Physics with Calculus/Mechanics/Linear-Rotational Analogs

We have seen that replacing distance with angular displacement, velocity with angular velocity, and acceleration with angular acceleration, all kinematic equations will hold. This leads us to consider rotational analogs of force, momentum, and energy.

Torque

What is the rotational analog of force? From common experience, we know that it is easier to rotate a handle farther from the axis than closer to it. We also know that only the component of force that is perpendicular to the handle does any work, for if we push it along the handle, we would not rotate the handle. We thus expect that the rotational analog of force, called torque, is given by:

$\tau=rF\sin \theta$

or, in vector notation,

$\mathbf{\tau}=\mathbf{r} \times \mathbf{F}$

where $\mathbf{r}$ is the vector pointing perpendicularly away from the axis towards the point of application of the force.

We can generalize the above formula, which is for rigid rotation about an axis. Rotations are more generally about a point than an axis. For rotations about a point, $\mathbf{r}$ is the vector pointing from the rotation center to the point of application of the force. The torque is still given by $\mathbf{\tau}=\mathbf{r}\times \mathbf{F}$. For a body revolving around an axis, it can be verified that any rotation center chosen on the axis of rotation would give the same torque according to either definition of $\mathbf{r}$.

Let us examine in more detail the torque equation for rigid objects rotating around an axis. We have that

$\tau=\sum rF_\perp=\sum r(ma)=\sum mr(r \alpha)=\alpha \sum mr^2$

We define

$I=\sum mr^2$ to be the moment of inertia. Note that r is taken to be the distance from the axis, not from the origin.

We can rewrite the torque equation as

$\tau=I \alpha$

Note the similarities between this equation and Newton's Second Law. Newton's Second Law states that a force would cause an acceleration, and the magnitude of the acceleration is inversely proportional to the mass. This law, which we call Newton's Second Law for rotational motion, states that a torque would cause an angular acceleration inversely proportional to the moment of inertia. The moment of inertia depends on how mass is distributed around an axis. A mass farther from the axis has a higher moment of inertia than the same mass closer to the axis. The moment of inertia is not unique for any object; in general, the moment of inertia is different for every axis of rotation. For a finite number of objects, we can simply evaluate the sum. For a continuous distribution of matter, we use

$I=\int r^2 dm=\int r^2 \rho dV$

where $\rho$ is the density. This integral is generally quite messy to evaluate. The moment of inertia for simple geometric shapes such as spheres and cylinders are given in various tables around commonly-used axis.

Suppose we know the moment of inertia about an axis through the center of mass, $I_o$. Then, the moment of inertia about any other axis parallel to the first is given by

$I=I_{o}+Md^2$

where d is the distance between the two axis.

Angular Momentum

We know that $F=\frac{dp}{dt}$, where p is the momentum. Similarly, we would think that $\tau=\frac{dL}{dt}$ where L is the angular momentum, given by $L=I \omega$ for rigid bodies rotating around an axis. Like linear momentum, angular momentum is conserved. The Conservation of Angular Momentum is one of the most fundamental laws of physics and is experimentally verified to astounding accuracy. While Newton's Law of Gravity and his second law of motion do not hold in relativity and quantum mechanics, the conservation of angular momentum holds in every physical theory.

Angular momentum is used just as linear momentum would. In situation where we do not know and are not interested in the specific torques involved, angular momentum can increase our understanding of the situation. For instance, suppose a star was rotating at speed $\omega_0$, when it suddenly shrunk (due to gravitational collapse) to half its size. The moment of inertia is given by $I=\frac{2}{5}MR^2$. What is its final rotational speed?

In this case we are not interested in and have no clue about the torques that got it into the final state. We do know that angular momentum is conserved. Thus, $\frac{2}{5}MR^2 \omega_0=\frac{2}{5}M(\frac{1}{2}R)^2 \omega_f$, so the final rotational speed is four times its original.

In general, the angular momentum with respect to any given point is given by $\mathbf{L}=\mathbf{r}\times\mathbf{p}$. This is more general than the above formula, which only holds for rigid rotations about an axis. The first equation can be derived from the more general formula.