you exert a torque on something, it exerts an equal and opposite torque back on you. We have seen how this works for forces, and it works the same for torque. Here's a similar pair of videos for torque.

In this video, I am waving a pair of weights left and right with my upper body at arm's length. The upper body is rotating back and forth to make it happen. That implies torque. Where is the torque coming from? The oblique muscles in the middle of the body are applying the torque, coming from a stationary lower body. The hips are providing a stable linkage from the obliques to the upper body. Continuing down the chain, the legs are applying force to the ground in order to keep the hips in place as they torque the upper body back and forth. And of course the ground ain't goin' nowhere.

But look what happens when the lower body is not held in place. I'll answer that question by sitting on a swivel stool while I wave the same pair of weights. If the lower body is not actively held stationary (as it was above), then it will twist back and forth, exactly opposite to the upper body. That is the equal and opposite torque in action! As the obliques drive the upper body from below, they also drive the lower body from above. Equal and opposite! And I can't accurately control what my arms are doing with the weights because my lower body is twisting back and forth without any ground reaction forces to stabilize it.

This hints at what will happen if you try to swing a golf club without any ground reaction forces. But there are videos around that show it pretty graphically.

Here's a video of a golfer trying to hit a ball standing on a frozen lake. Note how the feet spin in the opposite direction from the golf swing. Not a pretty picture.
There is also a rather well known video of golf coach, biomechanist, and TV commentator Chris Como diving into a pool from a high platform while trying to swing a golf club. (The video was created by my friend Russ Ryden.).It shows how the unsupported lower body twists in the opposite direction from the golf club and the upper body.

Rotation and angular units

Rotation is a change of angle over time, so we should expect rotational velocity and acceleration to use units we associate with angles. We are all familiar with measuring angles in degrees. There are 360° (three hundred sixty degrees) in a circle, and 90° (a quarter of that) in a right angle (a quadrant of a circle).

But there are other not uncommon ways of expressing the size of an angle.

The most familiar one for most of us is the revolution. You may not think of this as a measure of angle, but it is. One revolution is 360°, and we usually use revolutions when describing angular velocity. That unit is RPM, or revolutions per minute. But if we can use it for velocity, we can also use it for position or acceleration.
There is a unit of rotation called the radian. It is derived from trigonometry, and its size is such that 2π radians is a complete circle, same as 360°. When you do the math, one radian is approximately equal to 57.3 degrees.
There are others that do not concern us in golf, so I won't discuss them further. For instance, the military describes angles in "mils", because that makes arithmetic easier when aiming artillery.

Newton's laws of angular motion, and other formulae derived from them, require angles to be measured in radians in order to work. (At least if you are not going to use unit conversion factors, which tends to be tricky if you don't understand what you are doing.)

So now we know how to express an angle -- a rotational position. Rotational kinematics and kinetics also requires expressing angular vlocity (position change per unit time) and angular acceleration (angular velocity per unit time). Here are some frequently encountered examples:

Angular Position	Angular Velocity	Angular Acceleration
radians (This is fundamental for basic physical formuas without having to do unit conversion)	Radians per second	Radians per second per second or Radians per second squared
Degrees	Degrees per second	Degrees per second per second or Degrees per second squared
Revolutions	RPM or Revolutions per minute	RPM per second

I sometimes encounter the reaction, "That's just plain silly," if I talk about, say, a clubface rate of closure of 2400 degrees per second at impact. "2400 degrees is more than 6 full revolutions. The clubhead isn't going to do 6 complete revolutions in a swing. That's ridiculous! Besides, the swing doesn't last a full second."

If that is your reaction, you need to get comfortable with the notion of an instantaneous rate. That might be 2400 degrees per second at the moment of impact, but it is much lower than that even ten milliseconds before impact or after impact. The unit doesn't require the quantity to be the same for a full second.

In order to get a better feel for it, let's think about a much more familiar unit: linear velocity in miles per hour (mph). You are driving on a road with a posted speed limit of 65 miles per hour. You don't say, "That's ridiculous! I'm not going 65 miles on this trip, and I'm certainly not holding this speed for an hour." You are comfortable with MPH as a description of your speed right now -- your instantaneous speed -- and you don't feel it has to be maintained for an hour nor for 65 miles in order to apply.

Now apply that same reasoning to units like degrees per second. What 2800 degrees per second means is:
That is the speed it is going "right now", this very instant. If it kept going at that speed for a full second, then it would indeed rotate more than 6 full revolutions. But it doesn't have to in order for its rate right now to be 2800 degrees per second.

Relating angular to linear motion

Just as angular kinetics (force & torque) can be related to linear kinetics, we can relate angular kinematics (motion) to linear kinematics.
Remember Τ=Fr? (Torque, force, and radius)
Well there is a similar relationship for motion: V=ωr. (Linear velocity, angular velocity, and radius)
We have to be careful about the units -- they must be compatible, and the angular velocity must be in radians per second. From time to time, we will encounter examples where we have to convert to radians per second from, for instance, RPM.

Let's try an example. I am riding my old road bike; it has 27" tires. I am pedaling at 120rpm, and my 42:14 sprockets (a high gear) gives me a wheel speed rotation of 360rpm. How fast am I going in miles per hour?

First let's choose an approach to the problem. The bike wheel is spinning about the hub at Xmph. If the bike is placed -- with its spinning wheel -- on the ground, the point on the wheel where it touches the ground isn't moving -- but it is still moving at Xmph with respect to the bike itself. That means the bike itself must be going at Xmph. So all we have to do is find out the speed the wheel is spinning around the hub, and we have the answer. Let's start...
The radius r is 27/2=13.5 inches. (The green measurement is diameter. We need to halve it to get the radius.) That is 1.125 feet.
The angular velocity has to be converted from RPM to radians per second. My favorite tool for unit conversions like this is the free Windows application calc98. It tells me that 360rpm is 37.7rad/s.
Now just apply V=ωr.
1.125*37.7 = 42.4 ft/sec = 28.9mph.

That is a sane answer. I know from my bicycle experience that a pedal cadence of 120 is typical for a road racer on level ground, I am in a high-speed gear, and road racers go about 25-30mph when they are not attacking. And by now, you should realize that a lot of these principles are applicable to other sports as well. It's physics, it isn't just golf.

Moment of inertia

The "moment of inertia" functions in rotational kinetics exactly as mass does in linear kinetics. For a given acceleration, you need either:

more force (linear) or more torque (rotational), or...
less mass (linear) or less moment of inertia (rotational).

Intuitively, it would seem that if you add mass to an object, that ought to increase its moment of inertia. And that is true. But how much does it increase it? That depends on the distance from the axis of rotation. Wait! Does that mean that MOI depends on the axis? If you choose a different axis, does that change the object's MOI? The answer is a resounding "yes!"

Before you can talk about the moment of inertia of an object, you have to specify MOI about which axis. The moment of inertia is calculated based on how far from the axis each particle of mass is located.

In particular, the MOI is calculated by adding together each tiny chunk of mass times the square of its distance from the axis. In the diagram, we have isolated a little chunk of mass m₄₃. It lies at a distance (a radius) r₄₃ from the axis of rotation. (We are taking the moment of inertia from that axis.) Mass m₄₃ is so small in size that its size is negligible compared to the radius. So if each tiny chunk of mass m_i is accounted for, the formula for moment of inertia I is:

I =

∑
i

(m_i * r_i²)

For those readers who are comfortable with calculus, it isn't hard to see where that equation goes if you shrink the tiny chunks of mass down to infinitesimal chunks of size dm.

I = ∫ r² dm

I show only one integration, but use as many as you need to account for every particle of mass.

Why r squared?

OK, so each chunk of mass is more important the farther it is from the axis; that's why r is a factor. But why is the radius squared?
This is one of those questions that you don't actually need to know the answer to, but moment of inertia might be a little more intuitive to you if you knew it. The math itself is pretty easy, so hang with it if you can.

Remember that moment of inertia is the measure of a body's resistance to turning when a torque is applied. It is analogous to mass, the resistance of a body to moving when a force is applied. Actually, the linear formulation is more fundamental; we are going to explore how angular motion is a combination of a coordinated linear motion of a lot of chunks of mass.

We know that r² means we multiply by r twice. It turns out that each time has a different rationale, and let's look at both. Each comes directly from Newton's second law, F=ma, from linear kinetics.

In the diagram, the gray rod is rotating through an angle A. I have marked two particles of mass, one blue and one red. The red one is twice as far from the axis as the blue one. In order for the object (the gray rod) to rotate through angle A, the red particle has to go twice as far as the blue particle. In general, the distance the particle has to go will be proportional to r, its distance from the axis.

Let's see how each particle contributes to the total moment of inertia. We know that T=αI so clearly

I = T/α

We have just seen on the diagram that linear acceleration a is proportional to the radius r. In fact, if we express angular acceleration in radians per second, a=αr so

α = a/r

Plugging this back into our expression above for I:

I =

a/r

So that's the first r.

But that linear acceleration a requires a force. Where does this force come from? It has to come from the torque T.

Let's look again at the wrench we originally used to relate force to torque. But this time, instead of the force producing the torque, we'll apply a torque at the axis (the crossed centerlines) and see what forces the torque can produce.

Remember, torque is force times a moment arm, and the moment arm is r. The red force has twice the moment arm of the blue force. Since torque is driving things, a shorter moment arm means a bigger force. So the blue force produced by the driving torque is twice as large as the red force produced by the same torque. In general, the force is inversely proportional to r. That's a strong hint that we need proportionally more torque for a given force as r gets larger. Let's do the math.

Our basic equation for torque is:

T = Fr

Plugging this back into what we found above as an equation for I:

I =

(Fr) r

Fr²

But we also know -- very basic -- that F=ma. We can express that as m=F/a. And we already have an F/a in our latest expression for I. So:

I =

F r²

r² = m r²

And there you have it!

Example: Bicycle wheel

Let's try a simple example of computing moment of inertia. Here is a bicycle racing wheel. We want to find its moment of inertia around its axle, the way it normally rotates in use. There are three major subassemblies that contribute to I, The rim, the spokes, and the hub. The moment of inertia of the whole wheel is the sum of those three components. Let's analyze each in turn.

I have chosen a 700C road touring bike wheel, because data is relatively easy to come by. I'm using components from an old Gene Portuesi catalog. Let's look at each component in turn:

Rim
The rim is a thin ring of mass 400g and a diameter of 700mm. We will do our computation in grams and centimeters, so that's 70cm diameter or 35cm radius (R in the diagram).

Suppose we take the "tiny chunk of mass" to be a cross-section through the rim. Then every chunk has the same R, 35cm. When we add together all the Rm² fragments, we get:

I_rim = 400g * 35cm * 35cm = 490,000 g cm²

Spokes
A typical bike will have 36 spokes, each of roughly 6 grams mass. We will make the approximation that they are the length of the radius, 35cm. To compute the I of a spoke, we will cheat a little; instead of doing the calculus here, we'll go to one of the many engineering reference tables. (BTW, this table shows the formula we just used for the rim as well; it is decribed as "thin ring or hollow cylinder".) We are interested in the "thin rod about end"; the spoke ends at the axis in our simplifying approximation. (It is not hard to make a more realistic approximation, but requires physics we are not going to cover -- and it won't matter all that much.) We will compute the I of a single spoke and multiply it by 36, the number of spokes.)

I_spokes = (1/3 * 6g * 35cm * 35cm) * 36 = 88,200 g cm²

Hub
Let's appoximate the hub, obviously contrary to fact, as a simple hollow cylinder with an outside diameter (OD) of 3cm and an inside diameter (ID) of 1cm; that would be radii of 0.5 and 1.5. The formula for a hollow cylinder is in another engineering table. The mass of such a hub is usually something like 300g.

I_hub = 1/2 * 300g ((0.5cm * 0.5cm) + (1.5cm * 1.5cm)) = 375 g cm²

Total
So the total moment of inertia of the bicycle wheel is 490,000+88,200+375=578,575gcm²

There are some lessons to be learned from this exercise:

There are engineering tables available in textbooks and on the web that give simple formulas for the moment of inertia of geometric shapes.
If we can break an object down to an assembly of simple shapes, we can use the formulas to compute the moment of inertia for each piece. We have to be careful we are using the formula for the axis we want. If we get that right, then just add the MOI for each piece to get the MOI of the entire object.
Mass close to the axis has a lot less effect on the total MOI than mass further from the axis (larger R). We can see this very dramatically when we look at the contribution of the hub. It contains almost a third of the mass (300g out of just over 900g), but contributes less than 0.1% to the moment of inertia.That is because its mass averages roughly 1cm from the axis while the rim's mass averages 35cm. That is a factor of about 35, which is greatly amplified because the distances are squared. 35 squared is over 1200, so each gram of mass in the hub contributes only 1/1200 of what a gram of mass in the rim does. This is an important lesson!

Demonstration

Let's reinforce that last lesson with a demonstration.

Here's a pair of videos of two hanging assemblies with the same mass but different moments of inertia. The axis we are talking about is the vertical axis through the center of mass of the assembly. We will nudge each one with a finger, trying to spin it around that axis and see what happens. (I made a casual attempt to use the same impulse with my finger on both assemblies, but I don't think I was very good at it. And I don't think it would be easy to be good at it, without creating a mechanism to start the spin.)

The two assemblies are exactly the same -- same PVC tube, same light gauge nylon line, and same two weight plates -- except for the position of the weight plates. The one on the left has the plates together touching the spin axis; the one on the right has the plates as far apart, and as far from the spin axis, as possible. From what we saw in doing the calculations above, the one with the weight far from the axis should have the higher moment of inertia.

And it does! Look how much slower the assembly on the right spins back and forth. The torque trying to make them spin is the same, but the angular acceleration is much lower because the I is much higher on the right. If you look closely, you can see that the maximum deflection for the first cycle is larger on the left; that would be due to the lower I, but it is also dependent on the impulse from the finger and I was not too careful to control that. The very different length of the cycle is a much better demonstration of the difference in moment of inertia.

Last modified -- Oct 10, 2024

Physical principles for the golf swing

Torque