# How can a particle in circular motion be in translational motion?

I came across this:

If a particle is moving in a circle it is in pure rotational motion about the centre of the circle, while for a moment it may be in pure translational motion about some other point.

• How can a rotating particle be in translational motion with respect to another observer?

• Does this observer have to be on the circumference?

• And why does it say that the particle may be in pure translational motion?

• Are there any other possibilities?

• Where did you run across that? What's the context? Commented Jan 14, 2016 at 16:28
• I came across it while reading up angular velocity, under mechanics of rotational motion.
– Isha
Commented Jan 14, 2016 at 16:36
• Where? And was there any further explanation? I ask because it is not strictly correct, but it might be correct if there were further conditions or explanations provided. Commented Jan 14, 2016 at 16:37
• I read it under 'important points in angular velocity' in the book 'Understanding Physics for the IITJEE' , which is a reputed indian text. The only other point mentioned was that angular velocity about the centre of the circle for a moving point, is twice the angular velocity about any other point on the circumference.
– Isha
Commented Jan 14, 2016 at 16:47
• In this thread: yet another example of the terrible quality of Indian physics textbooks. Commented Dec 27, 2017 at 6:38

This may be a result of bad wording and/or misunderstanding. Let me try to clarify - suppose a particle is tied to a string of length $l$ and orbits around its center with some angular velocity, $\omega$. That is, in polar coordinates, its position is given by $(r,\theta) = (l, \omega t)$. From this, we can find the velocity in this coordinate system - $$(\dot{r}, \dot{\theta}) = (0, \omega).$$ However, this is not the only coordinate system we can use for this system - we can also use Cartesian coordinates! Doing a coordinate transformation, we would find that the position of the particle is now given by $$(x,y) = (l \cos{\omega t}, l \sin{\omega t}).$$ But in this coordinate system, we can also find the velocity: $$(\dot{x}, \dot{y}) = (-l \sin{\omega t}, l \cos{\omega t}).$$

Now we get to what it means for the object to be considered to be moving either circularly or translationally. It is clear that polar coordinates are the most intuitive coordinates for this system, so we describe the entire motion with a single velocity - angular velocity, $\omega$. However, if we decide to be perverse, we can instead look at the translational velocity at any given time, i.e. $(\dot{x}, \dot{y})$, which, at say, $t = 0$, we get $(0,l)$.

Do note that there is a good relationship between the magnitude of the translational velocity, $v$, and the angular velocity, $\omega$: $$v = r \omega.$$

Now, obviously while the position in Cartesian coordinates is observer dependent (it will given by the above $\pm$ a constant), once we take the time derivative, the linear (translational) velocity is as written above, as long as the observer is in an inertial frame.

There is also the case of a non-inertial observer frame - for example, if the observer is located at the same point as the particle and moves with it, then it will observe (trivial) translational motion - no motion at all. I'm almost certain that there is no (inertial or otherwise) frame where you get non-trivial constant translational motion from circular motion in an inertial frame, leading me to believe that the statement is really about viewing angular velocity as a time-dependent translational velocity.

If you move along a circle with a constant speed and your friend oscillates along the diameter simple harmonically, your friend would see you performing translational motion. This answers your first question.

If the observer is on the circumference, he will see you as stationary. You arent moving with respect to him.

There may be other possibilities. You may think up such other scenarios.