Centripetal acceleration

An object executing a circular orbit of radius $r$ with uniform tangential speed $v$ possesses a velocity vector ${\bf v}$ whose magnitude is constant, but whose direction is continuously changing. It follows that the object must be accelerating, since (vector) acceleration is the rate of change of (vector) velocity, and the (vector) velocity is indeed varying in time.

Figure 58: Centripetal acceleration.

Suppose that the object moves from point $P$ to point $Q$ between times $t$ and $t+\delta t$, as shown in Fig. 58. Suppose, further, that the object rotates through $\delta\theta$ radians in this time interval. The vector $\stackrel{\displaystyle \rightarrow}{PX}$, shown in the diagram, is identical to the vector $\stackrel{\displaystyle \rightarrow}{QY}$. Moreover, the angle subtended between vectors $\stackrel{\displaystyle \rightarrow}{PZ}$ and $\stackrel{\displaystyle \rightarrow}{PX}$ is simply $\delta\theta$. The vector $\stackrel{\displaystyle \rightarrow}{ZX}$ represents the change in vector velocity, $\delta{\bf v}$, between times $t$ and $t+\delta t$. It can be seen that this vector is directed towards the centre of the circle. From standard trigonometry, the length of vector $\stackrel{\displaystyle \rightarrow}{ZX}$ is

$$\delta v = 2 v \sin(\delta\theta/2).$$ (256)

However, for small angles $\sin\theta \simeq \theta$, provided that $\theta$ is measured in radians. Hence,

$$\delta v\simeq v \delta\theta.$$ (257)

It follows that

$$a = \frac{\delta v}{\delta t} = v \frac{\delta\theta}{\delta t} = v \omega,$$ (258)

where $\omega = \delta \theta/\delta t$ is the angular velocity of the object, measured in radians per second. In summary, an object executing a circular orbit, radius $r$, with uniform tangential velocity $v$, and uniform angular velocity $\omega=v/r$, possesses an acceleration directed towards the centre of the circle--i.e., a centripetal acceleration--of magnitude

$$a = v \omega = \frac{v^2}{r} = r \omega^2.$$ (259)

Figure 59: Weight on the end of a cable.

Suppose that a weight, of mass $m$, is attached to the end of a cable, of length $r$, and whirled around such that the weight executes a horizontal circle, radius $r$, with uniform tangential velocity $v$. As we have just learned, the weight is subject to a centripetal acceleration of magnitude $v^2/r$. Hence, the weight experiences a centripetal force

$$f = \frac{m v^2}{r}.$$ (260)

What provides this force? Well, in the present example, the force is provided by the tension $T$ in the cable. Hence, $T=m v^2/r$.

Suppose that the cable is such that it snaps whenever the tension in it exceeds a certain critical value $T_{\rm max}$. It follows that there is a maximum velocity with which the weight can be whirled around: namely,

$$v_{\rm max} = \sqrt{\frac{r T_{\rm max}}{m}}.$$ (261)

If $v$ exceeds $v_{\rm max}$ then the cable will break. As soon as the cable snaps, the weight will cease to be subject to a centripetal force, so it will fly off--with velocity $v_{\rm max}$--along the straight-line which is tangential to the circular orbit it was previously executing.