Circular Motion

An object is moving in a circle. What are the vector expressions for its velocity and acceleration?

For any object in motion, the velocity vector always measures the rate of change of the position vector. Since all vectors are characterized by magnitude and direction we must first understand how either property can be changed. Let's think of a vector as a material arrow. The only way the length of the arrow can be changed is by stretching or contracting it. Either procedure involves a change along the arrow. If only the direction of the arrow is modified, this change occurs perpendicular to the arrow. These simple results explain how vectors change, as we now show with actual mathematical derivation.

Consider an object moving along a circle of radius R:

The object's position vector maintains the same length, R, but continuously changes its direction. So we expect that its velocity, the rate of change of the position vector, will be perpendicular to the radius vector. The position vector r is (Rcos q, Rsin q), which we can also write as


	r = R u(q), where u(q) = (cos q, sin q)	(1)

Since the magnitude of the vector u(q) is 1, this is a unit vector pointing in the direction that makes an angle q with the x axis:

Any direction in space can be specified by its unit vector. Examples of unit vectors we already encountered are the unit vectors along the coordinate axes i = u(0) and j = u(90°). Imagine yourself sitting at the origin of the coordinate system on a rotating chair. The direction you are facing at every instant can be specified by the angle q it makes with the x axis. This angle continuously changes, the rate of its change is the angular speed of your rotation, denoted w:


	w = dq/dt	(2)

Your rotation corresponds to the unit vector u(q) rotating at the same rate w as you. What is the vector du/dt? Because the length of u remains fixed at unity, only its direction changes and from the simple discussion we gave above we can expect that du/dt, which measures the change in u, will be perpendicular to u. Indeed,


du/dt	= (du/dq)´(dq/dt)
	= w du/dq
	= w (-sin q, cos q)	(3)

In the first step we employed the rules for taking derivatives, in the second the definition of w in equation (2) and in the last the definition of u in equation (1). Observe that the new vector (-sin q, cos q) is also a unit vector, therefore the magnitude of du/dt is simply w, the rotation angular speed. What is the direction of this vector? Recall a previous discussion where we figured the coordinates of a vector after rotation by 90°. The rule we found was that the x and y components are switched and the x component changes its sign — precisely the relation between u and the unit vector along du/dt. As we anticipated, the vector specifying the change in u is perpendicular to u.

Now let's get back to our object moving along the circle. Its position vector is given in equation (1) and since R is constant, the object's velocity vector is


v	= dr/dt
	= R du/dt
	= wR (-sin q, cos q)	(4)

In the last step we substituted the expression for du/dt from eq. (3). This gives us the velocity vector, both magnitude and direction.

The direction of v is perpendicular to the radius vector which means that it is aligned with the tangent. In fact, this is a general result. Irrespective of the type of motion and trajectory followed, the instantaneous velocity always points along the local tangent (you can understand this general result by thinking about the geometrical meaning of the vector dr used in the derivative procedure v = dr/dt ).

The magnitude of v obeys


	v = wR	(5)

The magnitude of the linear and angular speeds are always related this way. For a given angular speed, the magnitude of the velocity increases with distance from the rotation center; think of different points along a radial ray on a rotating wheel.

Uniform Rotation
Now that we have derived the velocity vector we can easily calculate the acceleration vector a by taking the next derivative, dv/dt. The calculation is done in complete analogy to the steps taken in equation (3). For simplicity, let us consider first the case of uniform rotation so that w is constant. Then


a	= dv/dt
	= (dv/dq)´(dq/dt)
	= w dv/dq
	= -w²R(cos q, sin q)
	= -w²r	(6)

In the last step we inserted the expression for r from eq. (1), in the previous one the expression for v from eq. (4). The acceleration is aligned with the radius vector but has the opposite direction. Since it points toward the center it is called centripetal acceleration. Even though the object's speed is constant, its velocity continuously changes its direction, which is why there is an acceleration. In fact, the statements "an object moves in a circle" and "an object's motion is subject to centripetal acceleration" are completely equivalent; each one implies the other. The centripetal acceleration magnitude a_c is always obtained from


	a_c = w²R = v²/R

whatever the speed (in the second equality we utilized eq. 5).

Non Uniform Rotation
Finally, what happens if the angular speed w is varying? Then dw/dt ¹ 0 and we must take that into account when taking derivative. But when we recall the rules for derivative of a product we immediately realize that the only difference from the uniform rotation case is that the acceleration gets an additional term in the direction (-sin q, cos q), namely parallel to the velocity (and the tangent). So the acceleration now has an additional tangential component whose magnitude is


	a_t = R dw/dt = dv/dt

where in the second equality we utilized eq. (5).

All Together
Our results can be summarized in a figure:

Since the object moves in a circle, the velocity vector changes its direction and there is always centripetal acceleration whose magnitude is v²/R.
If the object's speed is varying then there must be also tangential acceleration whose magnitude is dv/dt, just as in the case of linear motion. And just as in that case, the tangential acceleration points in the direction of rotation if the object speeds up and in the opposite direction if it slows down, the case depicted in the figure.
In the uniform rotation case the acceleration vector points toward the center, in the most general case it is pointed at a direction somewhere between radial and tangential.

Observe that the results conform precisely to what we figured considering a material arrow in the beginning of this note. It's a good way to understand what's happening in circular motion.

[Lecture Notes] [PHY231] [Physics Dept] [UK]