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Projecting the Wheel

Figure: Projected motion of a point on the rim of a wheel.
\epsfig{file=PS/,height=4.0in} }\end{center}\end{figure}

Imagine a rigid wheel rotating at constant angular velocity about a fixed central axle. A bolt screwed into the rim of the wheel executes uniform circular motion about the centre of the axle.13.3 For reference we scribe a line on the wheel from the centre straight out to the bolt and call this line the radius vector. Imagine now that we take this apparatus outside at high noon and watch the motion of the shadow of the bolt on the ground. This is (naturally enough) called the projection of the circular motion onto the horizontal axis. At some instant the radius vector makes an angle   $\theta = \omega t$  with the horizontal, where  $\omega$  is the angular frequency of the wheel ($2 \pi$ times the number of full revolutions per unit time) and we measure the time  t  from the instant when the radius vector is horizontal. From a side view of the wheel we can see that the distance  x  from the shadow of the central axle to the shadow of the bolt [i.e. the projected horizontal displacement of the bolt from the centre, where  x=0] will be given by trigonometry on the indicated right-angle triangle:

\begin{displaymath}\cos(\theta) \; \equiv \; {x \over r}
\quad \Rightarrow \quad
x \; = \; r \, \cos(\theta) \; = \; r \, \cos(\omega t)
\end{displaymath} (13.1)

The resultant amplitude of the displacement as a function of time is shown in Fig. 13.3.

Figure: The cosine function.
\epsfig{file=PS/,height=2.15in} }\end{center}\end{figure}

The horizontal velocity  vx  of the projected shadow of the bolt on the ground can also be obtained by trigonometry if we recall that the vector velocity  $\Vec{v}$  is always perpendicular to the radius vector  $\Vec{r}$. I will leave it as an exercise for the reader to show that the result is

\begin{displaymath}v_x \; = \; - \, v \, \sin(\theta)
\; = \; - \, r \, \omega \, \sin(\omega t)
\end{displaymath} (13.2)

where   $v = r \omega$  is the constant speed of the bolt in its circular motion around the axle. It also follows (by the same sorts of arguments) that the horizontal acceleration  ax  of the bolt's shadow is the projection onto the  $\hat{x}$  direction of  $\Vec{a}$,  which we know is back toward the centre of the wheel -- i.e. in the  $-\hat{x}$  direction; its value at time  t  is given by

\begin{displaymath}a_x \; = \; - \, a \, \cos(\theta)
\; = \; - \, r \, \omega^2 \, \cos(\omega t)
\end{displaymath} (13.3)

where   $a = {\displaystyle {v^2 \over r}} = r \omega^2$  is the magnitude of the centripetal acceleration of the bolt.

next up previous
Next: Simple Harmonic Motion Up: Sinusoidal Motion Previous: Sinusoidal Motion
Jess H. Brewer - Last modified: Sun Nov 15 13:34:45 PST 2015