How does a rotating object keep rotating?

Question

When an object rotates, every particle of the object experiences acceleration continuously. Doesn't that mean some force is being applied? Why doesn't it then stop when no force is applied? I have read that 'angular momentum' needs to be conserved. But giving it a name doesn't explain what happens.

Answer 1

You're right that every particle in the body experiences a force as the body rotates. But the key realization is that this force is exerted by other particles in the body. In fact, the conservation of angular momentum of a system of particles can be viewed as a consequence of the following three assumptions:

1. Newton's Laws hold.
2. The system is isolated: no forces on any particle are exerted by objects outside of the system.
3. The force between any two particles in the system is parallel to the line connecting them.

The derivation of this can be found in most calculus-based classical mechanics textbooks, but let's go through it carefully.

Suppose we have a system of particles numbered 1, 2, 3, ..., each with its own position $\mathbf{r}_i$, and momentum $\mathbf{p}_i$. Consider the quantity $$ \mathbf{L} = \sum_i \mathbf{r}_i \times \mathbf{p}_i. $$ The rate of change of this quantity is $$ \frac{d\mathbf{L}}{dt} = \sum_i \left[\frac{d\mathbf{r}_i}{dt} \times \mathbf{p}_i + \mathbf{r}_i \times \frac{d\mathbf{p}_i}{dt} \right]. $$ The first term vanishes for each term in the sum, since $d\mathbf{r}_i/dt = \mathbf{v}_i$, which is parallel to $\mathbf{p}_i$. Using Newton's Second Law, we have $d\mathbf{p}_i/dt = \mathbf{F}_i$, the total force on particle $i$. We now apply our second assumption, that this force is solely due to the other particles in the system: $$ \mathbf{F}_i = \sum_j \mathbf{F}_{ij} $$ where $\mathbf{F}_{ij}$ is the force on particle $i$ due to particle $j$. So the rate of change of $\mathbf{L}$ becomes $$ \frac{d\mathbf{L}}{dt} = \sum_{i,j} \mathbf{r}_i \times \mathbf{F}_{ij}. $$ Now, we also know from Newton's Third Law that $\mathbf{F}_{ij} = - \mathbf{F}_{ji}$ (every action has an equal and opposite reaction.) We can use this to rewrite this sum; in particular, for any term that contains $\mathbf{F}_{ji}$ with $j > i$, we can rewrite this in terms of $\mathbf{F}_{ij}$ with $i < j$. Thus, this becomes $$ \frac{d\mathbf{L}}{dt} = \sum_{i<j} \left[ \mathbf{r}_i \times \mathbf{F}_{ij} - \mathbf{r}_j \times \mathbf{F}_{ij}\right] = \sum_{i < j} \left( \mathbf{r}_i - \mathbf{r}_j \right) \times \mathbf{F}_{ij}. $$ But the vector $\mathbf{r}_i - \mathbf{r}_j$ points along the line from particle $i$ to particle $j$. So if we make our third assumption above, that $\mathbf{F}_{ij}$ is parallel to this line, then this whole sum vanishes and we have $d\mathbf{L} /dt = 0$.

It's also possible to show that a for a rigid body, there is a linear relationship between the object's angular velocity and its angular momentum; thus, if $\mathbf{L} \neq 0$ initially for an isolated body, then $\mathbf{L} \neq 0$ for all time, and thus $\pmb{\omega} \neq 0$ for all time as well.

Note that Assumption #3 above is independent of Newton's Laws. Newton's Laws do not necessarily imply conservation of angular momentum; you have to make an additional assumption about the directions of the forces between two particles, and in fact a Universe in which angular momentum is not conserved is entirely consistent with Newton's Laws. One way to get this additional assumption is to assume that the Universe has rotational symmetry; in this case, there is no "preferred direction" between two particles other than the axis between them, and so the force between them must point along this axis. (There's also a deep connection between rotational symmetry and angular momentum via a wonderful mathematical result called Noether's Theorem, which I highly recommend you look into.)

Answer 2

Edit: Wrote this answer while Michael Seifert was writing his, so forgive overlap.

"When an object rotates, every particle of the object experiences acceleration continuously. Doesn't that mean some force is being applied?"

Yes. That force is applied by the radially adjacent particles.

"Why doesn't it then stop when no force is applied?"

"It" does stop, if by "it" you mean circular motion. Imagine the particle breaks off. There is now "no force applied" so it flies off in a straight line tangent to its motion when it broke off.

Finally, angular momentum is conserved because the loss of angular momentum of the remaining rotating object (by virtue of losing a small mass) is made up for by the angular momentum of the flying particle, which is a sum of: angular momentum due to its own rotation (if it continues to rotate, which depends on the conditions under which it broke free); plus angular momentum due to the product of its linear momentum and the perpendicular distance between its line of motion and the center of mass of the remaining rotating object.

Answer 3

The force that is applied is called a centripetal force. Centripetal means "center-seeking", because that is the direction of the force: point towards the center. Since the force is always pointing towards the center, it is always perpendicular to the velocity, and thus changes the direction of the velocity without changing its magnitude (the speed). Since this force always rotates with the particles of the object, it always is changing the direction of the motion but never the speed.