Lecture 13

Newton's Law for Rotational Dynamics

We are now ready to make the last step, i.e. to examine the dynamics of rotational motions for the non-equilibrium case. And to do that, we will find out how Netwton's "F=ma" law looks like when expressed in term of angular quantities [it is useful to remind ourselves why we want to use angular variables when dealing with rotations: the reason is that, for a rotating body, angular quantities like angular velocity and accelerations are the same for each point of the body, while the linear (or tangential) quantities do vary according to the radius; this will also mean that quantities that depend upon the linear velocity, like kinetic energy and momentum, are not easily defined for a rotating rigid body, unless we succeed in expressing them in terms of angular variables].

So let us consider a body free to rotate around an axis, and let us apply to it a force F perpendicular to the axis and at a certain distance from it (so that the lever arm is not zero). For an arbitrary very small volume of mass m_i and located at a distance r_i from the rotation axis we can write

F_i= m_ia_i

F_ir_i = m_ia_ir_i

$\tau_{i}~=~m_{i}r^{2}_{i}\alpha_{i}$

where we have remembered the definitions of torque and of angular acceleration. This represents the effect of the elementary torque felt by an (arbitrarily small) section of our rotating object. To find the expression for the whole body, we would have to sum all the individual elementary contributions, i.e.

$\tau~=~\Sigma_{i}\tau_{i}~=~\Sigma_{i}m_{i}r^{2}_{i}\alpha_{i}~=~I\alpha$

with $I=\Sigma_{i}m_{i}r^{2}_{i}$ and taking advantage of the fact that $\alpha$ is the same for all the elements of the rotating body, therefore it can be factored out of the sum.

The expression $\tau~=~ I\alpha$ is exactly what we were looking for, as it relates torques to angular accelerations.The expression is in fact nothing but F=ma, but written in a more convenient form. The quantity $I~=~\Sigma_{i}m_{i}r^{2}_{i}$ is called moment of inertia, and it plays for rotations a role analogous to that of the mass for translations, since it represents a measure of the resistance (i.e. of the inertia) that a body opposes to undergo angular acceleration when subjected to a torque. But note that the moment of inertia I is not an intrinsic attribute of a given body, since it is defined with respect to a given axis of rotation. For different choices of axis, the same body will have different values of I.

Techniques of Integral Calculus allow to compute the Moment of Inertia for bodies of regular shape (see Table 9.1). For a body of irregular shape, it can be determined by measurement.
How ?? Obviously by applying to the body a known torque, and measuring its (angular) acceleration (see e.g. Problem 9.29).

Rotational Energy

We have learnt how a body moving with velocity v has a kinetic energy given by E=1/2mv². And we have also seen that to give an object initially at rest a certain amount of kinetic energy, one has to perform an equivalent amount of work, the work being given by the product F x s. If we now consider a rotating body, it is obvious that it possesses kinetic energy (every point of the body is moving with some velocity), and that work is required to transfer that amount of energy to it. You could probably guess what will be the epressions for rotational work and energy, when described in terms of angular quantities:

an object rotating with angular velocity $\omega$ has a rotational kinetic energy given by

$E^{rot}_{kin}~=~1/2I\omega^2$

where I is the body's moment of inertia; and this energy could be gained by doing an amount of work given by

$W~=~\tau\theta$

What about the sign of W ? Probably the best way to determine the sign of the work done by a torque is to think that positive work will increase the kinetic energy of a body, while negative work will decrease it (e.g. if friction brings a rotating flywheel to a stop, friction has done negative work).

We can then conclude that a translating object has kinetic energy 1/2mv², while for a rotating one $E=1/2I\omega^2$. But what if the object is both translating and rotating (e.g. a spinning football, or a gymnast doing somersaults, or even the wheel of a moving car)? Well, in this case the object will have both translational and rotational kinetic energy, and if a certain amount of total work is done on the body, this work in general will go to modify both quantities. Moreover, the principle of mechanical energy conservation for a body subjected only to conservative forces will have to include rotational energies. Specifically, for a body moving under the sole influx of gravity, we will have

$mgh~+~1/2mv^{2}~+~1/2I\omega^{2}$ = constant

(A legitimate question : if a body has extended dimensions, what is the correct choice for "h" ? Any guess ? )

The fact that rotational energy has to be included when dealing with falling (and rolling) objects has some, maybe surprising, consequences, e.g.

• rolling bodies fall "more slowly" than sliding bodies. And, not surprisingly, the rate of "rolling down" will depend upon the value of I, the moment of inertia.
• not all all parts of a rotating falling body fall at the same rate: according to its relative location, a point can have acceleration larger or smaller than "g".

Angular Momentum

You can guess the next step : as a translating body has a linear momentum given by p=mv, so a rotating body has an angular momentum given by L=I$\omega$. And, with arguments similar to the ones we applied to the linear case, we could also show that, when a system is not affected by external forces, its total angular momentum remains constant. Simple demonstrations can verify the validity of the principle of conservation of angular momentum.

Conceptual Questions 9.20, 9.21