ANGULAR ACCELERATION:

For this topic, you need to know about radians for measuring angles. Draw a circle and two radii, like cutting a piece of pie. The angle between the two in radians is the arc length divided by the radius. Note that this is a dimensionless ratio. To see how a radian compares to other angle measures, consider an arc that is the full circle. Its arc length is 2p r, so when you divide this by r you get 2p . Hence 2p = 360 degrees. (Ironically, in a sense, one pie = 360 degrees, so 1 pie = 2p.)

Recall Newton's 2nd law, net force = mass times acceleration, valid as written if certain units are used. The rotational version of this is: net torque = inertia times angular acceleration, and again it is for certain units. We define things in such a way that the units for angular acceleration must be radians per second squared, because this turns out to be the most convenient. When we finally communicate with beings elsewhere in the universe, we will find that all units will be different except for the radian and revolution (and certain natural units like electron charge), and they will probably use radian per (time unit)² for angular acceleration.

What is that inertia thing in the equation above? It is dependent on mass and where the mass is located. Mass far from the rotation axis has a greater effect on the behavior of a rotator than mass close to the axis. It turns out that inertia I is the sum of mass times the square of the distance to the axis. This is a calculus problem in general, and the sum becomes the integral of r²dm. A disc has
I_disc = (¹/₂)mr², a solid sphere has I = (²/₅)mr², a thin spherical shell has I = (²/₃)mr², a rectangular solid of length, width = a, b has
I = (¹/₁₂)m(a²+b²). These formulas are for the case of rotations about the center of mass. There is a parallel axis theorem for an axis located distance h from the center of mass: I = I_cm + mh² .

Here is another way of finding the I value of body with respect to any point other than the center of mass: 1) Let h = the distance from the point to the center of mass. You can find the center of mass easily: the balance point.

2) Let r_cp = the distance to the center of percussion, or sweet spot. When your object is swung as a small amplitude pendulum, a simple pendulum of length r_cp will have the same period. This is a good way to find r_cp experimentally.

3) I = mhr_cp.

Now if you want the inertia with respect to the center of mass, find I at some other point, as above, then I_cm = I - mh² . Want a derivation of equation 3 and some other stuff about inertia?

In rotation problems, you may run into angular speeds in revolutions per minute. If you do not need to apply Newton's 2nd law, you may not need to change units. Of course if you do need to change to rad/s, just multiply the number of rev/min by
(2p rad/rev)(1 min/60 s).

For each linear motion equation there is a corresponding rotational equation. For example x = v_ot + (1/2)at² relates distance traveled x to initial v, acceleration, and time of travel. For rotation, the corresponding relationship is among angle turned, initial angular velocity, angular acceleration and time: q = w _o t + (1/2)a t². Just replace distance with angle, velocity with angular velocity, acceleration with angular acceleration.

In applying net torque = I times alpha, (S t = Ia ) you may find that the hard part is the net torque.

1. Identify the pivot point, or center of rotation.

2. Identify all forces which exert torques. Ask yourself which way each is trying to rotate the body, clockwise or counterclockwise. Choose a sign convention. If you choose to make counterclockwise positive, then clockwise torques are negative.

3. Calculate the torques. If your pivot is not at the center of mass of the system, one torque may be due to gravity. Keep in mind that the torque due to gravity on a rigid body is the same as if the entire weight of the body acts at its center of mass. (Really, the center of gravity. In everyday life, the center of mass and the center of gravity are the same.)

ANGULAR MOMENTUM:

We define this as a vector quantity, the vector pointing along the rotation axis. Something rotating clockwise as viewed from above has its angular momentum vector pointing away from the viewer, the way a screw advances when rotated clockwise. Here are some examples you won't find in your book. Linear momentum of a system is conserved (constant) if the net force on the system is zero. Similarly, angular momentum is conserved if the net torque on a system is zero. Further discusssion of this and examples are here.

KINETIC ENERGY:

Due to rotation, kinetic energy is ¹/₂Iw ². This is the same beast as other kinds of energy, so it adds to the others. This is unlike all the other rotational stuff. You cannot add angular momentum and linear momentum, for example.

So the problem of a ball rolling down a hill: say it has v_o and w _o at the top of the hill of height h and v & w at the bottom. Then mgh + ¹/₂mv_o² + ¹/₂Iw _o² = ¹/₂mv² + ¹/₂Iw ², if no energy is "lost" (it does not disappear, it becomes heat if it seems to disappear).

Get the heck back. Now. to mechanics, that is.

Other main pages: fluid/heat/e&m, vibrations/waves, quantum, alphabetical index.

email me at fredrick.gram@ tri-c.edu (remove the space).