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8.3: Another Special Case: Circular Motion

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The other special case we examine in this chapter is that of the motion of an object moving in a circular path. Because the path is continually changing direction, there is always an acceleration, even if the speed is constant.

First we review the variables that describe rotational motion and define rotational acceleration.

Rather than using rectangular coordinates to describe the position $$P$$ of a particle moving in a circle, we find it convenient to use polar coordinates, $$r$$ and $$\Theta$$. The coordinate r is the distance of the point from an axis of rotation (the origin  $$\vartheta$$ ); $$\Theta$$ is the angular displacement from an arbitrarily chosen axis that defines zero. As in the figure 8.3.1 at left, $$\Theta$$ is frequently measured from the positive x axis, but it could be measured from any reference line as well.

Figure 8.3.1

When $$\Theta$$ changes by an amount $$\Delta \Theta$$ , the particle moves an amount $$\Delta s$$ along the circumference of the circle defined by the radius $$r$$. The arc length, $$\Delta s$$ is simply the product of $$r$$ and $$\Delta \Theta$$ :

$\Delta s = r \Delta \Theta$

The instantaneous velocity of the point P is always tangential, if the point moves in a circle. If we differentiate the displacement with respect to time to get the tangential velocity of this object, we get an expression that depends only on the time derivative of $$\Theta$$ :

$\dfrac{ds}{dt} = v_{tangential} = r \dfrac{d\Theta}{dt} .$

The time rate of change of the angular position, $$\Theta$$ , is called the angular velocity or rotational velocity and is usually represented by the Greek letter $$\omega$$ ("omega").

$\omega = \dfrac{d\Theta}{dt} .$

The rotational velocity and tangential velocity are related by:

$v_{tangential }= r \omega .$

Likewise, the angular acceleration or rotational acceleration is the time rate of change of angular velocity and is usually represented by the Greek letter $$\alpha$$ ("alpha").

$\alpha = \dfrac{d\alpha}{dt} .$

The rotational acceleration and tangential acceleration are related by:

$a_{tangential} = r\alpha.$

The units of $$\Theta$$, $$\omega$$ and $$\alpha$$ are respectively an angle unit, an angle unit divided by time, and an angle unit divided by time squared. We can use any units we want and that are useful for a particular application for $$\Theta$$, $$\omega$$ and $$\alpha$$ . Typical units are degrees, degrees/second, degrees/second2 ; revolutions, revolutions/second or rpm or revolutions/hour , and revolutions/second2 , etc. The "natural" units, are, however, radians, radians per second, and radians per second-squared.

Note

We must use radians, radians per second, and radians per second squared when we use the relations connecting x, v, and a to $$\Theta$$, $$\omega$$ and $$\alpha$$ .

Look back at the last expression for translational acceleration. This is the change in speed along the direction the particle is moving as it goes around the circle. But suppose the particle is not speeding up. That is, it goes at a continuous angular velocity. Is there still an acceleration associated with this particle? Indeed there is an acceleration, if the velocity vector changes. Now, the length of the velocity vector is constant, because the speed is constant, but the direction continually changes as the particle goes around. So there definitely is an acceleration! Which way does it point?

Figure 8.3.2

The velocity is purely tangential with magnitude $$r\omega$$. The figure 8.3.2 shows the direction of $$v$$ changing, but not its magnitude as the particle moves counterclockwise around the circle.

$v_{tangential} = r\omega ,$

$v_{radial} = 0.$

Figure 8.3.3

If the tangential component of the velocity vector is not changing, it must be only the radial component that changes. We state without deriving it that that the magnitude of the acceleration in the radial direction, that is, toward the center of the circle (Figure 8.3.3), is

$a_{radial} = -r\omega^ 2 = \dfrac{v^2}{ r}$

$a_{tangential} = 0$

There is only a radial acceleration of a particle moving at constant speed in a circle! Because the acceleration points toward the center of the circle, it is often referred to as the centripetal ("center-seeking") acceleration.

By Newton’s 2nd law, $$\sum F = ma$$, so the net force, $$\sum F$$, points in the same direction as a. Therefore, the net force must be pointing toward the center of the circle and must have, for a particle moving in a circle at constant speed, the magnitude:

$\sum F_{radial} = ma_{radial} = -mr\omega^ 2 = -\dfrac{mv^2}{ r} .$

where the negative sign tells us that this force points inwards towards the center of the circle. This is the net force $$\sum F$$ exerted on an object in order to make it move in a circle with constant speed. For this particular situation, this net force is sometimes called the centripetal force, but remember that this is just a label for whatever forces actually provide the center-seeking force (e.g., string tension, || contact force, gravity, electric, etc.) to make an object move in a circle with constant speed.