Rotational Dynamics

wheel in wheelIn the previous section of this course we established certain quantities useful in describing rotation. Angular displacement q measured in radians is analogous to linear displacement measured in meters. Angular velocity w measured in radians per second is analogous to linear velocity measured in meters per second. Angular acceleration a measured in radians per second squared is analogous to linear acceleration measured in meters per second squared. Torque t measured in Newton meters is analogous to force measured in Newtons. These angular quantities are not equal to their linear counterparts either numerically or in units but they play similar roles in rotational dynamics.

In linear dynamics, we noticed that the ratio of force to acceleration had a physical significance. It was the mass of an object. In rotational dynamics, the ratio of torque to angular acceleration has a similar significance. It is the moment of inertia of the object. The mass of an object is simple. It depends only of the amount of stuff in the object. The moment of inertia depends not only on how much mass an object has but also the shape of the object and the choice of the axis about which the moment of inertia is measured.

Consider the ratio of torque to angular acceleration for a single particle in circular motion of radius r about the origin of our reference frame. The particle is subject to a radial force which keeps it in the circular path, and a tangential force ft which provides the angular acceleration. The torque resulting from ft is just ft * r since the angle between a tangential force and the radius is 90 degrees so the sine factor in the cross product is 1.

The force ft may be written as the product of the tangential acceleration, at and the mass, m, of the particle according to Newton's second law. But we know already that at is ra, so
t = ft * r = (m*at) * r = (m * ra) * r = (m * r2) * a .

The ratio of t / a then is the quantity (m * r2). This is the moment of inertia, I, of a particle. It plays the role of mass in rotational dynamics.

For objects made up of many particles, the moment of inertia is the sum of all the moments of inertia for the individual particles. Real objects are indeed made up of many particles, so many that treating them individually is a hopeless task. We can imagine the object to be divided up into small volumes which can be treated as particles and then sum the moments of inertia over these small volumes. Mathematically this summing up is accomplished through integral calculus.

odd object The general approach to calculating a moment of inertia for an object of complicated geometry about an arbitrarily chosen line is an exercise in integral calculus which we do not need to go into at this point. In fact for most real objects the calculus is too hard for anyone. If a moment of inertia must be determined, either a computer technique called finite element analysis is used or if the object is not too large, we may measure the moment of inertia, I, by applying a known torque and measuring the angular acceleration.

For certain very simple objects we may determine the moment of inertia by reasoning alone, based on the m*r2 moment of a single particle. In the following examples we are taking the axis of rotation to be the axis of symmetry perpendicular to the plane in which the figure is defined.

For a hoop or cylindrical shell where all the bits of matter that make up the object may be taken to be at the same radius, the moment of inertia is just the total mass, M, times the radius squared. We arrive at that conclusion by mentally dividing the hoop up into little bits of mass Dm such that the moment of inertia of each slice is Dm * R2. Then the total moment of inertia is
I = sigmaDmi * ri2 ,
where n is the number of mass elements. The fact that ri = R for every i, allows us to factor R2 out of the sum giving us

I = R2 * sigmaDmi .
But sigmaDmi is just the total mass M, leaving the result
I = M * R2 .

For a hollow cylinder where the walls are of significant thickness, you may imagine that half the mass is in a hoop at the outer radius and the other half the mass in a hoop at the inner radius so that
I = M/2 * (R12 + R22) .

It might help you to visualize this situation by starting with the two hoop configuration and imagining taking a bit of mass from the outer hoop and forming a new hoop slightly inside the outer. This would decrease the moment of inertia. Then take an equal amount of mass from the inner loop and form a loop outside the inner, just far enough to restore original moment of inertia. If you continue this process you will find that you can fill the space between the inner and outer radius and maintain the original moment of inertia. A solid cylinder is just a hollow cylinder where the inner radius is zero so

I = M/2 * R2

for the solid cylinder or disk. Notice that in our discussion of cylinders, the height of the cylinder does not have any effect, except to the extent that the mass is effected by a tall cylinder as opposed to a flat disk.

A thin rod of length L rotated about one end is just a collection of particles each of mass DM at radii ranging from zero to the length of the rod. The contribution to the moment of inertia of a particle at distance r, from the origin is

DI = DM*r2 .

But DM can be expressed as the mass per unit length of the rod, m, times a little length element, Dr. so

DI = m * r2 * Dr , or DI / Dr = m * r2 .

This means that the rate of change of I with respect to r is m * r2. Remember that the rate of change in the limit as Dr becomes very small is the derivative of I with respect to r. To find the function I(r) which describes the moment of inertia as a function of the variable r, we need to find a function whose derivative is m * r2.

Thinking back to our discussion of the derivative of the quadratic function, in the section on rates of change , we found that if y=a*xn, the derivative of the y with respect to x, Dy / Dx, was n*a*x(n-1). In this case
DI / Dr = m * r2 .

To pick out the function which has this expression as its derivative, just increase the exponent by 1 and divide by the new exponent. This trick is called taking the anti-derivative of m * r2 and gives us m/3 * r3. So
I = m/3 * r3
for any r. Notice that the total mass of the rod is m * r, so
I = M/3 * r2
and for a rod of length L,
I = M/3 * L2 .

Now what would happen if we rotated the rod about its center rather than about one end. That would be equivalent to having two rods rotated about their ends, each of half the length of the original. The moment of inertia of a rod of length L/2 rotated about its end would be (1/2*M)/3 * (L/2)2, based on the result above. This reduces to 1/24 * M * L2. Of course there are two such half rods, so the total moment of inertia for a rod rotated about its center is
I = M/12 * L2 .

In finding the moment of inertia of a rod about two different axes, we have happened upon an important principle called the parallel axis theorem. The theorem states that if an object has moment of inertia Icm about its center of mass, like the rod in the second case, then its moment of inertia about any axis parallel to the original axis of rotation is Icm + M*D2, where D is the distance between the parallel axes,

I = Icm + M * D2 .

Let's check this with the rod example. D in this case was L/2. Icm was M/12 * L2. So

I = M/12 * L2 + M * (L/2)2 = M*(1/12 + 1/4)*L2 = M*(1/12 + 3/12)*L2 = M/3 * L2 .

A rectangular plate of mass M and dimensions a by b has a moment of inertia about its center of mass the same as if all its mass were located in a rod whose length was the diagonal of the plate. The length of the diagonal is the square root of (a2 + b2), so the moment of inertia is

I = M/12 * (a2 + b2) .

Without going through the mathematics for the hollow sphere and solid sphere, I will just state the moment of inertia about a center of mass axis for each. For the hollow sphere,
I = 2*M/3 * L2 .
For the solid sphere it is
I = 2*M/5 * L2 .

Now if we have an object of one of the shapes we know about, we can calculate the angular acceleration about any axis parallel to the axis of symmetry, from the torque about the given axis. All we need to do is divide by the moment of inertia. In analyzing the motion of solid objects, those with significant dimensions as opposed to a particle, we need to pay attention not only to the magnitude and direction of the forces, but also to where on the object they are applied. On the Disk Research display at the end of this section, we will demonstrate this. Before we get to that however, we need to develop some additional rotational dynamics ideas.

straight line angular momentumGetting back to our particle in the (x,y) plane at position r. Suppose that the particle has linear momentum in our reference frame of pkgm/s. The "angular momentum" of the particle about the origin is defined as L = r X p. Only the component of p perpendicular to r contributes to the angular momentum. Let's consider two particular instances of angular momentum about the origin of a particle.

If a particle is moving at a constant velocity in a straight line in our reference frame, can it have an angular momentum? Look at the Straight Line Angular Momentum display for an illustration.

The other particular instance of angular momentum of a particle about the origin we should discuss is that of a particle in uniform circular motion around the origin. In this case the linear momentum p is always perpendicular to the radius vector r so

L = |r| * |p| .
But |p| is m * |v| so
L = m * |r| * |v| .
In terms of angular quantities, |v| = |r|*w so we now have
L = m * |r| * |r| * w .
So for circular motion of a particle, the angular momentum is
L = I * w ,

which is exactly analogous to the linear momentum being mass times velocity.

Now I want to derive a connection between the torque and the angular momentum. Remember that the force f=Dp/Dt where p is the linear momentum. Let's take the cross product of r with both sides of that equation so that

rXf = rX(Dp/Dt) .
But rXf is just the torque t so
t = rX(Dp/Dt) .

Next, go back to the definition of angular momentum, L = rXp and take the derivative with respect to time of both sides. Remember that the derivative is just the rate of change of the variable where the time interval over which the change happens approaches zero. That is the change in the variable divided by the tiny time over which the change took place. So
DL/Dt = D(rXp)/Dt .

OK so how do we get the rate of change in the cross product rXp in terms of the individual factors r and p. We might be tempted do distribute the D among the factors, making D(rXp)/Dt = Dr/Dt X Dp/Dt . That makes the assumption that the derivative of a cross product is the cross product of the derivatives, which turns out to be a bogus assumption. Just look at the units on each side of this expression. We have m2kg/s2 on the left and m2kg/s3 on the right. Evidently we need to look at this in more detail.

The change in the cross product with respect to time is

D(rXp)/Dt = ((r + Dr)X(p + Dp) - rXp) / Dt .

Next we carry out the multiplications indicated by the cross products remembering that the order in which we place the factors matters. That gives us

D(rXp)/Dt = (rXp + DrXp + rXDp + DrXDp - rXp) / Dt .

The rXp and the -rXp terms cancel out and the DrXDp term, being the product of two very small numbers may be considered zero as Dt approaches zero. That leaves us with
D(rXp)/Dt = (DrXp + rXDp) / Dt ,
D(rXp)/Dt = Dr/DtXp + rXDp/Dt .

But Dr/Dt is just the particle velocity v and p is m*v so
Dr/DtXp = vXm*v = 0

because the v and m*v point in the same direction and parallel vectors have a zero cross product.

This leaves us with the rate of change of the particles angular momentum,

DL/Dt = rXDp/Dt = t .

The rate of change of angular momentum equals the torque just as the rate of change of linear momentum equals the force. This tells us that if the torque on an object is zero, then the angular momentum remains constant. This is the principle of the conservation of angular momentum which has far reaching consequences, extending to quantum mechanics even though we discovered it in the realm of classical mechanics.

raza To calculate the total angular momentum of a system of particles we must add up the angular momentum of each of them about the same point. For that point we choose the origin of our reference frame. The angular momentum of each is a vector so the total angular momentum is the vector sum. For a system of n particles we have

L = L1 + L2 + ... + Ln = sigma Li .

As time passes, the total angular momentum about the origin may change if a net torque exists. We know that the forces experienced by the particles in an object may be internal forces, forces among the particles in the object or external forces, those applied to the particles in the object by some outside agent. Newton's third law tells us that the force between any pair of particles is equal and opposite on each of them and is directed along the line joining them. In this case the internal torques are all zero since the forces are collinear. So

DL/Dt = t ,
where t is the net external torque.

Let's take another look at the Rotation About the Z Axis display.

Since each particle moves around its orbit with constant speed, the angular momentum of each particle and therefore of the system as a whole is constant. Constant angular momentum means that the net external torque on the system is zero. Any external force then must not have a tangential component so the net external force points toward the z axis. This is the same conclusion we reached in talking about angular acceleration, but in this instance we arrived at the conclusion from the standpoint of angular momentum.

The analogy between linear dynamics and rotational dynamics extends to the ideas of work and energy also. Kinetic energy of rotation, ker, is the sum linear kinetic energy of all the particles involved. Suppose we have a solid object rotating with angular velocity w. Any particle in the object, say the ith one, will have kinetic energy
kei = 1/2 * mi * vi2 .
The vi factor can be expressed as vi = ri * w so
kei = 1/2 * mi * ri2 * w2 .
For the whole object the rotational kinetic energy is
ker = 1/2 * sigmami * ri2 * w2 .
But sigma mi * ri2 is just the moment of inertia, I, so
ker = 1/2 * I * w2 ,
just as you would expect from your knowledge of linear dynamics.

rotational workLet's use this picture to understand how a force applied to a solid object free to rotate about a fixed axis, results in work being done. Here we see a solid blue disk with its center of mass at the origin in the (x,y) plane. A white radius vector is painted on the disk. A force is applied at the circumference of the disk, indicated by a yellow vector. The force vector makes an angle f with the radius vector. After a short time, the applied force will rotate the disk to a new position, moving the radius vector from position r1 to position r2. The angular displacement is Dq. The distance through which the force f was applied is the arc length Ds.

The bit of work DW done in angular displacement Dq is f · Ds. The arc length Ds may be replaced by the quantity |r|*Dq as long as Dq is small. The dot product of the force vector with the vector |r|*Dq is the product of the magnitude of |r|*Dq and the magnitude of the component of f in the direction of |r|*Dq. That component of f is |f|*sin(f). So we have
DW = |f|*sin(f)*|r|*Dq .

But we know that the quantity |f|*sin(f)*|r| is the torque produced by f about the origin, on the disk. So

DW = t*Dq .

This result is consistent with the expression for work in linear motion, being the product of force times displacement. In rotation torque plays the role of force and Dq plays the role of displacement.

In linear motion we found a work-energy theorem which proved useful in predicting the future of a moving object. In rotation we can get a similar relationship. We will begin with Newton's second law expressed for rotation of a symetrical solid body rotating around its axis of symmetry,

t = I * a = I * Dw / Dt .

Now we are going to use a trick that tends to make students say, "How did you know to do that?". We are looking for a relationship between work and energy and we have just discovered that work involves torque and displacement, like this:

DW = t*Dq .

But our expression for Newton's second law does not involve displacement, Dq. To introduce Dq into our equation we can multiply the right side of the equation by Dq / Dq, in effect multiplying the right side of the equation by 1, which leaves it unchanged. The result is,

t = I * (Dw / Dt) * (Dq / Dq) .

Now looking at the product of the two fractions, (Dw / Dt) * (Dq / Dq), notice that we can associate the denominators with the opposite numerators so we get,
t = I * (Dw / Dq) * (Dq / Dt) .
But Dq / Dt is just w, so
t = I * w * Dw / Dq .
Now multiplying both sides of this equation by Dq we get
t * Dq = I * w * Dw = DW .
Then dividing by Dw we find
DW / Dw = I * w ,

which says that the rate of change of work with respect to angular velocity is the moment of inertia times the angular velocity.

To get work as a function of angular velocity from the rate of change we use the anti-derivative trick introduced earlier.

W = 1/2 * I * w2 .

So as w goes from and initial to final value, the amount of work done is

W = 1/2 * I * wf2 - 1/2 * I * wi2 ,

meaning that the net work done by external forces in rotating an object is equal to the change in the objects rotational kinetic energy.

Although the measures of rotational dynamics were carefully developed to parallel their counterparts from linear dynamics, the analysis of rotary motion in the most general case is beyond the scope of this course. We have to limit our analysis to a few simple situations. We can only handle the rotation of solid objects which have an axis of symmetry, rotating about an axis that is parallel to one of the primary axes of the reference frame. In addition the axis of rotation must be fixed in space (spinning motion) or moving in a straight line as in the case of a cylinder rolling along a surface (rolling motion). Even with these restrictions you will find many commonly occuring examples of rotation posed as problems in your physics textbook.

I suspect that at this point you have had enough of algebra and anti-derivatives. To bring these ideas together, take a look at the Disk Research display.

The point of discussing this different way of formulating the laws of motion is to help you visualize the motion a system of particles might undergo. To completely describe its motion we might want to treat its rotation using some of the concepts here. There is one additional kind of motion we will cover and that is vibration. Not only might a system be translating and rotating but also quivering in some fashion. In the next section of the course we will look at this.

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