We have covered almost everything we need to make a general description of motion. However, many motions can not be adequately described by either translational or uniform circular motion. By extending the ideas of circular motion, however, we can form a completely general description of rotational motion.
The figure above shows a point-like particle of mass m in motion
around a circular path of radius r. The particle has, at any
instant, an instantaneous tangential velocity,
vt,
which takes it around the path. Unlike the case for uniform
circular motion, this velocity
vt, could possibly
be not constant. From our previous examination of this
problem in the
Maple file on uniform circular
motion, we found that one way of characterizing the connection
between distance traveled and angular displacement, was to note
that, if q is defined in radians, then
the arclength of the distance traveled by the particle is given by
or more precisely by s = r(qf - q0) where f indicates the final angular position and 0 indicates the initial position. The connection between the tangential velocity and the angular velocity is
with w = d(q)/dt. If vt is NOT constant with time, then the rate of change of vt is at, the tangential acceleration. We can relate the tangential acceleration to the angular acceleration, a, by noting that at = dvt/dt = r d(w)/dt = ra if we define angular acceleration as the time-rate of change of the angular velocity. The equations for rotational motion can be derived by the rules of calculus, just as for translational motion. So we have
| qf | = q0 + w0t + ½ at2 |
| wf | = w0 + at |
| wf2 | = w02 + 2a(q - q0) |
We can now look at how a force might cause the tangential acceleration (and hence angular acceleration) to occur. Consider the figure below which shows a force F acting on the particle.
Since the string tying the particle to the center constrains its motion, only the component of the force which lies along the tangential direction at any instant can cause any change to the tangential velocity, i.e. only Fy affects vt. The component Fx has no affect. From the figure, Fy = F sin(g - q). Newton's Second Law, assuming the mass of the particle does not change with time, is Fy = mat = mar. We note that, although at does not depend on the radius, r, a does. In fact, a decreases linearly with r for fixed values of Fy and m. This is counterintuitive if we think about the action of one of the simplest of tools, the lever.
If there are two forces acting on the lever above, the forces have to act in the way shown above to maintain balance. If a force F acts a distance d away from the center-of-gravity, then you need a force ½ F acting on the other side of the center-of-gravity if it acts a distance 2d away. Hence we expect the effect of forces on rotation to increase with distance away from the center. We define the product of a force perpendicular to the lever arm, r, about the axis of rotation, to be the torque. The torque increases as r increases. This is consistent with our experience with rotational motion.
So why do we see that angular acceleration goes down with increasing distance of the force from the axis of rotation in our single particle in a circular path? The missing piece of information has to do with the inertia or opposition to change in angular motion associated with rotational motion.
We can understand better what the observable consequences of changing r are for angular displacement by applying the work-energy theorem. First, we note that the kinetic energy of the particle can be expressed in terms of angular motion variables, K = ½ mvt2 = ½ mr2w2. We can also express the change in kinetic energy by applying the work-energy theorem. Over a small change in position of the particle, the force is approximately constant so
Therefore, the work-energy theorem says that the change in kinetic energy is (using one of our equations for rotational motion)
We see the same term mr2 on both sides of the equation. For rotational motion, this term appears to serve the same role as mass does for translational motion. For that reason, we give it a specific name, the rotational inertia. The general symbol for rotational inertia is I.
Once the idea of rotational inertia is in place, then we can understand the effect of changing the radius at which a force works to change the angular motion of a particle since its opposition to change in angular motion is proportional to r2. Therefore, as we increase r for a fixed Fy and fixed mass, the force becomes more effective at producing torque (rFy), but the rotational inertia of the particle goes up as mr2, hence the angular acceleration actually goes down by rFy/(mr2) = 1/r. Furthermore, we have a description of rotational motion which is completely in form of angular variables, the torque, t, is equal to the rotational inertia times the angular velocity, i.e.
This looks remarkably like F = ma and is, in fact, a statement of Newton's Second Law for angular motion. Although we can apply it directly for point particles, it's more interesting to generalize it to three-dimensional, rigid objects. That's easy to do since a rigid object can be thought of as a large number of infinitesimal point masses. All the point masses rotate at the same angular acceleration (except those points on the axis of rotation) or else the object is not rigid. Therefore, the net force on the system of point masses produces a net torque whose effect on the angular acceleration of the system depends on the rotational inertia of the system. To find the rotational inertia for a three-dimensional object, we note that each point mass has rotational inertia dI = r2dm, so the rotational inertia of all of them is an integral of dI over the whole object. The textbook does an example of calculating I for a cylindrical disk. From a similar calculation of I for a cylindrical shell, you can easily extend to the calculation of I for any cylindrically shaped object, including the flat disk. We'll do the same for spherical shapes by calculating I for a thin spherical shell of radius R (shown below).
To find the rotational inertia, we divide the shell up into rings. Each ring has a radius r with respect to the axis of rotation (which could have been along any diameter of the shell) and lying in the plane perpendicular to this axis. The ring has a width which is given by R d(q), therefore the area of the ring is dA = 2pr (R d(q)). If the shell has mass M, then the ratio of mass in the ring, dm, to the mass of the whole shell is given by the ratio of areas,
So, dm = M r d(q)/(2R). We can r as a function of q, by noting that trigonometry applied to the figure above shows that r = R cos(q). The rotational inertia of each ring is (as shown in the text) dm r2. Therefore, to get the rotational inertia of the whole spherical shell, we integrate over all the rings
where we used the substitutions u = sin(q) ==> du = cos(q) d(q) to evaluate the integral.
We can easily extend this result to get the rotational inertia of a full sphere of radius R by breaking it into an infinite number of infinitesimally thin spherical shells of radius r < R and thickness dr. For each shell the mass, dm, is given by the ratio of the volume of the shell to the volume of the whole sphere:
The rotational inertia of each shell is (2/3)dm r2 as we just showed. Therefore, the rotational inertia of all the shells is
