Angular momentum is the consequence of a torque in the same way that linear momentum is a consequence of a force. To remind you how force relates to linear momentum, here it is:[br][br][center][math][br]\vec{F}_{net}=m\vec{a} \\[br]\vec{F}_{net}=m\tfrac{d\vec{v}}{dt} \\[br]\vec{F}_{net}\;dt=md\vec{v} \\[br]\int\vec{F}_{net}\;dt=\int md\vec{v} \\[br]\int\vec{F}_{net}\;dt=m\Delta\vec{v} \\[br]\int\vec{F}_{net}\;dt=\Delta\vec{p} \\[br][/math][/center]All this means is that [b]force acting over time alters the linear momentum[/b]. Since force is a vector, however, the effect can be self cancelling. For instance, consider earth orbiting the sun. The sun always exerts a force on the earth toward itself. Since the direction of that gravitational force varies, over any complete orbit, there is no change in linear momentum of the earth. In other words, when that force pulls earth toward one direction in the heavens, at another time of year the force is directed opposite to that direction in the heavens. This cancels any change of momentum over an orbital period (year). Because of this cancellation, the planet has the same momentum one year after we first observe it. This means it moves at the same speed and in the same direction.[br][br]Over short periods, however, the earth's momentum obviously changes... not due to a change in orbital speed, but rather due to change in orbital direction.

I wish to do an analogous derivation to the one above, except with torque rather than force. Please take the time to digest the steps in the derivation above as well as this one. This one is slightly harder only in that we need to keep track of the direction of the result rather carefully since there will be two cross products. Other than that it is straightforward. The idea is that we assume a torque is applied to a mass that is part of a rotating, rigid object, and we wish to see the outcome.[br][br][center][math][br]\vec{r}\times\vec{F}_{net}=\vec{r}\times m\vec{a} \\[br]\vec{\tau}_{net}=\vec{r}\times m\tfrac{d\vec{v}}{dt} \\[br]\vec{\tau}_{net}\;dt=\vec{r}\times md\vec{v} \\[br]\int\vec{\tau}_{net}\;dt=\int \vec{r}\times md\vec{v} \\[br]\int\vec{\tau}_{net}\;dt=\int \vec{r}\times d\vec{p} \\[br]\text{Here we define $\vec{l}\equiv\vec{r}\times\vec{p}$.} \\[br]\int\vec{\tau}_{net}\;dt=\int d\vec{l} \\[br]\int\vec{\tau}_{net}\;dt=\Delta\vec{l} \\[br][/math][/center][br]The term [math]\vec{l}[/math] is the angular momentum associated with an individual mass 'm'. What this equation tells us is: [b]Torque acting over time serves to change angular momentum[/b] of an object. [br][br]What's interesting (or odd) about this result is that an angular momentum is associated with [i]an object that isn't necessarily going around a curved path[/i]! It also means that one and the same object has an angular momentum when observed from one vantage point (origin of [math]\vec{r}[/math]) but not from another. [br][br]If that sounds confusing, maybe consider the following: You are sitting in an office chair that can rotate while wearing a baseball mitt. At first the chair is not rotating. Now someone from across the office pitches a baseball at you. If the ball's velocity vector is directed right toward your center, you will catch the ball and not rotate. If the chair has wheels, you will recoil and roll slowly backwards to conserve momentum, but friction (an external force) will quickly put a stop to that. The reason you will not rotate is the force exerted on the mitt is directed through the rotation axis of the chair, and therefore generates no torque.[br][br]On the other hand, if the ball's velocity is such that you have to reach out to the side to catch it, and if you have your feet off the ground, you will begin rotating and will also roll backwards as in the previous case. The reason for the rotation is that the force on the mitt does not pass through the rotation axis of the chair, and therefore exerts a torque on you. Another way of seeing it is that the ball has angular momentum from your vantage point (and the definition above), as we’ll see below.[br][br]That torque will serve to change your angular momentum and cause you to spin. Another way of seeing it is that angular momentum is conserved for the two-body system of you and the ball. Since you have angular momentum after catching the ball, then the ball had angular momentum before you caught it. But realize that the ball was thrown along a linear path. It wasn't spinning nor curving. Had you been sitting in a different location, the ball could have had more angular momentum, angular momentum in the other direction, or perhaps none at all. [b]In this sense we can see that angular momentum is a relative quantity just as kinetic energy and linear momentum are[/b]. It depends on the reference frame of the observer.[br][br]Realize that in both cases the ball was traveling in an assumed straight line, but in one case it had angular momentum associated with it and in another it didn't.

The angular momentum of a rigid object like a ball in rotation or a spinning flywheel in an engine can be seen as the sum of the angular momenta [math]\vec{l}[/math] of its parts. When we denote the angular momentum of a rigid body we use a capital L.[br][br][center][math][br]\vec{L}=\sum_i\vec{l}_i \\[br]\vec{L}=\sum_i\vec{r}_i\times\vec{p}_i \\[br]\text{Separating the magnitudes from the directions of both vectors gives: } \\[br]\vec{L}=\sum_ir_i m_iv_i\hat{r_i}\times\hat{v_i} \\[br]\text{Writing the speed as $v=\omega r$ and shuffling the terms gives: } \\[br]\vec{L}=\sum_im_ir_i^2\omega_i\hat{r_i}\times\hat{v_i} \\[br]\text{On rigid objects, the angular speed $\omega$ is the same for all parts, so no subscript is needed. } \\[br]\vec{L}=\sum_im_ir_i^2\omega\hat{r_i}\times\hat{v_i} \\[br]\text{Recall that $I=\sum_im_ir_i^2$, and that direction for $\vec{\omega}$ is from the curly-fingered right hand rule: } \\[br]\vec{L}=I\vec{\omega}. \\[br][br][/math][/center][br][br]What we should understand from this is that an object made of masses that are subject to torques will experience a change in angular momentum. Stated differently: [b]Torque acting over time serves to change angular momentum of a rigid body[/b]. Depending on whether we have a rigid object rotating or a mass traveling with a velocity, we may use one of the two angular momentum expressions seen above, but no worry. They really describe the same thing in different terms. While we distinguish with either an upper or lower case letter, that's just to remind ourselves of the origin and not because it's different in any fundamental way. To summarize, the options are:[br][br][center][math]\vec{l}=\vec{r}\times\vec{p} \\[br]\vec{L}=I\vec{\omega}.[/math][/center]

[color=#1e84cc][br]EXAMPLE: Suppose a torque of 50Nm acts along the rotation axis of a disk of mass 4kg and radius 50cm for a duration of 10s. How fast will it be rotating after 10s if it started from rest?[br][br]SOLUTION: The general expression we need is:[math]\int\vec{\tau}\;dt = \Delta\vec{L}.[/math] Since torque is constant and we know the directions of the torque and acquired angular momentum match, we will relate magnitudes. The math looks like this:[br][center][math]\int\vec{\tau}\;dt = \Delta\vec{L} \\[br]\tau\hat{\tau}\Delta t=\Delta \vec{L} \\[br]\text{Here the unit vector $\hat{\tau}$ just indicates the direction of the torque. } \\[br]\tau\hat{\tau}\Delta t=I\Delta\vec{\omega} \\[br]\text{The inertia of a disk is $I=\tfrac{1}{2}mr^2$.} \\[br]\tau\hat{\tau}(\Delta t)=\tfrac{1}{2}mr^2(\vec{\omega_f}-\vec{\omega_i)} \\[br]\text{Plugging in numbers and using initial angular speed of zero gives: } \\[br]\vec{\omega_f} = 1000 rad/s\hat{\tau}.\\[br]\text{The final angular velocity is in the direction of the applied torque.}[br][/math][/center][br][/color]