Rotation
from Physclips:
Mechanics with animations and film.
This page gives background material for the multimedia tutorials on rotation from the Physclips main site.
(Some of the sections are not yet finished, for which we apologise.) |
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This demonstration, also shown in the multimedia tutorial on rotation, illustrates rotational kinetic energy. The brass object initially rolls on two inclined rails that support it by contacting its shaft. At the end of the incline, it reaches a horizontal track on which it rolls on its edge. This happens at about t = 4s on the clip.
The vertical time markers show how far it travels in each second. Going down the ramp, it accelerates smoothly. Notice that it suddenly accelerates when it reaches the horizontal track. This is not surprising: relative to the centre of the object, the edge is moving faster than the surface of the shaft (see rolling). So, when the edge contacts the horizontal track, it exerts a frictional force against it. The tract exerts an equal and opposite force, which accelerates it.
This sudden acceleration required energy: where was it stored?
When the rapidly moving edge contacted the horiztonal track, some of the rotational kinetic energy is converted into translational kinetic energy: on the level track, it is travelling faster but rotating less rapidly. (Some kinetic energy is also lost, however, because there would be some skidding during this process, so some energy lost. See the section on contact forces.)
Now let's get quantitative.
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Rotational kinetic energy
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Let us picture a rigid body turning with angular velocity ω, like the Earth in this picture. Mentally, let's divide it up into a collection of small masses. With respect to the axis of rotation, a single mass m at radius r is travelling at speed of
(You may wish to revise circular motion at this point.) Its kinetic energy is ½ mv2. So
let's imagine the dividing the object up into many masses mi at distances ri from the axis. Each has vi = riω, where ω has the same value for all the masses because the object is (by assumption) rigid. So the total rotational kinetic energy is
Krot = Σ Ki = Σ ½ miri2ω2
where the summation is over all of the i. ½ ω2 is a common factor in every term of the sum, so
This is the result for a collection of discrete masses, mi. For a continuous body, we should normally divide it up into small elements of volume, dV. (You may wish to revise calculus.) From the definition of density ρ, each has mass
Instead of an ordinary summation, we do an integral (the equivalent of summation for very small divisions), and we have
and where the integration is over the whole volume occupied by the rigid body in question.
Students often ask at this stage: 'I know how to integrate over x, y etc, but how do I integrate over mass?' The answer is via the density, ρ. To know the mass distribution, you need to know ρ(r) or ρ(x,y,z). So we consider a small element of volume dV, and write dm = ρ.dV. If the object had rectangular symmetry, we might choose dV to be a cube with sides dx, dy and dz and write, for that example,
dm = ρ.dV.z = ρ.dx.dy.dz. Then we just integrate over the limits of x, y and z that define the object being studied. For a sphere, however, the volume elements would be hollow cylinders about the axis, and the integration would go from zero radius to that of the sphere.
In the examples below, the value for the hoop is obvious: all of the mass is at distance r, so I = mr2. The disc about its axis can then be considered as a set of hoops, each of thickness t and width dr, and having mass ρdV = ρ.2πt.r.dr. The sphere may be considered as a set of discs with different radii. The rectangle and the disc about its diameter are both analysed as sets of rods. There are examples in most textbooks. I'll put some up here soon, but shall postpone it because it's hard to write maths in html!
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Moment of inertia
Here are some useful common cases that result from the integrals referred to above.
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Rolling problems
This is one of the puzzles introduced in the multimedia tutorial. Two similar jars, one full of water (low viscosity) and one full of honey (high viscosity). Which one rolls faster? Before your run the film, ask yourself these questions: Assuming their weights are the same, what can you say about their initial potential energies? When they reach the bottom, which will have more rotational kinetic energy? And therefore, which will have more translational kinetic energy?
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You should be able to use similar arguments, and the values of moments of inertia given above, to predict the results of most of the 'races' shown below. But first we might ask whether size comes into it, either via radius, or via mass. The clip below shows two aluminium discs of different sizes but the same mass.
The next race is between a disc (a solid cylinder) and a hollow cylinder.
A solid sphere (a billiard ball) and a solid disc (of aluminium).
And, finally, spheres of different sizes and masses. (two steel balls)
The graphs above show displacement, velocity and acceleration for linear motion with constant acceleration (at left) and for circular motion with constant angular acceleration. Just for practice, let's derive the new equations (and revise the kinematics section if this looks difficult!) If we consider motion with constant acceleration, and remember that α = dω/dt, we have
And from ω = dθ/dt, we can integrate again to get:
θ = ∫ ω dt = ½αt2 + ωt + θ0
From the two equations above, we can eliminate t to get
So we have equations completely analogous to those of linear kinematics:
ω = ω0 + αt and θ = θ0 + ω0t + ½αt2 and ω2 − ω02 = 2α(θ − θ0)
v = v0 + at and s = s0 + v0t + ½at2 and v2 − v02 = 2a(s − s0).
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As we saw in an earlier section, forces cause accelerations. To make something turn, we apply a torque. We shall define if first, and then explain why this defintion is logical. Later we shall see the complete analogy with Newton's laws for linear motion.
The torque τ is defined by
where force F acts at a point displaced by r from the axis. The magnitude of the torque is given by
where θ is the angle between r and F. (You may need to look at the cross product section of the support page on vectors.)
We shall discuss the magnitude first, then the direction.
The photos at right show three ways of using a spanner. In the first pair, we compare a small value of r (small torque) with a large r and large τ. In the second, we compare θ = zero and θ = 90°. In the former case, the torque is zero. From experience, you know that you need large r, θ = 90° and large F to obtain the maximum torque. |
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 This upper set of diagrams at right shows the dependence of torque on the angle θ. Maximum torque occurs when the component of F at right angles to r is maximum, i.e. when θ = 90°. The central figure shows the tangential component of F, which is F sin θ.
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 The equation
can be interpreted in two different ways, as shown in these sketches:
τ = r (F sin θ) or τ = F (r sin θ).
We can think of it as r times the tangential component of F (left sketch and equation) or as F times the shortest distance (r sin θ) between the axis and the line along which F acts (right sketch and equation). |
 Torque is a vector
The definition τ = r X F gives the direction of τ . It is at right angles to both r and F is in the right-handed sense: if you put your right thumb in the direction of r and your forefinger in the direction of F, your right middle finger points in the direction of τ. The second photograph shows the torque τ produced by the tension in the string about the axis of the pulley. |
In an earlier tutorial, we saw that Newton's first and second laws for linear motion are combined in the equation Force tends to produce linear acceleration, and mass resists linear acceleration.
For rotation, torque tends to produce angular acceleration, and moment of inertia resists angular acceleration . Considering just rotation about a fixed axis, we write Newton's law for angular motion as
In these film clips, we see different torques τ giving rise to different angular accelerations α for objects with the same moment of inertia I. Although the same mass is attached to the string, the forces are only approximately equal: the force in the example at right is a little less than that at left. (Can you see why? Think of the equation of motion for the descending mass.)
Despite the somewhat smaller force in the second case, the larger displacement of the point of application from the axis means that the torque is greater in this case, and so it produces a greater angular acceleration.
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In these film clips, we see the effect of changing the moment of inertia. Again, although the same mass is attached to the string, the forces are only approximately equal, but the approximation is better in this example. In these three case, the radius about which the force acts is the same, so the torques are approximately equal.
In the first film, an aluminium tube rotates. In the second, masses are attached to it, increasing its moment of inertia. Larger mass, larger I, smaller α. Comparing the second and third films, we see that it is not just the mass, but also the distribution of mass that determine the moment of inertia: when the masses are at large radii, I is larger and α is smaller.
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You have caught up with the author: this page is still being made and there is a lot more to write. Apologies.
This experiment shows the importance of the axis in determining the moment of inertia.  |
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An interesting demonstration. Why does the rod fall faster than the ball? Or is that a trick question> |
Gyroscopes
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More to write about gyroscopes, too. Come back later. |
Precession
Here we see that the applied torque about the centre of mass does indeed change the angular momentum in a nearly horizontal direction. More to write here, please come back later. |
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